Answer:
The final pressure of the carbon dioxide gas will 11.84 atm.
Explanation:
Moles of carbonic acid = [tex]\frac{60.0 g}{62 g/mol}=0.9677 mol[/tex]
[tex]H_2CO_3(aq)\rightarrow H_2O(l) + CO_2(g)[/tex]
According to reaction, 1 mol of carbonic acid gives 1 mole of carbon dioxide gas.
Then 0.9677 moles of carbonic acid will give :
[tex]\frac{1}{1}\times 0.9677 mol=0.09677 mol[/tex] of carbon dioxide
Moles of carbon dioxide gas = n = 0.09677 mol
Volume of soda bottle = [tex]V_1=2 L[/tex]
Pressure of the carbon dioxide gas = P
Temperature of the carbon dioxide gas = T = 298 K
[tex]PV=nRT[/tex] (ideal gas law)
[tex]P=\frac{nRT}{V}=\frac{0.9677 mol\times 0.0821 atm L/mol K\times 298 K}{2 L}=11.84 atm[/tex]
The final pressure of the carbon dioxide gas will 11.84 atm.
By using the ideal gas law equation PV = nRT, calculating the number of moles of CO₂ formed from 60.0 grams of decomposed carbonic acid, and inputting the given values for the volume (V) and temperature (T), we calculated that the final pressure of the CO₂ in the soda bottle would be approximately 1.18 atm.
Explanation:To calculate the final pressure of carbon dioxide in the soda bottle we must utilize the ideal gas law equation, which is PV = nRT. Where P stands for pressure, V is volume, n is the number of moles of gas, R is the gas constant and T is the temperature.
First, we need to determine the number 'n' of moles of CO₂ formed. Knowing that 60.0 grams of carbonic acid decompose completely into water and CO₂, and knowing that the molar mass of carbonic acid H₂CO₃ is approximately 62.03 g/mol, we can find that the number of moles (n) is equal to the mass (60g) divided by the molar mass (62.03 g/mol), so n equals about 0.967 moles.
Because it specifies that we should assume CO₂ to 'have the full 2.00 L in which to expand', the volume V in the ideal gas law equation is 2.00 L. Similarly, the temperature T is given as 298 K. We will now plug these values into the ideal gas law equation, using 0.0821 as the value for R (the ideal gas constant in units of L*atm/(mol*K)). This gives us P = (nRT) / V, so P is about 1.18 atm, which is the final pressure of CO₂.
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When 1 mole of H2CO(g) reacts with O2(g) to form CO2(g) and H2O(l) according to the following equation, 563 kJ of energy are evolved. H2CO(g) + O2(g)CO2(g) + H2O(l) Is this reaction endothermic or exothermic? What is the value of q? Write a balanced thermochemical equation
Final answer:
The reaction where 1 mole of H2CO(g) reacts with O2(g) to form CO2(g) and H2O(l) is exothermic, evolving -563 kJ of energy. This is represented in the balanced thermochemical equation with a negative enthalpy change (ΔH = -563 kJ).
Explanation:
When 1 mole of H2CO(g) reacts with O2(g) to form CO2(g) and H2O(l), the reaction is exothermic because energy is released. The value of q in this context represents the amount of energy evolved during the reaction, which is -563 kJ (negative sign indicates energy is released by the system). Therefore, the balanced thermochemical equation incorporating the heat of reaction would be as follows:
H2CO(g) + O2(g) → CO2(g) + H2O(l) ΔH = -563 kJ
An exothermic reaction is characterized by the release of energy to the surroundings, with a negative enthalpy change (ΔH), indicating that the products are at a lower energy level than the reactants. Conversely, an endothermic reaction absorbs energy, with a positive enthalpy change, indicating that the products are at a higher energy level than the reactants.
Final answer:
The reaction in question is exothermic, with the value of q is -563 kJ of energy being evolved. The balanced thermochemical equation reflects this by including a negative ΔH value.
Explanation:
The reaction of [tex]H_2CO[/tex] (g) with [tex]O_2[/tex] (g) to form [tex]CO_2[/tex] (g) and [tex]H_2O[/tex] (l) is an exothermic reaction because energy is released in the process.
The value of q for this reaction is -563 kJ, indicating that 563 kJ of energy are evolved, where the negative sign signifies the direction of energy flow out of the system. The balanced thermochemical equation for the reaction would be:
[tex]H_2CO[/tex] (g) + [tex]O_2[/tex] (g) → [tex]CO_2[/tex] (g) + [tex]H_2O[/tex] (l) ΔH = -563 kJ/mol
Here, the ΔH value represents the enthalpy change for the reaction, and since it is negative, the reaction is considered exothermic.
The last step of glycolysis converts phosphoenolpyruvate to pyruvate. phosphoenolpyruvate + ADP pyruvate kinase −−−−−−−−−→ pyruvate + ATP Δ G = − 33 kJ / mol ( − 7.5 kcal / mol ) The Δ G ° ' of the reverse reaction is + 31 kJ / mol ( + 7.5 kcal / mol ) . Instead of reversing the pyruvate kinase reaction, the step is bypassed in gluconeogenesis. Several steps for the conversion of pyruvate to phosphoenolpyruvate via gluconeogenesis are given. Place the steps in the correct order. You will not place all of the steps.
Answer:
Steps
i) In thee mitochondrion, pyruvate carboxylase converts pyruvate to oxloacetate.
ii) Malate dehydrogenase in the mitochondrion reduces oxaloacete to to malate.
iii) Malate dehydrogenase in the cytoplasm oxidizes malate to oxaloacetate.
iv) Phosphoenolpyruvate carboxykinase decarboxylates and phosphorylates oxaloacetate.forming phosphoenolpyruvate.
Explanation:
Gluconeogenesis is the synthesis of glucose from non - carbohydrate compounds. The substrates for gluconeogenesis are lactate, pyruvate, amino acids, propionate and glycerol.
Gluconeogenesis occurs only in cytosol but the precursor is produced in mitochondria. In the conversion of pyruvte to phosphoenolpyruvate occur in mitochondria and cytosol.
Step -1:
Pyruvate carboxylase is a biotin dependent enzyme located in mitochondria. It converts pyruvate to oxlaoacetate and carbondioxide in the presence of ATP.Oxlaocetate synthesized in mitochondrial matrix has to be transported to cytosol for gluconeogenesis. Oxaloacetate is impermeble, cannot be sent out of mitochondria. So it has to be converted to malate.
Step -2:
Malate dehydrogenase in mitochondria converts oxaloacetate synthesized in mitochondrial matrix to malate. And then it is transported to cytosol.
Step 3:
Malate dehydrogenase responsible for reversible reaction in cytosol converts malate to oxaloacetate.
Step -4
Phosphoenolpyruvate carboxy-kinase in cytosol converts oxaloacetate to PEP. The enzyme transfer high energy phosphate bond from GTP to oxaloacetate to from PEP and liberated carbondioxide.
Therefore, the steps of glycolysis converts phosphoenolpyruvate to pyruvate are as follows.
i) In thee mitochondrion, pyruvate carboxylase converts pyruvate to oxloacetate.
ii) Malate dehydrogenase in the mitochondrion reduces oxaloacete to to malate.
iii) Malate dehydrogenase in the cytoplasm oxidizes malate to oxaloacetate.
iv) Phosphoenolpyruvate carboxykinase decarboxylates and phosphorylates oxaloacetate.forming phosphoenolpyruvate.
Gluconeogenesis uses pyruvate as a starting point to generate glucose, going through several steps that largely mirror glycolysis in reverse order. However, the first step in gluconeogenesis varies, converting pyruvate into oxaloacetate before continuing with the conversion into phosphoenolpyruvate. It's a complex process that allows the cell to regulate the production and use of glucose efficiently.
Explanation:The process of converting pyruvate to phosphoenolpyruvate via gluconeogenesis is not a simple reversal of the glycolysis steps. It starts with the conversion of pyruvate to oxaloacetate. Oxaloacetate serves as a substrate for the enzyme phosphoenolpyruvate carboxykinase (PEPCK), which transforms oxaloacetate into phosphoenolpyruvate (PEP). From this step, gluconeogenesis nearly mimics glycolysis, but in reverse. PEP is converted back into 2-phosphoglycerate, then into 3-phosphoglycerate. Then, 3-phosphoglycerate is converted into 1,3 bisphosphoglycerate and then into glyceraldehyde-3-phosphate. Next, two molecules of glyceraldehyde-3-phosphate combine to form fructose-1-6-bisphosphate, which is then converted into fructose 6-phosphate and then into glucose-6-phosphate. Finally, a series of reactions generates glucose. This sequence enables glycolysis and gluconeogenesis to be regulated independently of each other.
