Answer: pH of the solution after the addition of 600.0 ml of LiOH is 7.
Explanation:
The given data is as follows.
Volume of [tex]HClO_{4}[/tex] = 900.0 ml = 0.9 L,
Molarity of [tex]HClO_{4}[/tex] = 0.18 M,
Hence, we will calculate the number of moles of [tex]HClO_{4}[/tex] as follows.
No. of moles = Molarity × Volume
= 0.18 M × 0.9 L
= 0.162 moles
Volume of NaOH = 600.0 ml = 0.6 L
Molarity of NaOH = 0.27 M
No. of moles of NaOH = Molarity × Volume
= 0.27 M × 0.6 L
= 0.162 moles
This shows that the number of moles of [tex]HClO_{4}[/tex] is equal to the number of moles of NaOH.
Also we know that,
[tex]HClO_{4} + NaOH \rightarrow NaClO_{4} + H_{2}O[/tex]
As 1 mole of [tex]HClO_{4}[/tex] reacts with 1 mole of NaOH then all the hydrogen ions and hydroxide ions will be consumed.
This means that pH = 7.
Thus, we can conclude that pH of the solution after the addition of 600.0 ml of LiOH is 7.
Calculate the [H+] in 1.0 M solution of Na2CO3 (for H2CO3, Ka1 = 4.3 × 10–7; Ka2 = 5.6 × 10–11). 7.5 × 10–6 M 1.3 × 10–2 M 7.5 × 10–13 M 6.6 × 10–4 M None of these choices are correct.
Answer:
7.5x10⁻¹³M = [H⁺]
Explanation:
When a solution of Na₂CO₃ is dissolved in water, the equilibrium produced is:
Na₂CO₃(aq) + H₂O(l) ⇄ HCO₃⁺(aq) + OH⁻(aq) + 2Na⁺
Where Kb is defined from equilibrium concentrations of reactants, thus:
Kb = [HCO₃⁺][OH⁻] / [Na₂CO₃] (1)
It is possible to obtain Kb value from Ka2 and Kw thus:
Kb = Kw / Ka2
Kb = 1x10⁻¹⁴ / 5.6x10⁻¹¹
Kb = 1.8x10⁻⁴
Replacing in (1):
1.8x10⁻⁴ = [HCO₃⁺][OH⁻] / [Na₂CO₃]
The equilibrium concentrations are:
[Na₂CO₃] = 1.0M - X
[HCO₃⁺] = X
[OH⁻] = X
Thus:
1.8x10⁻⁴ = [X][X] / [1-X]
1.8x10⁻⁴ - 1.8x10⁻⁴X = X²
X² + 1.8x10⁻⁴X - 1.8x10⁻⁴ = 0
Solving for X:
X = -0.0135 → False answer, there is no negative concentrations
X = 0.0133
As [OH⁻] = X;
[OH⁻] = 0.0133
From Kw:
Kw = [OH⁻] [H⁺]
1x10⁻¹⁴ = 0.0133[H⁺]
7.5x10⁻¹³M = [H⁺]
The concentration of [H+] in a 1.0M solution of Na2CO3 is approximately 1.2 x 10-4M, closest to 6.6 × 10^-4 M
Explanation:To solve this problem, we will use the dissociation constants (Ka1 and Ka2) provided for the diacid H2CO3, as well as the fact that Na2CO3 will dissociate completely in water to form 2Na+ and CO32-. The CO32- anion can then react with water (H2O) to produce HCO3- and OH-, the latter of which will increase the pH of the solution. However, the HCO3- anion is amphiprotic and can further react with water to produce H2CO3 and OH-, again increasing the pH.
The calculations necessary to solve this question require solving the equilibrium problems for two equations: HCO3- + H2O <=> H2CO3 + OH- and H2CO3 + H2O <=> H3O+ + HCO3-. The concentrations at equilibrium are given as [H2CO3] = 0.033M, [HCO3-] = 1.2 × 10-4 M, [CO32-] = 4.7 x 10-11M, [H3O+] = 1.2 × 10-4M. Hence, [H+] = [H3O+] = 1.2 x 10-4 M. Given the multiple choice options, the closest is 6.6 × 10-4M.
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The formation of tert-butanol is described by the following chemical equation:
(CH3)3CBr (aq) + OH- (aq) → Br- (aq) + (CH3)3COH (ǎq )
Suppose a two-step mechanism is proposed for this reaction, beginning with this elementary reaction
(CH3)3CBr (aq) → (CH3)3C+ (aq) + Br- (aq)
Suppose also that the second step of the mechanism should be bimolecular. Suggest a reasonable second step. That is, write the balanced chemical equation of a bimolecular elementary reaction that would complete the proposed mechanism
Answer:
Second step
(CH3)3C+ (aq) + OH^-(aq) ------->(CH3)3COH(aq)
Explanation:
This reaction involves;
First the ionization of the tertiary halide to firm a carbocation
Secondly the attack of the hydroxide ion on the carbocation to form tert-butanol
First step;
(CH3)3CBr (aq) → (CH3)3C+ (aq) + Br- (aq)
Second step
(CH3)3C+ (aq) + OH^-(aq) ------->(CH3)3COH(aq)
This second step completes the reaction mechanism.
Final answer:
The second step of the tert-butanol formation mechanism involves the tert-butyl cation reacting with a hydroxide ion to form tert-butanol, exemplifying a bimolecular reaction in a SN1 mechanism.
Explanation:
The question pertains to the formation of tert-butanol from tert-butyl bromide in a two-step mechanism. Given that the first step involves the formation of a tert-butyl cation and bromide ion, a reasonable second step in this bimolecular reaction would involve the tert-butyl cation reacting with hydroxide ion (OH-). The second and final step in the mechanism, thus, can be written as:
(CH3)3C+ (aq) + OH- (aq) → (CH3)3COH (aq)
This step involves the nucleophilic attack of the hydroxide ion on the carbocation, leading to the formation of tert-butanol. It exemplifies the bimolecular nature (involving two reactant species) of the second step, consistent with a SN1 mechanism where the first step is the rate-determining step.
Treatment of a cyclic ketone with diazomethane is a method for accomplishing a ring-expansion reaction. The reaction involves the initial nucleophilic attack by diazomethane on the carbonyl carbon to form a tetrahedral intermediate. Collapse of this intermediate is accompanied by bond migration and loss of N2. For example, treatment of cyclohexanone with diazomethane yields cycloheptanone. Draw the structure of the organic product(s) of the ring expansion of this compound:
Answer:
Explanation:
find the solution below
what is the main difference between a physical reaction and a
chemical reaction?
Answer:
physical reactons dont change the substance
Explanation:
physical changes are things like shape and color it doesnt change the base material but a chemical change changes the base aterial baking soda and vinegar create co2 but cutting wood still results in wood hope this helps god bless
The standard change in Gibbs free energy is ΔG°′=7.53 kJ/mol . Calculate ΔG for this reaction at 298 K when [dihydroxyacetone phosphate]=0.100 M and [glyceraldehyde-3-phosphate]=0.00100 M .
Answer:
ΔG = -3879.6 J/mol = -3.88 kJ/mol
Explanation:
Step 1: Data given
The standard change in Gibbs free energy is ΔG°′=7.53 kJ/mol
Temperature = 298 K
[dihydroxyacetone phosphate]=0.100 M
[glyceraldehyde-3-phosphate]=0.00100 M .
Step 2: Calculate ΔG for this reaction
ΔG = ΔG° + RT ln ([glyceraldehyde-3-phosphate]/ [dihydroxyacetone phosphate])
⇒with ΔG° = 7.53 kJ/mol = 7
⇒with R = 8.314 J/mol*K
⇒with T = 298 K
⇒ with [glyceraldehyde-3-phosphate]=0.00100 M
⇒ with [dihydroxyacetone phosphate]=0.100 M
ΔG = 7530 J/mol + 8.314 * 298 * ln(0.001/0.1)
ΔG = -3879.6 J/mol = -3.88 kJ/mol
The Gibbs free energy change ΔG is approximately -3882 J/mol (or -3.88 kJ/mol).
To calculate the change in Gibbs free energy (ΔG) for the given reaction at 298 K, we use the relationship:
ΔG = ΔG°' + RT ln(Q)
where:
ΔG°' is the standard Gibbs free energy change, which is 7.53 kJ/mol (or 7530 J/mol since 1 kJ = 1000 J).R is the gas constant, 8.314 J/(mol·K).T is the temperature, 298 K.Q is the reaction quotient, given by the ratio of the concentrations of products to reactants.For the reaction, Q is calculated as:
Q = [glyceraldehyde-3-phosphate] / [dihydroxyacetone phosphate]Given:
[dihydroxyacetone phosphate] = 0.100 M[glyceraldehyde-3-phosphate] = 0.00100 MTherefore:
Q = 0.00100 / 0.100 = 0.01Now, substituting the values into the equation:
ΔG = 7530 J/mol + (8.314 J/(mol·K) * 298 K * ln(0.01))Calculating the term involving the natural logarithm:
ln(0.01) ≈ -4.605Thus, the calculation is:
ΔG = 7530 J/mol + (8.314 J/(mol·K) * 298 K * -4.605)ΔG ≈ 7530 J/mol - 11412 J/molΔG ≈ -3882 J/molTherefore, the Gibbs free energy change ΔG for the reaction under the given conditions is approximately -3882 J/mol (or -3.88 kJ/mol).