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Draw the major product formed when the following epoxide reacts with methanol in the presence of sulfuric acid. Use wedge/dash bonds, including H\'s at each stereogenic center, to show the stereochemistry of the product.
Answer:
The major product of the reaction is (2S,3R)-3-methoxy - 3-methylpentan-2-ol.
Explanation:
Epoxide reacts with methanol in the presence of sulfuric acid.
In this chemicl reaction methanol act as nucleophile and sulfuric acid act as catalyst.
The chemical reaction is as follows.
An epoxide is a ring compound in which the oxygen atom closes the ring.
What is epoxide?An epoxide is a ring compound in which the oxygen atom closes the ring. The reaction of the epoxide with methanol in the presence of acid is an example of a ring opening reaction.
The product of this reaction is shown in the image attached to this answer. The acid attacks the epoxide and opens the ring then te methanol attacks subsequently to yield the product.
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If we start with 1.000 g of strontium-90, 0.953 g will remain after 2.00 yr. This means that the half-life of strontium-90 is ________ yr. If we start with 1.000 g of strontium-90, 0.953 g will remain after 2.00 yr. This means that the half-life of strontium-90 is ________ yr. 28.8 41.6 2.10 1.91 1.45
Answer:
The half-life of strontium-90 is 28.81 years
Explanation:
Step 1: Data given
Mass of strontium 90 = 1.000 grams
After 2 years there remain 0.953 grams
Step 2: Calculate half-life time
k = (-1/t) * ln (Nt/N0)
⇒ with k = rate constant for the decay
⇒ with N0 = the mass of strontium at the start (t=0) = 1.000 grams
⇒ with Nt = the mass of strontium after time t (2 years) = 0.953 grams
k = (-1/2) * ln(0.953/1)
k = (-1/2) * (-0.0481) = 0.02405 /yr
t1/2 = 0.693/k= 0.693/ 0.02405 = 28.81 years
The half-life of strontium-90 is 28.81 years
The given data does not provide a clear half-life for strontium-90 since the mass of the substance did not halve in 2 years. Thus, none of the options provided fits the scenario.
Explanation:The half-life of a radioactive substance is the time taken for half of the atoms in a sample to decay. Here, we have the sample of strontium-90 that has not halved in 2 years since we still have 0.953g out of 1.000g. Given the available options, none of them matches with the provided data because if a substance's mass did not reduce to half in 2 years then its half-life should be greater than 2 years. Possibly, there might be a calculation or data collection error in the problem scenario provided.
The half-life of strontium-90 can be calculated using the given information. If we start with 1.000 g of strontium-90 and after 2.00 years, 0.953 g remains, we can calculate the percentage of strontium-90 remaining: (0.953 g / 1.000 g) x 100% = 95.3%. Since half-life is the time required for half the atoms in a sample to decay, we can conclude that it took 2.00 years for 50% of the strontium-90 to decay. Therefore, the half-life of strontium-90 is 2.00 years.
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At 25.0 ⁰C the henry's law constant for hydrogen sulfide(H2S) gas in water is 0.087 M/atm. Caculate the mass in grams of H2S gas that can be dissolved in 400.0 ml of water at 25.00 C and a H2S partial pressure of 2.42atm.
Answer: The mass of hydrogen sulfide that can be dissolved is 2.86 grams.
Explanation:
Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the gas.
To calculate the molar solubility, we use the equation given by Henry's law, which is:
[tex]C_{H_2S}=K_H\times p_{liquid}[/tex]
where,
[tex]K_H[/tex] = Henry's constant = [tex]0.087M/atm[/tex]
[tex]p_{H_2S}[/tex] = partial pressure of hydrogen sulfide gas = 2.42 atm
Putting values in above equation, we get:
[tex]C_{H_2S}=0.087M/atm\times 2.42atm\\\\C_{H_2S}=0.2105M[/tex]
To calculate the mass of solute, we use the equation used to calculate the molarity of solution:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
We are given:
Molarity of solution = 0.2105 M
Molar mass of hydrogen sulfide = 34 g/mol
Volume of solution = 400.0 mL
Putting values in above equation, we get:
[tex]0.2105M=\frac{\text{Mass of hydrogen sulfide}\times 1000}{34g/mol\times 400.0mL}\\\\\text{Mass of }H_2S=\frac{0.2105\times 34\times 400}{1000}=2.86g[/tex]
Hence, the mass of hydrogen sulfide that can be dissolved is 2.86 grams.
All of the following statements correctly describe a contemporary approach to create the Work Breakdown Structure (WBS) for a project EXCEPT:
a. Project teams may use a template or a WBS from a previous project as a starting point.
b. Project teams may jump start the WBS construction with a brainstorming session.
c. Project teams may adopt a top-down approach to construct the WBS.
d. Project teams may rely upon the project manager to create all levels of the WBS.
Answer: option d
Explanation:
Project teams must be proactive. Each member of the team should empowered and each member can share ideas and experience in order to make the WBS more precise.
Dry ice is solid carbon dioxide. Instead of melting, solid carbon dioxide sublimes according to the following equation: CO2(s)→CO2(g). When dry ice is added to warm water, heat from the water causes the dry ice to sublime more quickly. The evaporating carbon dioxide produces a dense fog often used to create special effects. In a simple dry ice fog machine, dry ice is added to warm water in a Styrofoam cooler. The dry ice produces fog until it evaporates away, or until the water gets too cold to sublime the dry ice quickly enough. Suppose that a small Styrofoam cooler holds 15.0 liters of water heated to 90 ∘C. Calculate the mass of dry ice that should be added to the water so that the dry ice completely sublimes away when the water reaches 10 ∘C. Assume no heat loss to the surroundings.
To calculate the mass of dry ice, we need to consider heat transfer and use the formula q = m × c × ΔT. The mass of dry ice can be found using the equation mdry ice = (q / ΔHsublimation). By plugging in the given values, we can find the mass of dry ice that should be added to the water.
Explanation:In order to calculate the mass of dry ice that should be added to the water, we need to consider the heat transfer that occurs during the process. The heat gained by the water is equal to the heat lost by the dry ice. We can use the formula:
q = m × c × ΔT
where q is the heat gained by the water, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
By rearranging the equation, we can solve for the mass of dry ice:
mdry ice = (q / ΔHsublimation)
where mdry ice is the mass of dry ice, q is the heat gained by the water, and ΔHsublimation is the heat of sublimation of CO2.
In this case, the temperature change is the difference between the initial temperature of the water (90°C) and the final temperature (10°C). The specific heat capacity of water is approximately 4.18 J/g°C, and the heat of sublimation of CO2 is 571 J/g. By plugging in the values and solving the equation, we can find the mass of dry ice that should be added to the water.
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A fossil was analyzed and determined to have a carbon-14 level that is 70 % that of living organisms. The half-life of C-14 is 5730 years. How old is the fossil? Express your answer with the appropriate units. View Available Hint(s) t t t = nothing nothing
Answer: 2948
Explanation:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]
[tex]k=\frac{0.69}{t_{\frac{1}{2}}}=\frac{0.693}{5730}=1.21\times 10^{-4}years^{-1}[/tex]
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = [tex]1.21\times 10^{-4}years^{-1}[/tex]
t = age of sample = ?
a = let initial amount of the reactant = 100
a - x = amount left after decay process = [tex]\frac{70}{100}\times 100=70[/tex]
[tex]t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{100}{70}[/tex]
[tex]t=2948years[/tex]
Thus the fossil is 2948 years old.
The age of the fossil with 70% of the Carbon-14 level of living organism is 2948 years old
To determine the age of the fossil, we need to consider the decay of Carbon-14.
Given that the Carbon-14 level in the fossil is 70% that of living organisms, we can calculate the age using the concept of half-life.
Half life is the amount of time taken by a radioactive material to decay to half of its original value.t{(1)/(2)}=(0.693)/(k)k= {0.69}{t{(1)/(2)}}=(0.693)/(5730)=1.21* 10⁻⁴years⁻¹Expression for rate law for first order kinetics is given by:
t=(2.303)/(k)\log(a)/(a-x)where,
k = rate constant = 1.21* 10⁻⁴years⁻¹t = age of sample = ?a = let initial amount of the reactant = 100a - x = amount left after decay process = (70)/(100)* 100=70t=(2.303)/(1.21* 10⁻⁴)\log(100)/(70)t=2948yearsThus the fossil is 2948 years old.
below regarding an electrochemical cell in an automotive lead-acid battery. The cell's anode is made of lead and the cathode is made of lead(IV) oxide. Both are submerged in 4.30 M sulfuric acid. The half-reactions are: PbO_2(s) + 3H^+ (aq) + HSO_4^- (aq) + 2e^- rightarrow PbSO_4(s) + 2H_2O(l) E degree = 1.685 V PbSO_4(s) + H^+(aq) + 2e^- rightarrow Pb(s) + HSO_4^- (aq) E degree = -0.356 V (a) Calculate the value of E degree. (b) Determine the initial value of E_cell. Assume that the first ionization of H_2SO_4 is complete and that [H^+] almostequalto [HSO_4^-]. (c) Find E_cell when the H^+ concentration has dropped by 76.00%. Again, assume [H^+] almostequalto [HSO_4^-].