Please circle the process which does not involve irreversibility: (A) The conversion of mechanical work to heat by friction. (B) Conversion of electrical energy to thermal energy through a resistor heater. (C) Mixing of O2 and N2 for the production of air. (D) The theoretical conversion from electricity to mechanical work.
Answer:
Mixing of O2 and N2 for the production of air.
Explanation:
An irreversible process is defined as any process which cannot return both the system and its surroundings to their original conditions. That is, the system and its surroundings would not return to their original conditions if the process was reversed. Irreversible Processes usually increase the entropy of the universe.
Processes that involve evolution of heat are usually irreversible processes since heat is lost to the surrounding. Hence the answer
The partial pressure of N2 in the air is 593 mm Hg at 1 atm. What is the partial pressure of N2 in a bubble of air a scuba diver breathes when he is 132 ft below the surface of the water where the pressure is 5.00 atm?
Answer: Partial pressure of [tex]N_{2}[/tex] at a depth of 132 ft below sea level is 2964 mm Hg.
Explanation:
It is known that 1 atm = 760 mm Hg.
Also, [tex]P_{N_{2}} = x_{N_{2}}P[/tex]
where, [tex]P_{N_{2}}[/tex] = partial pressure of [tex]N_{2}[/tex]
P = atmospheric pressure
[tex]x_{N_{2}}[/tex] = mole fraction of [tex]N_{2}[/tex]
Putting the given values into the above formula as follows.
[tex]P_{N_{2}} = x_{N_{2}}P[/tex]
[tex]593 mm Hg = x_{N_{2}} \times 760 mm Hg[/tex]
[tex]x_{N_{2}}[/tex] = 0.780
Now, at a depth of 132 ft below the surface of the water where pressure is 5.0 atm. So, partial pressure of [tex]N_{2}[/tex] is as follows.
[tex]P_{N_{2}} = x_{N_{2}}P[/tex]
= [tex]0.78 \times 5 atm \times \frac{760 mm Hg}{1 atm}[/tex]
= 2964 mm Hg
Therefore, we can conclude that partial pressure of [tex]N_{2}[/tex] at a depth of 132 ft below sea level is 2964 mm Hg.
The reaction C4H10 ---> C2H6 + C2H4 has activation energy (Ea) of 450 kJ/mol, and the Ea of the reverse reaction is 250 kJ/mol. Estimate ΔH, in kJ/mol, for the reaction as written above.
The enthalpy change (ΔH) of the reaction C4H10 ---> C2H6 + C2H4 can be calculated as the difference between the activation energies of the forward and reverse reactions, giving a result of 200 kJ/mol. Activation energy, the minimum energy that reactants need to react, influences the reaction rate.
Explanation:The question pertains to the activation energy and enthalpy change in the chemical reaction C4H10 ---> C2H6 + C2H4. The activation energy (Ea) is the minimum energy that reactants need to undergo a reaction, and it varies depending on the direction of the reaction. Using the given activation energies, the enthalpy change (ΔH) can be estimated as ΔH = Ea(forward) - Ea(reverse). Substituting the given values, ΔH = 450 kJ/mol - 250 kJ/mol = 200 kJ/mol.
Activation energy is highly significant to the rate of a chemical reaction. If the activation energy is larger than the average kinetic energy of the reactants, the reaction will occur slowly as only a few high-energy molecules can react. Conversely, if the activation energy is smaller, more molecules can react and the reaction rate is higher.
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Which of the following are the bark, roots, seeds, buds or berries of an aromatic plant?
-Spices
-Herbs
-Vegetables
-Fruits
Answer:
Spices
Explanation:
Herbs are the leaves.
Vegetables are the seeds.
Fruits are the seeds.
So spices are the only option left.
The correct term for the aromatic parts of plants such as bark, roots, seeds, buds, or berries is Option a i.e, spices. Spices are used for their flavor and aroma and come from various dried parts of plants. Examples include black pepper from the fruit of Piper nigrum and cinnamon from tree bark.
When referring to the aromatic parts of plants such as bark, roots, seeds, buds, or berries, the correct term is spices. Spices are aromatic substances derived from various dried parts of plants including roots, shoots, fruits, bark, and leaves. They are often used in cooking to add flavor and aroma and are sold in forms such as whole spices, ground spices, or seasoning blends. The correct option is a i.e, Spices.
An example is black pepper, which comes from the fruit of the Piper nigrum plant, and cinnamon, which is derived from the bark of a tree in the Laurales family.
The decomposition of nitryl chloride is described by the following chemical equation: 2NO2C1(g) → 2NO2 (g)-C12 (g) Suppose a two-step mechanism is proposed for this reaction, beginning with this elementary reaction NO, Cl(g)→NO2(g)+Cl(g) Suppose also that the second step of the mechanism should be bimolecular. Suggest a reasonable second step. That is, write the balanced chemical equation of a bimolecular elementary reaction that would complete the proposed mechanism.
Answer:
Second reaction
NO2 + F -------> NO2F
Rate of reaction:
k1 [NO2] [F2]
Explanation:
NO2 + F2 -----> NO2F + F slow step1
NO2 + F -------> NO2F fast. Step 2
Since the first step is the slowest step, it is the rate determining step of the reaction
Hence:
rate = k1 [NO2] [F2]
Answer:
Cl(g)+NO_2Cl(g)-->NO_2(g)+Cl_2(g)
Explanation:
Acetylene C2H2 gas is often used in welding torches because of the very high heat produced when it reacts with oxygen O2 gas, producing carbon dioxide gas and water vapor. Calculate the moles of oxygen needed to produce 1.5mol of water. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.
Answer:
the water evaporates into the carbon dioxide
Explanation:
i just know by heart
For Na2HPO4:(( (Note that for H3PO4, ka1= 6.9x10-3, ka2 = 6.4x10-8, ka3 = 4.8x10-13 ) a) The active anion is H2PO4- b) The active anion is HPO4 3- c) is basic d) Is acidic
Answer:
Check the explanation
Explanation:
Answer – Given, [tex]H_3PO_4[/tex] acid and there are three Ka values
[tex]K_{a1}=6.9x10^8, K_{a2} = 6.2X10^8, and K_{a3}=4.8X10^{13}[/tex]
The transformation of [tex]H_2PO_4- (aq) to HPO_4^2-(aq)[/tex]is the second dissociation, so we need to use the Ka2 = 6.2x10-8 in the Henderson-Hasselbalch equation.
Mass of KH2PO4 = 22.0 g , mass of Na2HPO4 = 32.0 g , volume = 1.00 L
First we need to calculate moles of each
Moles of KH2PO4 = 22.0 g / 136.08 g.mol-1
= 0.162 moles
Moles of Na2HPO4 = 32.0 g /141.96 g.mol-1
= 0.225 moles
[H2PO4-] = 0.162 moles / 1.00 L = 0.162 M
[HPO42-] = 0.225 moles / 1.00 L = 0.225 M
Now we need to calculate the pKa2
pKa2 = -log Ka
= -log 6.2x10-8
= 7.21
We know Henderson-Hasselbalch equation
pH = pKa + log [conjugate base] / [acid]
pH = 7.21 + log 0.225 / 0.162
= 7.35
The pH of a buffer solution obtained by dissolving 22.0 g of KH2PO4 and 32.0 g of Na2HPO4 in water and then diluting to 1.00 L is 7.35
The energy per ion in an ionic crystal can be modeled by . For a crystalline sample, the cohesive energy, defined as the absolute value of the total position-dependent interaction energy, per ion is 9.1 eV, the lattice constant is 1.7 nm, and the exponent is determined to be 7. What is the bulk modulus of this crystal?
The bulk modulus of the crystal is -27.3 eV or [tex]\( -4.37 \times 10^{-9} \) J/m³.[/tex]
The cohesive energy per ion in an ionic crystal is given by the expression:
[tex]\[ E = \frac{k}{r^n} \][/tex]
Where:
E is the cohesive energy per ion.
k is a constant.
r is the interionic distance (lattice constant).
n is the exponent.
Given that E = 9.1 eV, r = 1.7 nm, and n = 7 , we can solve for k :
[tex]\[ 9.1 \text{ eV} = \frac{k}{(1.7 \times 10^{-9} \text{ m})^7} \][/tex]
[tex]\[ k = 9.1 \text{ eV} \times (1.7 \times 10^{-9} \text{ m})^7 \][/tex]
Now, the bulk modulus B of the crystal can be calculated using the relationship between cohesive energy and bulk modulus:
[tex]\[ B = -V \frac{dE}{dV} \][/tex]
Given that [tex]\( V = r^3 \)[/tex], we can find [tex]\( \frac{dE}{dV} \)[/tex] by differentiating E with respect to V , then substituting the values:
[tex]\[ B = -3E \][/tex]
[tex]\[ B = -3 \times 9.1 \text{ eV} \][/tex]
[tex]\[ B = -27.3 \text{ eV} \text{ (or } -4.37 \times 10^{-9} \text{ J/m}^3 \text{)} \][/tex]
Thus, the bulk modulus of the crystal is -27.3 eV or [tex]\( -4.37 \times 10^{-9} \) J/m³.[/tex]
What volume of oxygen at 423.0 K and a pressure of 127.4 kPa is produced by the decomposition of 129.7 g of BaO2 to BaO and O2?
The volume of oxygen produced by the decomposition of 129.7 g of BaO2 at 423.0 K and a pressure of 127.4 kPa is 10.37 L.