Answer:
a. 2.041 V
b. 2.116 V
c. 2.043 V
Explanation:
a.
In an electrochemical cell;
reduction occurs at the cathode & oxidation at the anode.
Thus, from the question; reduction at cathode=
[tex]PbO_{2(s)} + 3H_{(aq)}^{+} +HSO_{4}_{(aq)} +2e^{-}[/tex] → [tex]PbSO_{4(s)} + 2H_{2} O_{(l)} E^{0}_{(cathode)} = 1.685V[/tex]
Oxidation at anode;
[tex]Pb_{s} + HSO^{-}_{4}[/tex] → [tex]PbSO_{4(s)}+H^{+}+2e^{-} E^{0}_{anode} =0.356V[/tex]
The net process in the Cell is obtained by the algebraic sum of the two half-cell reactions:
Overall reaction:
[tex]PbO_{2(s)}+Pb_{s}+2HSO^{-}_{4(aq)}+ 2 H^{+}_{(aq)}[/tex] → [tex]PbSO_{4(s)} + 2H_{2} O_{(l)}[/tex]
[tex]E^{0} _{cell} = E^{0} _{cathode}- E^{0} _{anode}[/tex]
= 1.685V - (- 0.356V)
= 2.041V
b.
To determine the initial value of E-cell :
In the electrochemical cell; both the anode and the cathode were submerged into 4.30M of Sulfuric acid; therefore:
[tex]H_{2} SO_{4}[/tex] → [tex]H^{+} + HSO_{4}^{-}[/tex]
Assuming:
[tex]H^{+} = HSO_{4}^{-}[/tex] such that (;) [tex]H_{2} SO_{4}[/tex] = 4.30M
let : a= [tex]H^{+}[/tex] & b= [tex]HSO_{4}^{-}[/tex]
[tex]E^{i} _{cell}[/tex] =
[tex]E^{0} _{cell}[/tex] - [tex]\frac{0.0591}{2}[/tex] × log [tex]\frac{1}{(a)^{2} (b)^{2} }[/tex]
=2.041V - [tex]\frac{0.0591}{2}[/tex] × log [tex]\frac{1}{(4.30)^{2} (4.30)^{2} }[/tex]
= 2.041V - 0.02955 × log [tex]\frac{1}{341.8801}[/tex]
= 2.041V - 0.02955 × log [tex]{0.0029250026}[/tex]
= 2.041V - 0.02955 × (-2.5339)
= 2.041V + 0.075V
=2.116V
c.
If concentration of [tex]H^{+}[/tex] is dropped by 76.00%
[tex]H^{+} = HSO_{4}^{-}[/tex] = 4.30M
= 4.30M - 4.30M [tex]\frac{76}{100}[/tex]
= 1.032M
[tex]E_{cell}[/tex] = 2.041V - [tex]\frac{0.0591}{2}[/tex] × log [tex]\frac{1}{(1.032)^{2} (1.032)^{2} }[/tex]
=2.041V - 0.02955 × log [tex]\frac{1}{1.13427612058}[/tex]
=2.041V - 0.02955 × log (0.88161954735)
=2.041V - 0.02955 (- 0.0547)
=2.041V + (0.0020)
= 2.043V
The standard electrode potential (E°) of the cell in the lead-acid battery is 2.041 V. The initial value of E_cell in the same condition is also 2.041 V. As [H+] falls by 76%, E_cell can be calculated with the Nernst Equation.
Explanation:The electrochemical cell in a lead-acid battery consists of an anode made of lead and a cathode made of lead(IV) oxide, both submerged in sulfuric acid. The half reactions are as follows:
PbO_2(s) + 3H+ (aq) + HSO_4- (aq) + 2e- → PbSO_4(s) + 2H_2O(l) with E° = 1.685 VPbSO_4(s) + H+(aq) + 2e- → Pb(s) + HSO_4- (aq) with E° = -0.356 V(a) You can calculate the cell's standard electrode potential (E°) by subtracting the anode potential from the cathode potential, as per the formula E°cell = E°cathode - E°anode. Therefore, E° = 1.685 V - (-0.356 V) = 2.041 V.
(b) The initial value of E_cell is the same as the standard electrode potential, since initially [H+] = [HSO_4-] = 1M as per the standard conditions, which means E_cell = E° = 2.041 V.
(c) If the H+ concentration drops by 76%, it becomes 0.24M. For calculating the new E_cell we have to utilize the Nernst Equation, which is E=E°-(0.059/n)log(Q), where n is the number of electrons transferred (in this case 2), and Q is the reaction quotient ([Pb][HSO4-])/([PbSO4][H+]^2). Plugging in the values, we can determine the new E_cell.
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530 mL of CO gas at a pressure of 1.2 atm and a temperature of 17ºC. What is the volume of the gas at 0ºC and 1 atm?
Answer:
598.74 mL
Explanation:
Using Ideal gas equation for same mole of gas as
[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}[/tex]
Given ,
V₁ = 530 mL
V₂ = ?
P₁ = 1.2 atm
P₂ = 1 atm
T₁ = 17 ºC
T₂ = 0 ºC
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (17 + 273.15) K = 290.15 K
T₂ = (0 + 273.15) K = 273.15 K
Using above equation as:
[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}[/tex]
[tex] \frac {{1.2}\times {530}}{290.15}=\frac {{1}\times {V_2}}{273.15}[/tex]
Solving for V₂ , we get:
V₂ = 598.74 mL
At 20°C, the ion-product constant of water, Kw, is 6.88 x 10-15 . What is the pH of pure water at 20°C?
A) 7.000
B) 6.501
C) 7.181
D) 7.081
E) none of these
Answer:
The correct answer is D) 7.081
Explanation:
The water equilibrium is the following:
H₂O ⇄ H⁺ + OH⁻ Kw
Where Kw= [H⁺] x [OH⁻]
In pure water: [H⁺] = [OH⁻]= C
If we introduce C in the expression for Kw:
Kw= [H⁺] x [OH⁻]= C x C= C²
⇒ C= [tex]\sqrt{Kw}[/tex]= [tex]\sqrt{6.88 x 10^{-15} }[/tex]= 8.29 x 10⁻⁸
pH= -log [H⁺] = -log C = -log (8.29 x 10⁻⁸) = 7.081
Final answer:
At 20°C, the pH of pure water is 7.000.
Explanation:
The ion-product constant of water (Kw) is the product of the concentrations of hydrogen ions (H3O+) and hydroxide ions (OH-) in water. At 20°C, the value of Kw is 6.88 x 10^-15. Since pure water is neutral, the concentration of hydrogen ions and hydroxide ions in pure water are equal. Therefore, at 20°C, the concentration of hydrogen ions and hydroxide ions in pure water is 6.88 x 10^-15 M.
The pH of a solution is a measure of the concentration of hydrogen ions. It is calculated as the negative logarithm (base 10) of the hydrogen ion concentration. Therefore, at 20°C, the pH of pure water is 7.000 (approximately) since the hydrogen ion concentration and the hydroxide ion concentration in pure water are both 6.88 x 10^-15 M.
Dry ice is solid carbon dioxide. Instead of melting, solid carbon dioxide sublimes according to the equation: CO2(s)→CO2(g). When dry ice is added to warm water, heat from the water causes the dry ice to sublime more quickly. The evaporating carbon dioxide produces a dense fog often used to create special effects. In a simple dry ice fog machine, dry ice is added to warm water in a Styrofoam cooler. The dry ice produces fog until it evaporates away, or until the water gets too cold to sublime the dry ice quickly enough. Suppose that a small Styrofoam cooler holds 15.0 L of water heated to 90 ∘C. Calculate the mass of dry ice that should be added to the water so that the dry ice completely sublimes away when the water reaches 28 ?
Final answer:
To calculate the mass of dry ice needed to completely sublime in water, we can use the concept of stoichiometry and the ideal gas law equation. The moles of CO2 can be determined using the ideal gas law equation, and then converted to moles of water using the stoichiometry of the balanced equation. Finally, the mass of dry ice can be calculated using the moles of CO2 and the molar mass of CO2.