Explanation:The decomposition of BaO2 (Barium peroxide) to BaO (Barium oxide) and O2 (Oxygen) is represented by the balanced chemical reaction: 2BaO2(s) → 2BaO(s) + O2(g). Using the molar mass of BaO2 (169.33 g/mol), we can calculate the number of moles of BaO2 in 129.7 g which is 0.766 moles. From the reaction stoichiometry, we can see that 2 moles of BaO2 produces 1 mole of O2. Therefore, 0.766 moles of BaO2 produces 0.766/2 = 0.383 moles of O2. Using the ideal gas law, PV=nRT, we can solve for volume (V) using n=0.383 moles, R=8.314 kPa L/mol K (universal gas constant) and T=423 K (Temperature), and P=127.4 kPa (Pressure) which gives us, V = (nRT)/P , V = (0.383 moles * 8.314 kPa L/mol K * 423 K) / 127.4 kPa = 10.37 L. Thus, the volume of oxygen produced by the decomposition of 129.7 g of BaO2 at 423.0 K and a pressure of 127.4 kPa is 10.37 L.
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The volume of oxygen produced by the decomposition of 129.7 g of BaO2 under the given conditions is approximately 13.2 liters.
Explanation:The decomposition of barium peroxide (BaO2) to barium oxide (BaO) and oxygen (O2) is a reaction that can be described by the equation:
2 BaO2(s) → 2 BaO(s) + O2(g)
From the molar mass of BaO2, which is 169.34 g/mol, we can calculate the number of moles of BaO2 that the 129.7 g represents:
moles of BaO2 = 129.7 g / 169.34 g/mol ≈ 0.766 moles
According to the balanced equation, 2 moles of BaO2 produce 1 mole of O2, hence:
moles of O2 produced = 0.766 moles BaO2 / 2 ≈ 0.383 moles
Using the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/K·mol), and T is the temperature:
We first convert the pressure from kPa to atm: 127.4 kPa = 1.258 atm (using the conversion 101.3 kPa = 1 atm).
Then, we solve for V using the ideal gas law equation:
V = nRT / P = (0.383 moles) × (0.0821 L·atm/K·mol) × (423.0 K) / 1.258 atm ≈ 13.2 L
Therefore, the volume of oxygen produced under the given conditions is approximately 13.2 liters.
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Mg(OH)2 + 2 HCl = MgCl2 + 2 H20
How many molecules of water are produced?
Answer:
2 molecules
Explanation:
From the stoichiometric equation, only 2-molecules of water is produced
A flexible container at an initial volume of 6.13 L contains 3.51 mol of gas. More gas is then added to the container until it reaches a final volume of 16.3 L. Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container. number of moles of gas: mol
Answer: The number of moles of gas added to the container are 5.82
Explanation:
Avogadro's law states that volume is directly proportional to number of moles at constant temperature and pressure.
The equation used to calculate number of moles is given by:
[tex]\frac{V_1}{n_1}=\frac{V_2}{n_2}[/tex]
where,
[tex]V_1[/tex] and [tex]n_1[/tex] are the initial volume and number of moles
[tex]V_2[/tex] and [tex]n_2[/tex] are the final volume and number of moles
Putting values in above equation, we get:
[tex]\frac{6.13}{3.51}=\frac{16.3}{n_2}\\\\n_2=9.33[/tex]
number of moles of gas added to the container = (9.33-3.51) = 5.82
Thus the number of moles of gas added to the container are 5.82
A hot metal plate at 150°C has been placed in air at room temperature. Which event would most likely take place over the
next few minutes?
A)Molecules in both the metal and the surrounding air will start moving at lower speeds.
B)Molecules in both the metal and the surrounding air will start moving at higher speeds.
C)The air molecules that are surrounding the metal will slow down, and the molecules in the metal will speed up
D)The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down
D) The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down.
This means the air around will heat up, and the plate will cool down. They are trying to reach a thermal equilibrium.
The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down is the event which would most likely to take place over next few minutes.
Hence, option (D) is correct answer.
What is Exothermic Reaction ?Exothermic reaction is a reaction that is chemical in nature and in which the energy is released in form of heat, electricity or light. Energy is released when the new bonds are formed and this bond making process is an exothermic process.
What is Exothermic Reaction ?Endothermic reaction is a chemical reaction that absorbs heat from the surroundings. Energy is absorbed when the bonds breaks and this bond breaking process is endothermic process.
Here in this process the heat is evolved so it is exothermic process.
Thus, from the above conclusion we can say that The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down is the event which would most likely to take place over next few minutes.
Hence, option (D) is correct answer.
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Determine the number of Ni2+ ions involved in metalization (Ni plating) of an electrode for 30 sec in an electrolysis cell operated at 0.020A at 78 % current efficiency? The Faraday constant, F = 96485 C/mol and Avogadro # NA = 6.023×1023. What will be the number of Ag+ ions if Ag is plated under the same experimental conditions instead of Ni2+?
Answer: The number of Ni2+ ions is 1.46 × 10^18 ions
The number of Ag+ ions is 2.92 × 10^18 ions
Explanation: Please see the attachments below
Why does an incomplete combustion reaction occur?
The temperature –60 °C is higher than –60 °F.
Answer:
false
Explanation:
it is MUCH lower in temperature
Temperature comparison on the Celsius and Fahrenheit scales are being compared, with a focus on -60 °C and -60 °F. -60 °C is colder than -60 °F because the Celsius scale has a smaller degree interval.
The question is about comparing temperatures in different units, specifically Celsius and Fahrenheit. In physics, temperature is measured using a scale known as the Celsius scale.
The Celsius scale is based on the freezing and boiling points of water, with 0 °C being the freezing point and 100 °C being the boiling point at standard atmospheric pressure. On the other hand, the Fahrenheit scale is another temperature scale commonly used in countries like the United States.
The freezing point of water on the Fahrenheit scale is 32 °F, while the boiling point is 212 °F at standard atmospheric pressure.
To answer the question, let's compare -60 °C and -60 °F. Since the Fahrenheit scale has a larger degree interval between freezing and boiling points, it means that each degree on the Fahrenheit scale is smaller than each degree on the Celsius scale.
So, -60 °C is actually colder than -60 °F. This is because -60 °C is closer to the freezing point of water (0 °C) than -60 °F is to the freezing point of water (32 °F).
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The probable question may be:
The temperature –60 °C is higher than –60 °F. Explain.
In cases of ethylene glycol poisoning, treatment involves administration of Ethanol (grain alcohol), which works by competitively inhibiting ADH, an enzyme that oxidizes ethylene glycol to organic acids. As a competitive inhibitor, ethanol: decreases apparent Km without affecting Vmax· increases apparent Vmax without affecting Km. decreases both apparent Vmax and apparent Km. increases apparent Km without affecting Vmax· decreases apparent Vmax without affecting Km.
Answer:
Option B is the correct option- Ethanol increases apparent Km without affecting Vmax.
Explanation:
Vmax remains the same, and Km increases in competitive inhibition. There is an increment in Km because competitive inhibitors interfere with substrate binding to the enzyme. Vmax is not affected because the competitive inhibitor cannot bind to ES and therefore does not alter the catalysis.
Option B is the correct option- Ethanol increases apparent Km without affecting Vmax.
In the attached image, the first picture is the Michaelis Menten Plot for competitive inhibition in which an increase in km but constant Vmax is observed.
In the images, ............................. represents the case with competitive inhibitor, while _______________ represents the case without competitive inhibitor.
Tritium 3 1H decays to 3 2He by beta emission. Find the energy released in the process. Answer in units of keV.
Answer:
The energy released in the decay process = 18.63 keV
Explanation:
To solve this question, we have to calculate the binding energy of each isotope and then take the difference.
The mass of Tritium = 3.016049 amu.
So,the binding energy of Tritium = 3.016049 *931.494 MeV
= 2809.43155 MeV.
The mass of Helium 3 = 3.016029 amu.
So, the binding energy of Helium 3 = 3.016029 * 931.494 MeV
= 2809.41292 MeV.
The difference between the binding energy of Tritium and the binding energy of Helium is: 32809.43155 - 2809.412 = 0.01863 MeV
1 MeV = 1000keV.
Thus, 0.01863 MeV = 0.01863*1000keV = 18.63 keV.
So, the energy released in the decay process = 18.63 keV.
How much heat in kJ is produced by the oxidation of 18.6 g of Mn?
3 Mn(s) + 2 O2(g) => Mn3O4(s) ΔHo = -1388 kJ
The oxidation of 18.6 g of manganese releases approximately -155.7 kJ of heat when reacting to form Mn₃O₄(s), using stoichiometry and the molar mass of manganese.
The student is asking about the heat produced by the oxidation of manganese (Mn) to form Mn₃O₄. The provided chemical equation indicates that the enthalpy change (ΔH⁰) for the reaction is -1388 kJ for the oxidation of 3 moles of manganese.
To find the heat released by the oxidation of 18.6 g of Mn, we must first calculate the number of moles of Mn in 18.6 g. Manganese has a molar mass of approximately 54.94 g/mol. Therefore:
Moles of Mn = 18.6 g ÷ 54.94 g/mol = 0.3386 mol Mn
Now, we will use the stoichiometry of the reaction to find the enthalpy change for 0.3386 mol Mn given that -1388 kJ is the enthalpy change for 3 moles of Mn:
Heat produced = 0.3386 mol Mn × (-1388 kJ ÷ 3 mol Mn) = -155.7 kJ (rounded to one decimal place)
The oxidation of 18.6 g of manganese produces approximately -155.7 kJ of heat.