Explanation:
In order to calculate the mass of dry ice that should be added to the water, we need to use the concept of stoichiometry. The equation for the sublimation of carbon dioxide is CO2(s) -> CO2(g). We can use the molar ratio between CO2(s) and CO2(g) to convert the volume of water to moles of CO2. From there, we can use the molar mass of CO2 to calculate the mass of dry ice needed.
First, we need to calculate the moles of CO2 that can be produced using the ideal gas law equation:
n = PV / RT
where P is the pressure, V is the volume, R is the ideal gas constant, and T is the temperature. Since temperature and pressure are not given, we cannot directly solve for moles of CO2. However, we can assume that the pressure and temperature remain constant throughout the process, so the ideal gas equation can still be used to find the moles of CO2.
The next step is to use stoichiometry to convert moles of CO2 to moles of water. The balanced equation is:
CO2(s) + H2O(l) -> H2CO3(aq)
From the balanced equation, we can see that the mole ratio between CO2 and water is 1:1. Therefore, if we know the moles of CO2, we also know the moles of water. The molar mass of CO2 is 44.01 g/mol, so we can calculate the mass of dry ice needed using the equation:
mass = moles * molar mass
Since we know the volume of water (15.0 L) and the final temperature (28°C), we can plug those values into the ideal gas equation and calculate the moles of CO2. Then, we can use the moles of CO2 to calculate the mass of dry ice needed. The final mass will depend on the pressure and temperature at which the dry ice is added, as well as the final temperature of the water.
Final answer:
The student's question involves calculating the mass of dry ice needed to produce fog in water until the temperature drops to 28°C. To perform this calculation, details such as the heat capacity of water, the heat of sublimation of CO2, and the complete temperature value are critical, but these are not provided in the question.
Explanation:
Dry ice, known chemically as solid carbon dioxide (CO2), sublimes at room temperature and pressure, meaning it converts directly from solid to gas. In the context of the question, when dry ice is added to warm water, the heat transfer from the water to the dry ice accelerates this sublimation process. The interaction cools the water down while providing a fog effect until either all the dry ice sublimes or the water temperature drops too much to sustain rapid sublimation.
To calculate the mass of dry ice required to create these special effects until the water cools to 28°C, one would need to determine the amount of heat the water can provide before reaching this temperature, and then use the heat of sublimation for CO2 to find the corresponding mass of dry ice. However, since the question doesn't provide specific heat capacity, heat of sublimation values, or the completion temperature value (presumably 28°C), an exact calculation cannot be performed without this information.
Handling dry ice requires caution due to its extremely cold temperature, and it is essential to be in a well-ventilated area since the resulting CO2 gas is heavier than air and can displace oxygen.
A solution is 0.10 M in Pb(NO3)2 and 0.10 M in AgNO3. Solid NaI is added until the second solid compound is on the verge of precipitating.
Which compound precipitates first and what is the I- concentration when the second compound begins to precipitate?Ksp for AgI is 1.5 x 10-16 Ksp for PbI2 is 8.7 x 10-9PbI_2,\:5.9\times10^{-4}\:M
PbI_2,\:4.4\times10^{-3}\:M
AgI,\:2.9\times10^{-4}\:M
AgI,\:9.3\times10^{-5}\:M
AgI,\:8.7\times10^{-8}\:M
AgI precipitates first due to its smaller Ksp. The I⁻ concentration for the second compound (PbI₂) to begin precipitating is 9.3 x 10⁻⁵ M.
Explanation:The compound that will precipitate first will be the one with the smallest solubility product constant, Ksp. The given Ksp values for AgI and PbI₂ are 1.5 x 10⁻¹⁶ and 8.7 x 10⁻⁹ respectively; thus, AgI has a smaller Ksp, indicating it is less soluble and precipitates first.
To find the I⁻ concentration for the second compound to precipitate, we set up the Ksp expression for PbI₂ (assuming AgI has already precipitated out, AgNO₃ is gone and the remaining AgI does not affect this calculation): [Pb²⁺][I⁻]² = Ksp. Pb(NO₃)₂ fully dissociates, so [Pb²⁺] is assumed to be 0.10 M, and we can substitute this and the Ksp into our expression, then solve for [I-]: (0.10 M)([I⁻])² = 8.7 x 10⁻⁹ M, from which we find [I-] = sqrt((8.7 x 10⁻⁹ M) / 0.10 M) = 9.3 x 10⁻⁵ M.
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Consider a general reaction A ( aq ) enzyme ⇌ B ( aq ) A(aq)⇌enzymeB(aq) The Δ G ° ′ ΔG°′ of the reaction is − 5.980 kJ ⋅ mol − 1 −5.980 kJ·mol−1 . Calculate the equilibrium constant for the reaction at 25 °C. K ′ e q = Keq′= What is ΔG for the reaction at body temperature (37.0 °C) if the concentration of A is 1.8 M 1.8 M and the concentration of B is 0.55 M ?
The equilibrium constant for the reaction at 25 °C, [tex]K_{eq}[/tex] is 11.2.
ΔG for the reaction at body temperature (37.0 °C) is -9.04 kJ/mol.
Relation between equilibrium constant and standard Gibbs free energy:[tex]\triangle G^o= -RT*lnK_{eq}[/tex]
where,
[tex]\triangle G ^o[/tex] = standard Gibbs free energy = -5.980 kJ/mol = -5980 J/mol
R = gas constant = 8.314 J/K.mol
T = temperature = 298K
[tex]K_{eq}[/tex] = equilibrium constant = ?
→ Calculation for [tex]K_{eq}[/tex] :
Now, substituting the values in the above formula:
[tex]\triangle G^o= -RT*lnK_{eq}\\\\-5980J/mol=-(8.314J/K.mol)*(298K)* lnK_{eq}\\\\K_{eq}=11.2[/tex]
Thus, the value of [tex]K_{eq}[/tex] is 11.2.
→ Calculation for [tex]\triangle G_{rxn}[/tex]:
Chemical reaction:
A(aq) ⇌B(aq)
The formula used is:
[tex]\triangle G_{rxn}=\traingle G^o+RT lnQ\\\\\triangle G_{rxn}=\traingle G^o+RT ln\frac{[B]}{[A]}[/tex]
where,
[tex]\triangle G_{rxn}[/tex] = Gibbs free energy for the reaction = ?[tex]\triangle G^o[/tex] = standard Gibbs free energy = -30.5 kJ/molR = gas constant = 8.314 J/K.molT = temperature = 310 KQ = reaction quotient[A] = concentration of A = 1.8 M[B] = concentration of B = 0.55 MOn substituting the values in the above formula:
[tex]\triangle G_{rxn}=\traingle G^o+RT ln\frac{[B]}{[A]}\\\\\triangle G_{rxn}=(-5980J/mol)+[(8.314J/mole.K)*(310K)*ln\frac{0.55}{1.8}] \\\\\triangle G_{rxn}=-9035.75J/mole=-9.04kJ/mol[/tex]
Therefore, the value of [tex]\triangle G_{rxn}[/tex] is -9.04 kJ/mol.
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While many metabolic pathways classify as catabolic or anabolic, the citric acid cycle is amphibolic Select statements that describe amphibolic characteristics of the citric acid cycle ? ? catabolic pathways for several macromolecules involve the citric acid cycle. Both oxidation and reduction reactions occur. -Ketoglutarate is a product of the citric acid cycle and an amino acid precursor. Both catabolic and anabolic processes occur. The citric acid cycle produces succiny/L-CoA, a requirement for heme group synthesis.
Answer:Answer: The statements that describe amphibolic characteristics of the citric acid cycle is that both catabolic and anabolic processes occur.
Explanation: the metabolic pathways includes; catabolic, anabolic and amphibolic processes.
A catabolic pathway is an exergonic system that produces chemical energy in the form of Adenosine Triphosphate (ATP) usually from energy containing sources such as carbohydrates, fats, and proteins.
An anabolic pathway is a biosynthetic pathway, whereby smaller molecules combine to form larger and more complex ones.[
An amphibolic pathway is one that can be either catabolic or anabolic based on the need for energy. The citric acid pathway contains both energy producing and utilizing pathway.
The citric acid cycle is amphibolic, participating in both catabolic and anabolic processes, producing intermediates for various biosynthetic pathways and involving oxidation and reduction reactions.
The citric acid cycle, also known as the Krebs cycle, is considered amphibolic because it has both catabolic and anabolic functions. It is a central metabolic pathway in the cell, involved in the oxidation of acetyl-CoA into CO2, and is crucial for energy production in the form of ATP. Furthermore, the cycle provides intermediates for various anabolic pathways, making it an integral part of cellular metabolism. For instance, it produces
-ketoglutarate, which is a precursor for amino acid synthesis, and succinyl-CoA, which is required for heme group synthesis. Also, the cycle involves both oxidation and reduction reactions, further asserting its amphibolic nature.