The product of the nitration reaction will have a nitro group at which position with respect to the methyl group? Group of answer choices ortho and para at a 50:50 ratio mostly ortho mostly para meta
Answer:
Mostly Para
Explanation:
First, let's assume that the molecule is the toluene (A benzene with a methyl group as radical).
Now the nitration reaction is a reaction in which the nitric acid in presence of sulfuric acid, react with the benzene molecule, to introduce the nitro group into the molecule. The nitro group is a relative strong deactiviting group and is metha director, so, further reactions that occur will be in the metha position.
Now, in this case, the methyl group is a weak activating group in the molecule of benzene, and is always ortho and para director for the simple fact that this molecule (The methyl group) is a donor of electrons instead of atracting group of electrons. Therefore for these two reasons, when the nitration occurs,it will go to the ortho or para position.
Now which position will prefer to go? it's true it can go either ortho or para, however, let's use the steric hindrance principle. Although the methyl group it's not a very voluminous and big molecule, it still exerts a little steric hindrance, and the nitro group would rather go to a position where no molecule is present so it can attach easily. It's like you have two doors that lead to the same place, but in one door you have a kid in the middle and the other door is free to go, you'll rather pass by the door which is free instead of the door with the kid in the middle even though you can pass for that door too. Same thing happens here. Therefore the correct option will be mostly para.
what substance is produced by the reaction: H+[aq]+OH-[aq]=?
Answer:
It produces water.
Explanation:
H+ + OH- produces H2O.
It is a type of Neutralization reaction.
Suppose the formation of nitrogen dioxide proceeds by the following mechanism
step elementary reaction rate constant
1 2NO(g) → N2O2(g) k1
2 NO2 (g) + O2 (g) → 2NO2 (g) k2
Suppose also k1<
Write the balanced chemical equation for the overall chemical reaction:
Write the experimentally- observable rate law for the overall chemical reaction rate:
Final answer:
The balanced chemical equation for the overall reaction is 2NO(g) + O2(g) → 2NO2(g). The experimentally-observable rate law, assuming the second step is rate-determining, is rate = k[NO]^2[O2], where k is the overall rate constant.
Explanation:
The balanced chemical equation for the overall reaction combining the two steps 2NO(g) → N2O2(g) and NO2(g) + O2(g) → 2NO2(g) is:
2NO(g) + O2(g) → 2NO2(g)
To write the experimentally-observable rate law for the overall reaction, we must identify the rate-determining step. Assuming the second step is the rate-determining step, and given that the first step is a fast equilibrium, the overall rate can be expressed as:
rate = k2[N2O2][O2]
Since [N2O2] is the intermediate formed in the first step and we know that the rate of formation of N2O2 is proportional to the square of [NO] concentration, we can express [N2O2] in terms of [NO]. Substituting into the rate law, we get:
rate = k2k1[NO]2[O2]
Here, k = k1k2 represents the overall rate constant for the reaction. Therefore, the rate law for the overall chemical reaction is:
rate = k[NO]2[O2]
In the redox persulfate-iodide experiment, a 100 mL reaction mixture is prepared for one of the runs as follows. 0.200 M KI 20 mL 0.200 M KNO3 15 mL 0.25% starch 5 mL 0.01 M Na2S2O3 10 mL 0.1 M EDTA 1 drop water to make 65 mL 0.200 M (NH4)2S2O8 35 mL a. How many moles of S2O32- would react when the solution turns dark blue? b. How many moles of S2O82- would react when the solution turns dark blue? c. What would be the initial rate of reaction if this reaction mixture took 200 seconds to turn dark blue?
Answer:
a) The number of moles of thiosulphate ( S₂O₃2-) that reacted to turn the solution dark blue will be 0.0001M.
b) The number of moles of S₂O₈2- that would react to produce the blue color will be 0.00005 M.
c) rate = 2.5 x 10-6M/L/s
Explanation:
a)
The reactions taking place in this experiment are represented by the ionic equations,
S₂O₈ 2- + 2I- ----------> 2SO₄2- + I₂ --------------------(1)
2 S₂O₃2- + I₂ -----------> S₄O₆ +2 I- -------------------(2)
The persulphate ions react with the iodide ions to produce free iodine which is in turn reduced by the thiosulphate ions to produce iodide ions again. This reaction proceeds till all the thiosulphate ions are used up. Therefore the rate of the reaction will be the rate at which iodine is formed and used up.
When there is free iodine the reaction mixture, the solution gives a dark blue coloration. This happens when all the thiosulphate ions are used up.
The volume of sodium thiosulphate ( Na₂S₂O3) solution added to the reaction vessel = 10ml
Molarity of sodium thiosulphate ( Na₂S₂O3) solution = 0.01M
Number of moles of ( Na2S2O3) = 0.01ml x 10M /1000ml = 0.0001M (molarity x volume in L)
The number of moles of thiosulphate reacted will be equal to the number of moles taken since the reaction proceeds till all the thiosulphate is consumed.