Anaplerotic and cataplerotic pathways are essential for maintaining the function of the citric acid cycle. Anaplerotic pathways replenish cycle intermediates that are removed for biosynthesis, whereas cataplerotic pathways describe the removal of intermediates from the cycle for the synthesis of other compounds. This balance allows the cycle to efficiently meet the fluctuating needs of the cell for both energy production and the synthesis of crucial molecules.
Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Supposed 0.481 g of methane is mixed with 0.54 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.
Answer:
1.08g
Explanation:
Like all other hydrocarbons, methane burns in oxygen to form carbon iv oxide and water. The chemical equation of this equation is shown below;
CH4 (g) + 2O2 (g) -----------> 2H2O (l) + CO2 (g)
From the reaction equation, we can see that one mole of methane gave 2 moles of water. This is the theoretical yield. We need to note the actual yield.
To get the actual yield, we get the number of moles of methane reacted. To get this, we divide the mass of methane by the molar mass of methane. The molar mass of methane is 16g/mol. The number of moles is thus 0.481/16 = 0.03 moles.
Since 1 mole methane gave 2 moles of water, this shows that 0.06 moles of water were produced. The mass of water thus produced is 0.06 multiplied by the molar mass of water. The molar mass of water is 18g/mol. The mass produced is 0.06 * 18 = 1.08g
Now, we do same for the mass of oxygen. From the reaction equation, 2 moles of oxygen produced two moles of water. Hence, we can see from here that the number of moles here are equal. We then proceed to calculate the actual number of moles of oxygen produced. This is the mass of the oxygen divided by the molar mass of molecular oxygen. The molar mass of molecular oxygen is 32g/mol. The number of moles thus produced is 0.54/32 = 0.016875 mole. The number of moles are equal, this means that the number of moles of oxygen produced is also 0.016875
Now, to get the mass of water produced, we multiply the number of moles by the molar mass of water. The molar mass of water is 18g/mol.
The mass is thus, 0.016875 * 18 = 0.30375g
1.08g is higher and thus is the maximum mass
A mixture of gaseous reactants is put in to a cylinder, where a chemical reaction turns them in to gaseous products. The cylinder has a piston that moves in or out as necessary to keep a constant pressure on the mixture of 1 atm. The cylinder is also submerged in a large insulated water bath.
From previous experiments, this chemical reaction is known to release 244. kJ of energy.
The temperature of the water bath is monitored, and it is determined from this data that 150. kJ of heat flows out of the system during the reaction.
1) is the reaction exothermic or endothermic?
2) Does the temperature of the water go up or down?
3) does the piston move in, out, or neither?
4) does The gas mixture do work, Or is work done on it?
5) how much work is done on (or by) the gas mixture?
Answer:
1) Exothermic.
2) It goes up.
3) It moves out.
4) It does work.
5) 94 kJ.
Explanation:
1) An endothermic reaction is a reaction that absorbs heat from the surroundings, and an exothermic reaction is a reaction that releases heat to the surroundings. So the reaction placed is exothermic.
2) Because the heat is flowing to the water, its temperature will go up.
3) By the first law of the thermodynamics:
ΔU = Q - W
Where ΔU is the total energy, Q is the heat, and W is the work. Because the energy and the heat are being released, they are both negative:
-244 = -150 - W
W = 94 kJ
The work is positive, so it's being doing by the system, it means that the system is expanding, and the piston moves out.
4) As explained above, the gas mixture (the system) does work.
5) As shown above, W = 94 kJ
For the reaction A + B − ⇀ ↽ − C + D , assume that the standard change in free energy has a positive value. Changing the conditions of the reaction can alter the value of the change in free energy ( Δ G ) . Classify the conditions as to whether each would decrease the value of Δ G , increase the value of Δ G , or not change the value of Δ G for the reaction. For each change, assume that the other variables are kept constant.
a. Adding a catalyst decrease the free energy value, increase the free energy value, or not change the free energyb. increasing [C] and [D] decrease the free energy value, increase the free energy value, or not change the free energyc. Coupling with ATP hydrolysis decrease the free energy value, increase the free energy value, or not change the free energyd.Increasing [A] and [B] decrease the free energy value, increase the free energy value, or not change the free energy
Final answer:
The conditions that can alter the value of ΔG for the reaction A + B ⇌ C + D are explained. Adding a catalyst does not change the value of ΔG. Increasing the concentrations of products decreases ΔG. Coupling with ATP hydrolysis decreases ΔG. Increasing the concentrations of reactants also decreases ΔG.
Explanation:
The reaction A + B ⇌ C + D has a positive standard change in free energy (ΔG°). Changing the conditions of the reaction can alter the value of ΔG. Let's classify the conditions:
a. Adding a catalyst: Adding a catalyst can decrease the activation energy and increase the rate of the reaction. However, it does not change the value of ΔG.
b. Increasing [C] and [D]: Increasing the concentrations of products (C and D) will shift the equilibrium towards the reactants and decrease the value of ΔG.
c. Coupling with ATP hydrolysis: Coupling the reaction with ATP hydrolysis, which is an exergonic reaction, can decrease the value of ΔG and make the overall reaction more spontaneous.
d. Increasing [A] and [B]: Increasing the concentrations of reactants (A and B) will shift the equilibrium towards the products and decrease the value of ΔG.
What is the wavelength (in nm) of the photon absorbed for a transition of an electron from ninitial = 2 that results in the least energetic spectral line in the visible series of the H atom?
Answer: 656.6 nm.
Explanation:
Using Rydberg's Equation for hydrogen atom:
[tex]\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )[/tex]
Where,
[tex]\lambda[/tex] = Wavelength of radiation
[tex]R_H[/tex] = Rydberg's Constant
[tex]n_f[/tex] = Higher energy level = 3 (least energetic for visible series)
[tex]n_i[/tex]= Lower energy level = 2
We have:
[tex]n_f=3, n_i=2[/tex]
[tex]R_H=1.097\times 10^7 m^{-1}[/tex]
[tex]\frac{1}{\lambda}=1.097\times 10^7 m^{-1}\times \left(\frac{1}{2^2}-\frac{1}{3^2} \right )[/tex]
[tex]\frac{1}{\lambda}=1.097\times 10^7 m^{-1}\times \frac{5}{36}[/tex]
[tex]\frac{1}{\lambda}=0.1523\times 10^{7} m[/tex]
[tex]\lambda=6.566\times 10^{-7}m=656.6nm[/tex]
([tex]1 m= 10^9nm[/tex])
The wavelength of the photon emitted when the hydrogen atom undergoes a transition from n = 2 to n = 3 is 656.6 nm.
Consider the reaction directly below and answer parts a and b. C6H4(OH)2 (l) + H2O2 (l) à C6H4O2 (l) + 2 H2O (l)
(a) Use the following information to calculate DH° for the above reaction. Show all work. DH° C6H4(OH)2 (l) à C6H4O2 (l) + H2 (g) +177.4 kJ H2 (g) + O2 (g) à H2O2 (l) -187.8 kJ H2 (g) + O2 (g) à H2O (l) -285.8 kJ
(b) Based on your answer to part a, above, would heat be gained or lost by the surroundings as the reaction at the very top of the page occurred at standard conditions?
(c) How many kilojoules of heat are given off when 20.0 g of H2O (l) are produced at standard conditions according to the reaction at the very top of the page? Show all work.
Answer:
a. -206,4kJ
b. Surroundings will gain heat.
c. -115kJ are given off.
Explanation:
It is possible to obtain ΔH of a reaction using Hess's law that consist in sum the different ΔH's of other reactions until obtain the reaction you need.
Using:
(1) C₆H₄(OH)₂(l) → C₆H₄O₂(l) + H₂(g) ΔH: +177.4 kJ
(2) H₂(g) + O₂(g) → H₂O₂ (l) ΔH: -187.8 kJ
(3) H₂(g) + 1/2O₂(g) → H₂O(l) ΔH: -285.8 kJ
It is possible to obtain:
C₆H₄(OH)₂(l) + H₂O₂(l) → C₆H₄O₂(l) + 2H₂O(l)
From (1)-(2)+2×(3). That is:
(1) C₆H₄(OH)₂(l) → C₆H₄O₂(l) + H₂(g) ΔH: +177.4 kJ
-(2) H₂O₂(l) → H₂(g) + O₂(g) ΔH: +187.8 kJ
2x(3) 2H₂(g) + O₂(g) → 2H₂O(l) ΔH: 2×-285.8 kJ
The ΔH you obtain is:
+177,4kJ + 187,8kJ - 2×285.8 kJ = -206,4kJ
b. When ΔH of a reaction is <0, the reaction is exothermic, that means that the reaction produce heat and the surroundings will gain this heat.
c. 20,0g of H₂O are:
20,0g×[tex]\frac{1mol}{18,01g}[/tex] = 1,11 mol H₂O
As 2 moles of H₂O are produced when -206,4kJ are given off, when 1,11mol of H₂O are produced, there are given off:
1,11mol H₂O×[tex]\frac{-206,4kJ}{2mol}[/tex] = -115kJ
I hope it helps!
Final answer:
To calculate the enthalpy change (ΔH°) for the given reaction, use Hess's Law and the enthalpy changes provided. Heat would be gained by the surroundings as the reaction occurs, based on the positive enthalpy change. When 20.0 g of H2O is produced, approximately -317 kJ of heat would be given off.
Explanation:
To calculate the enthalpy change (ΔH°) for the given reaction, we can use the Hess's Law and the enthalpy changes provided. We need to find a combination of reactions whose sum gives us the desired reaction. Here's an approach:
1. Reverse the first equation and multiply it by 2 to get 2CO(g) → 2C(s) + O2(g) with ΔH° = -442.0 kJ.
2. Use the second equation as it is: C(s) + O2(g) → CO2(g) with ΔH° = -393.5 kJ.
3. Add the two equations together: 2CO(g) + C(s) + O2(g) → 2CO2(g) with ΔH° = -835.5 kJ.
This combined equation is equivalent to the given reaction, but we need to multiply it by -2 to match the coefficients. Therefore, the enthalpy change for the given reaction is ΔH° = -2*(-835.5 kJ) = 1671 kJ.
For part (b), based on the positive enthalpy change, heat would be gained by the surroundings as the reaction occurs. In part (c), we can use the enthalpy change value from part (a) to calculate the heat produced. Since the reaction produces 2 moles of H2O, we can use the provided enthalpy change for H2O(l) to find the heat produced when 20.0 g of H2O is produced.
Using the given enthalpy change values:
ΔH° C6H4(OH)2(l) → C6H4O2(l) + H2(g) = +177.4 kJ
ΔH° H2(g) + O2(g) → H2O(l) = -285.8 kJ
We can set up the following equation to find the heat produced:
(20.0 g H2O) x (1 mol H2O / 18.015 g H2O) x (-285.8 kJ / 1 mol H2O) = -317 kJ.
Balance the following equation: K2CrO4+Na2SO3+HCl→KCl+Na2SO4+CrCl3+H2O Generally coefficients of 1 are omitted from balanced chemical equations. When entering your answer, include coefficients of 1 as required for grading purposes. Enter the coefficients for each compound, separated by commas, in the order in which they appear in the equation (e.g., 1,2,3,4,5,6,7).
Answer:
The coefficients found are: 2,3,10,4,3,2,5
The sum of the coefficients is: 2+3+10 + 4 + 3 + 2 + 5 = 29
Explanation:
K2CrO4+Na2SO3+HCl→KCl+Na2SO4+CrCl3+H2O
On the left side we 2 times K , on the right side we have 1 time K
So we have to multiply KCl by 2
K2CrO4+Na2SO3+HCl→2KCl+Na2SO4+CrCl3+H2O
On the left side wehave 1 Cl, on the right side we have 5 Cl
So the left side we have to multiply by 5
K2CrO4+Na2SO3+5 HCl→2KCl+Na2SO4+CrCl3+H2O
On the left side we have 5 times H, on the right side, we have 2 times H
So on the left side the HCl we should multiply by 10 (instead of 5)
On the right side, we should multiply H2O by 5
K2CrO4+Na2SO3+ 10HCl→2KCl+Na2SO4+CrCl3+ 5H2O
So on the left side we have 10 times Cl, on the right side KCl and CrCL3 should be doubled
K2CrO4+Na2SO3+ 10HCl→4 KCl+Na2SO4+ 2CrCl3+ 5H2O
On the right side we have 4 times K, and 2 times Cr. This mean on the left side K2CrO4 should be multiplied by 2
2K2CrO4+Na2SO3+ 10HCl→4 KCl+Na2SO4+ 2CrCl3+ 5H2O
On the left side we have 8 times O plus 3x times O
On the right side we have 5 times O plus 4x times O
To equal this we should multiply on the left side Na2SO3 by 3
and Na2SO4 on the right side also by 3
Now the entire equation is balanced
2 K2CrO4 + 3 Na2SO3 + 10 HCl → 4 KCl + 3 Na2SO4 + 2 CrCl3 + 5 H2O
The coefficients found are: 2,3,10,4,3,2,5
The sum of the coefficients is: 2+3+10 + 4 + 3 + 2 + 5 = 29
Final answer:
To balance the chemical equation, coefficients are placed in front of the compounds to ensure the same number and types of atoms on both sides. The balanced equation is: 3K2CrO4 + 3Na2SO3 + 14HCl → 3KCl + 3Na2SO4 + 2CrCl3 + 7H2O and the coefficients are 3,3,14,3,3,2,7.
Explanation:
To balance the given chemical equation, we must ensure that the number of atoms of each element is the same on both sides of the equation. Let's add coefficients to balance:
Start with Chrome atoms: There is 1 Cr on the left side and 3 on the right side, so place a coefficient 3 in front of K2CrO4.
Next, look at sulfur atoms: There are 3 S on the left and 3 S on the right, so we are now balanced for sulfur.
For potassium atoms, put a coefficient 3 in front of KCl on the product side.
Balance sodium by having a coefficient 3 in front of Na2SO4 since there are 6 Na on the reactant side.
Hydrogen atoms are balanced by placing a coefficient 6 in front of HCl to get 6 H atoms on both sides.
Now, balance the chloride ions: We have 6 Cl from HCl, but we need 9 Cl on the products side (3 from KCl and 6 from CrCl3). So the final coefficient for HCl is 14.
Finally, place a coefficient 7 in front of H2O to balance out the oxygen atoms.
The balanced equation is therefore: 3K2CrO4 + 3Na2SO3 + 14HCl → 3KCl + 3Na2SO4 + 2CrCl3 + 7H2O. The coefficients in the order that they appear in the equation are: 3,3,14,3,3,2,7.
UsingLewissymbolsandLewisstructures,diagramthefor- mation of SiCl4 from Si and Cl atoms, showing valence-shell electrons. (a) How many valence electrons does Si have ini- tially? (b) How many valence electrons does each Cl have initially? (c) How many valence electrons surround the Si in the SiCl4 molecule? (d) How many valence electrons sur- round each Cl in the SiCl4 molecule? (e) How many bonding pairs of electrons are in the SiCl4 molecule?
Answer:
(a) 4
(b) 7
(c) 8
(d) 8
(e) 4
Explanation:
In order to draw the Lewis structure, we need to take into account the octet rule: atoms will share electrons to have 8 electrons in their valence shell.
(a) How many valence electrons does Si have initially?
Si has 4 valence electrons. It has to share 4 pairs of electrons to reach the octet.
(b) How many valence electrons does each Cl have initially?
Cl has 7 valence electrons. Each Cl has to share 1 pair of electrons to reach the octet.
(c) How many valence electrons surround the Si in the SiCl₄ molecule?
Si is surrounded by 8 electrons in SiCl₄.
(d) How many valence electrons surround each Cl in the SiCl₄ molecule?
Each Cl is surrounded by 8 electrons in SiCl₄.
(e) How many bonding pairs of electrons are in the SiCl₄ molecule?
There are 4 bonding pairs between Si and Cl.
Silicon (Si) forms SiCl₄ by bonding with four chlorine (Cl) atoms. Si initially has 4 valence electrons, while each Cl has 7. In SiCl₄, Si has 8 valence electrons, and each Cl has 8. There are four bonding pairs of electrons in the molecule, formed by single bonds between Si and Cl.
Lewis Structure for the Formation of SiCl₄:
(a) **Valence Electrons in Si (Silicon):** Silicon is in Group 14 of the periodic table, so it has 4 valence electrons.
(b) **Valence Electrons in Each Cl (Chlorine) Atom:** Chlorine is in Group 17, so each Cl atom has 7 valence electrons.
(c) **Valence Electrons Surrounding Si in SiCl₄ Molecule:** In SiCl₄, Silicon forms four single bonds with four Cl atoms. Each bond involves two electrons. Therefore, around Silicon, there are 4 * 2 = 8 valence electrons.
(d) **Valence Electrons Surrounding Each Cl in SiCl₄ Molecule:** Each Cl atom is involved in one bond with Si, contributing one pair of electrons. Therefore, each Cl has a total of 8 valence electrons around it (7 from its own valence shell plus 1 from the shared bond).
(e) **Bonding Pairs of Electrons in SiCl₄ Molecule:** Since Si forms four single bonds with four Cl atoms, there are four bonding pairs of electrons in the SiCl₄ molecule.
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Use Appendix E in the textbook. Which of the following substances can oxidize RuO42−(aq) to RuO4−(aq) under standard conditions? Use Appendix E in the textbook. Which of the following substances can oxidize to under standard conditions? Ni2+(aq) I2(s) Cr2O72−(aq) ClO3−(aq) Pb2+(aq)
Answer:
Cr₂O₇⁻²(aq) and ClO₃⁻(aq)
Explanation:
At a redox reaction, one substance must be reduced (gain electrons) and others must be oxidized (lose electrons). To evaluate the potential of the substance to be reduced, it's placed a reaction, in standards conditions, with H₂.
The potential reduction is quantified by E°, and as higher is the value of E°, as easy is to the compound to be reduced. So, at a redox reaction, the compound with the greatest E° will be reduced, and the other will be oxidized, in a spontaneous reaction. The values of E° are:
RuO₄⁻(aq) to RuO₄²⁻(aq) E° = + 0.59 V (the reduction reaction is the opposite of the oxidation reaction).
Ni⁺²(aq) E° = -0.257 V
I₂(s) E° = +0.535 V
Cr₂O₇⁻²(aq) E° = +1.33 V
ClO₃⁻(aq) E° = +0.890 V
Pb²⁺(aq) E° = -0.125 V
So, the substances that have E° higher than the E° of the RuO₄⁻²(aq) are Cr₂O₇⁻²(aq) and ClO₃⁻(aq), which are the substances that can oxidize RuO₄⁻(aq) to RuO₄²⁻(aq).
The correct substances that can oxidize RuO42(aq) to RuO4(aq) under standard conditions are Cr2O72(aq) and ClO3(aq).
To determine which substances can act as oxidizing agents for RuO42−(aq), we need to look at the standard reduction potentials (E°) for the given half-reactions. An oxidizing agent is a substance that accepts electrons, and for a redox reaction to occur spontaneously, the substance with the higher reduction potential will oxidize the substance with the lower reduction potential.
The half-reaction for the reduction of RuO4’(aq) to RuO4(aq) is as follows:
[tex]\[ \text{RuO}_4(aq) + e^- \rightarrow \text{RuO}_4^{2-}(aq) \][/tex]
We do not have the standard reduction potential for this half-reaction, but we can compare the potentials of the given substances to determine if they can act as oxidizing agents.
From Appendix E, we can find the standard reduction potentials for the given substances:
1.[tex]\( \text{Ni}^{2+}(aq) + 2e^- \rightarrow \text{Ni}(s) \) with \( E^\circ = -0.25 \) V[/tex]
2.[tex]\( \text{I}_2(s) + 2e^- \rightarrow 2\text{I}^-(aq) \) with \( E^\circ = +0.54 \) V[/tex]
3.[tex]\( \text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6e^- \rightarrow 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l) \) with \( E^\circ = +1.33 \) V[/tex]
4. [tex]\( \text{ClO}_3^-(aq) + 6\text{H}^+(aq) + 6e^- \rightarrow \text{Cl}^-(aq) + 3\text{H}_2\text{O}(l) \) with \( E^\circ = +1.45 \) V[/tex]
5.[tex]\( \text{Pb}^{2+}(aq) + 2e^- \rightarrow \text{Pb}(s) \) with \( E^\circ = -0.13 \) V[/tex]
Comparing these potentials with the half-reaction for RuO42(aq), we can see that Ni2+(aq) and Pb2+(aq) have negative reduction potentials, which means they cannot oxidize RuO42(aq) because they are stronger reducing agents than RuO42(aq).
I2(s) has a positive reduction potential, but it is lower than the reduction potentials of Cr2O72(aq) and ClO3(aq). Therefore, I2(s) can oxidize RuO42’(aq), but there are stronger oxidizing agents available.
Cr2O72(aq) and ClO3(aq) have the highest reduction potentials in the list, which means they are the strongest oxidizing agents and can oxidize RuO4(aq) to RuO4(aq) under standard conditions.
Thus, the correct substances that can oxidize RuO42(aq) to RuO4(aq) under standard conditions are Cr2O72(aq) and ClO3(aq)
Using the standard enthalpies of formation for the chemicals involved, calculate the enthalpy change for the following reaction.
(note: show the math clearly and provide units in your set up) ( Hf values in kJ/mol are as follows: NO2 32, H2O 286, HNO3 207, NO 90.)
3NO2(g) H2O(l) 2HNO3(aq) NO(g) g
Final answer:
The enthalpy change (ΔHreaction) for the given reaction is calculated using Hess's law and is found to be 122 kJ/mol.
Explanation:
To calculate the enthalpy change (ΔHreaction) for the reaction 3NO₂(g) + H₂O(l) → 2HNO3(aq) + NO(g), we will apply Hess's law and use the standard enthalpies of formation (ΔHf) for each compound.
The formula to calculate ΔHreaction is:
ΔHreaction = ∑(ΔHf products) - ∑(ΔHf reactants)
Using the provided ΔHf values:
NO₂(g): 32 kJ/molH₂O(l): 286 kJ/molHNO3(aq): 207 kJ/molNO(g): 90 kJ/molWe can calculate ΔHreaction as follows:
ΔHreaction = [2 × ΔHf(HNO3) + ΔHf(NO)] - [3 × ΔHf(NO2) + ΔHf(H2O)]
ΔHreaction = [(2 × 207 kJ/mol) + (90 kJ/mol)] - [(3 × 32 kJ/mol) + (286 kJ/mol)]
ΔHreaction = (414 kJ/mol + 90 kJ/mol) - (96 kJ/mol + 286 kJ/mol)
ΔHreaction = 504 kJ/mol - 382 kJ/mol
ΔHreaction = 122 kJ/mol
Which of the following conclusions can be drawn from JJ Thomson's cathode ray experiments?A.) Atoms contain electronsB.) Practically all the mass of an atom is contained in its nucleusC.) Atoms contain protons, neutrons, and electronsD.) Atoms have positively charged nuclues surrounded by electron cloud
Answer:
A.) Atoms contain electrons
Explanation:
J.J. Thomson conducted an experiment in which he took a gas at low pressure in a discharge tube and applied high voltage current. When he did so, he noticed that there are some particles emitting from cathode going towards the anode. Then he concluded that they are negatively charged particles and coined them electrons.
Hence, out of the options:- A.) Atoms contain electrons is correct.
What is the percent ionization of a 1.8 M HC2H3O2 solution (Ka = 1.8 10-5 ) at 25°C?
A) 0.57 %
B) 0.32 %
C) 2.2 %
D) 0.18 %
E) 0.24 %
Answer:
B) 0.32 %
Explanation:
Given that:
[tex]K_{a}=1.8\times 10^{-5}[/tex]
Concentration = 1.8 M
Considering the ICE table for the dissociation of acid as:-
[tex]\begin{matrix}&CH_3COOH&\rightleftharpoons &CH_3COOH&+&H^+\\ At\ time, t = 0 &1.8&&0&&0\\At\ time, t=t_{eq}&-x&&+x&&+x\\ ----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:-&1.8-x&&x&&x\end{matrix}[/tex]
The expression for dissociation constant of acid is:
[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}[/tex]
[tex]1.8\times 10^{-5}=\frac{x^2}{1.8-x}[/tex]
[tex]1.8\left(1.8-x\right)=100000x^2[/tex]
Solving for x, we get:
x = 0.00568 M
Percentage ionization = [tex]\frac{0.00568}{1.8}\times 100=0.32 \%[/tex]
Option B is correct.
Final answer:
After setting up the equilibrium expression for acetic acid dissociation and solving the equation, the percent ionization of a 1.8 M HC2H3O2 solution is determined to be approximately 0.1%. The closest answer option given is D) 0.18%.
Explanation:
To calculate the percent ionization for a 1.8 M HC2H3O2 solution, we need to set up an equilibrium expression based on the acid dissociation constant (Ka) and initial concentration of acetic acid. We start with the dissociation reaction:
HC2H3O2 \<=> H+ + C2H3O2-
Let's assume that the concentration of H+ (x) is equal to the concentration of C2H3O2- (x) at equilibrium due to stoichiometry:
Ka = [H+][C2H3O2-]/[HC2H3O2]
If we let x represent the molarity of H+ and C2H3O2- ions at equilibrium and assuming that x is small enough that (1.8 M - x) ≈ 1.8 M, we can simplify the expression to:
Ka = x^2/1.8 M
The equation becomes:
1.8 \ 10^-5 = x^2/1.8
Solving for x gives us:
x ≈ \ 10^-5 \ 1.8)}\> ≈ 1.8 \ 10^-3 M
The percent ionization is then calculated by:
Percent ionization = (x / [HC2H3O2] initial) \ 100%
Percent ionization = (1.8 \ 10^-3 M / 1.8 M) \ 100% = 0.1%
Since none of the options given matches the calculated value strictly, and provided that we've made an approximation, we'll have to choose the closest answer, which is:
D) 0.18%
Identify which redox reactions occur spontaneously in the forward direction. Check all that apply.
a.Fe(s)+Mn2+(aq)→Fe2+(aq)+Mn(s)
b. 2Ag+(aq)+Fe(s)→2Ag(s)+Fe2+(aq)
c. Mg2+(aq)+Zn(s)→Mg(s)+Zn2+(aq)
d. 2Al(s)+3Pb2+(aq)→2Al3+(aq)+3Pb(s)
Answer:
For a: The reaction is not spontaneous.
For b: The reaction is spontaneous.
For c: The reaction is not spontaneous.
For d: The reaction is spontaneous.
Explanation:
For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.
Relationship between standard Gibbs free energy and standard electrode potential follows:
[tex]\Delta G^o=-nFE^o_{cell}[/tex]
For a reaction to be spontaneous, the standard electrode potential must be positive.
To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:
[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex] .......(1)
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
For a:The chemical reaction follows:
[tex]Fe(s)+Mn^{2+}(aq.)\rightarrow Fe^{2+}(aq.)+Mn(s)[/tex]
We know that:
[tex]E^o_{Fe^{2+}/Fe}=-0.44V\\E^o_{Mn^{2+}/Mn}=-1.18V[/tex]
Calculating the [tex]E^o_{cell}[/tex] using equation 1, we get:
[tex]E^o_{cell}=-1.18-(-0.44)=-0.74V[/tex]
As, the standard electrode potential is coming out to be negative. So, the reaction is not spontaneous.
For b:The chemical reaction follows:
[tex]Fe(s)+2Ag^{+}(aq.)\rightarrow Fe^{2+}(aq.)+2Ag(s)[/tex]
We know that:
[tex]E^o_{Fe^{2+}/Fe}=-0.44V\\E^o_{Ag^{+}/Ag}=0.80V[/tex]
Calculating the [tex]E^o_{cell}[/tex] using equation 1, we get:
[tex]E^o_{cell}=0.80-(-0.44)=1.24V[/tex]
As, the standard electrode potential is coming out to be positive. So, the reaction is spontaneous.
For c:The chemical reaction follows:
[tex]Zn(s)+Mg^{2+}(aq.)\rightarrow Zn^{2+}(aq.)+Mg(s)[/tex]
We know that:
[tex]E^o_{Zn^{2+}/Zn}=-0.76V\\E^o_{Mg^{2+}/Mg}=-2.37V[/tex]
Calculating the [tex]E^o_{cell}[/tex] using equation 1, we get:
[tex]E^o_{cell}=-2.37-(-0.76)=-1.61V[/tex]
As, the standard electrode potential is coming out to be negative. So, the reaction is not spontaneous.
For d:The chemical reaction follows:
[tex]2Al(s)+3Pb^{2+}(aq.)\rightarrow 2Al^{3+}(aq.)+3Pb(s)[/tex]
We know that:
[tex]E^o_{Al^{3+}/Al}=-1.66V\\E^o_{Pb^{2+}/Pb}=-0.13V[/tex]
Calculating the [tex]E^o_{cell}[/tex] using equation 1, we get:
[tex]E^o_{cell}=-0.13-(-1.66)=1.53V[/tex]
As, the standard electrode potential is coming out to be positive. So, the reaction is spontaneous.
When 4.00 mol of each X(g) and Y(g) are placed in a 1.00 L vessel and allowed to react at constant temperature according to the equation below, 6.00 mol of Z(g) is produced. What is the value of the equilibrium constant, Kc?
X(g) + Y(g) ---> 2 Z(g)
8
16
36
6
Answer: The value of equilibrium constant for the above equation is 36
Explanation:
We are given:
Initial moles of X = 4.00 moles
Initial moles of Y = 4.00 moles
Equilibrium moles of Z = 6.00 moles
Volume of vessel = 1.00 L
Initial concentration of X = [tex]\frac{4}{1}=4[/tex]
Initial concentration of Y = [tex]\frac{4}{1}=4[/tex]
Equilibrium concentration of Z = [tex]\frac{6}{1}=6[/tex]
The given chemical equation follows:
[tex]X(g)+Y(g)\rightarrow 2Z(g)[/tex]
Initial: 4 4
At eqllm: 4-x 4-x 2x
Calculating for 'x', we get:
[tex]\Rightarrow 2x=6\\\\\Rightarrow x=3[/tex]
The expression of [tex]K_c[/tex] for above equation follows:
[tex]K_c=\frac{[Z]^2}{[X]\times [Y]}[/tex]
Putting values in above equation, we get:
[tex]K_c=\frac{6^2}{1\times 1}\\\\K_c=36[/tex]
Hence, the value of equilibrium constant for the above equation is 36
The equilibrium constant, Kc, for the given reaction is 36. This is calculated using the ratio of product concentrations to reactant concentrations at equilibrium, each concentration being raised to the power of its stoichiometric coefficient.
Explanation:The equilibrium constant, Kc, describes the ratio of product concentrations to reactant concentrations at equilibrium, with each concentration raised to the power of its stoichiometric coefficient. Here, the equation is:
X(g) + Y(g) ---> 2 Z(g)
In terms of moles to start, we have 4.00 mol of X and Y, and 0 mol of Z. At equilibrium, we end up with 6.00 mol of Z, which, since 2 mol of Z are generated for every 1 mol of X and Y, means we have lost 3.00 mol of X and Y.
As the volume is 1L, the molar concentrations are the same as the number of moles. So, the concentrations at equilibrium are: [X] = [Y] = 1.00 M and [Z] = 6.00 M.
Applying this to the Kc expression gives: Kc = ([Z]^2) / ([X] * [Y]) = (6.00^2) / (1.00 * 1.00) = 36. Therefore, the equilibrium constant, Kc, is 36.
Learn more about Equilibrium Constant here:https://brainly.com/question/32442283
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In NMR if a chemical shift(δ) is 211.5 ppm from the tetramethylsilane (TMS) standard and the spectrometer frequency is 556 MHz, how many Hz from TMS is the signal at? Give at least three significant figures.
Answer:
The answer is: 11759 Hz
Explanation:
Given: Chemical shift: δ = 211.5 ppm, Spectrometer frequency = 556 MHz = 556 × 10⁶ Hz
In NMR spectroscopy, the chemical shift (δ), expressed in ppm, of a given nucleus is given by the equation:
[tex]\delta (ppm) = \frac{Observed\,frequency (Hz)}{Frequency\,\, of\,\,the\,Spectrometer (MHz)} \times 10^{6}[/tex]
[tex]\therefore Observed\,frequency (Hz)= \frac{\delta (ppm)\times Frequency\,\, of\,\,the\,Spectrometer (MHz)}{10^{6}}[/tex]
[tex]Observed\,frequency= \frac{211.5 ppm \times 556 \times 10^{6} Hz}{10^{6}} = 11759 Hz[/tex]
Therefore, the signal is at 11759 Hz from the TMS.
According to Coulomb's Law, how does the force between a hydrogen nucleus and an electron in the n =2 level compare with that for one in the n = 1 level if the distance between the nucleus and the n = 2 level is twice as great as that between the nucleus and the n = 1 level?
one-quarter as strong
four times as strong
one-half as strong
twice as strong
Answer:
one-quarter as strong
Explanation:
Coulomb's law gives the mathematical expression to calculate the electrical force (F) between the charge of the nucleus (q₊) and the charge of an electron (q₋) separated by a distance (r).
[tex]F=\frac{k.q_{+}.q_{-}}{r^{2} }[/tex]
where,
k is the Coulomb's constant
The force between the nucleus and an electron in the level 1 is:
[tex]F_{1}=\frac{k.q_{+}.q_{-}}{r_{1}^{2} }[/tex]
Considering the distance to an electron from n = 2 is twice as great as the distance to an electron from n = 1, the force between the nucleus and an electron in the level 2 is:
[tex]F_{2}=\frac{k.q_{+}.q_{-}}{r_{2}^{2} }=\frac{k.q_{+}.q_{-}}{(2r_{1})^{2} }=\frac{1}{4} \frac{k.q_{+}.q_{-}}{(r_{1})^{2} }=\frac{1}{4}F_{1}[/tex]