Hence, the number of moles of thiosulphate ( S₂O₃2-) that reacted to turn the solution dark blue will be 0.0001M.
b)
To calculate S₂O₈ 2-
The permanent blue color is produced once all the thiosulphate ions are used up and persulphate reacts with iodide ions to produce iodine, so, the number of moles of persulphate ions will be equal to the number of moles of iodine formed
According to the stoichiometry of equation 1.
1 mole of S₂O₈ 2-produces 1 mole of iodine.
According to the stoichiometry of equation 2,
1mole of iodine produced consumes 2 moles of S₂O₃2-
The number of moles of S₂O₃2- taken = 0.0001M.
2 moles of S₂O₃2- is equivalent to 1 mole I₂
therefore
0.0001 mole of S₂O₃2- = 0.0001/2 = 0.00005M of I₂
Since stoichimetrically,
1 mole of S₂O₈2- is equivalent to 1 mole I2, the number of moles of S₂O₈2- that would react to produce the blue color will be 0.00005 M.
c)
The initial reaction rate is given by
rate =change in concentration of persulphate ion [S₂O₈2-] / time
rate = change in Concentraion of I₂ / t
since initial concentration of I₂ = 0.
rate = Concentraion of I₂/ t
The concentration of I₂ = number of moles of iodine / total volume of solution in L
= 0.00005M/ 0.1L = 0.0005M/L (Volume of the solution = 100ml = 0.1L)
rate = Concentraion of I₂ / t
= 0.0005 /200s
rate = 2.5 x 10-6M/L/s
Yousef measured the height of each seedling on day 1 and day 7. These are his results.
Group
Seedling
Increase in
height/mm
A no water
6.5
0.5
0.5
0.5
5.5
B 2 cm water
Height of seedling/mm
Day 1
Day 7
- 6.0
5.5
6.0
6.0
5.5
7.5
6.0
8.0
6.0
8.5
6.0
9.5
5.5
9.5
6.0
10.0
2.0
00 UN
2.0
2.5
C5cm water
a
b
C
Calculate the increase in height for each of the seedlings 7, 8 and 9.
Calculate the mean increase in height for each group of seedlings.
On graph paper, draw a bar chart to show Yousef's results. Put volume of
water on the x-axis, and mean increase in height on the s-axis.
Write a conclusion that Yousef could make from his results.
d
1 Plants
The main difference causing variation in mean stem length between the plants from the two dishes is the presence or absence of light, which affects photosynthesis and plant growth.
Explanation:The most probable cause for the difference in mean stem length between plants in dish A (no light) and dish B (light cycle) is the effect of light on plant growth. Plants in dish A, with no exposure to light, likely did not undergo photosynthesis effectively, resulting in shorter stems. On the other hand, the plants in dish B received a 14-hour light cycle, allowing them to photosynthesize and grow taller.
In terms of experimental variables, the key difference between the groups is the presence of light, a critical factor for photosynthesis, and consequently, plant growth. This is supported by the fact that other conditions were kept consistent for both dishes.
Elemental mercury is a silver liquid at room temperature. Its normal freezing point is –38.9 °C, and its molar enthalpy of fusion is ∆Hfusion = 2.29 kJ/mol. What is the entropy change of the system (in J/K) when 5.590 g of Hg(l) freezes at the normal freezing point?
Answer:
[tex]\Delta _fS=0.2724\frac{J}{K}[/tex]
Explanation:
Hello,
In this case, we define the entropy change for such freezing process as:
[tex]\Delta _fS=\frac{n_{Hg}\Delta _fH}{T_f}[/tex]
Thus, we compute the moles that are in 5.590 g of liquid mercury:
[tex]n_{Hg}=5.590 gHg*\frac{1molHg}{200.59gHg} =0.02787molHg[/tex]
Hence, we compute the required entropy change, considering the temperature to be in kelvins:
[tex]\Delta _fS=\frac{0.02787mol*2.29\frac{kJ}{mol} }{(-38.9+273.15)K}\\\\\Delta _fS=2.724x10^{-4}\frac{kJ}{K} *\frac{1000J}{kJ} \\\\\Delta _fS=0.2724\frac{J}{K}[/tex]
Best regards.
Answer:
ΔS = -0.272 J/K
Explanation:
Step 1: Data given
Its normal freezing point is –38.9 °C
molar enthalpy of fusion is ∆Hfusion = 2.29 kJ/mol
Mass of Hg = 5.590 grams
Step 2:
ΔG = ΔH - TΔS
At the normal freezing point, or any phase change in general,
ΔG =0
0 = Δ
Hfus
−
Tfus Δ
S
fus
Δ
S
fus = Δ
Hfus
/Tfus
Δ
S
fus = 2290 J/mol / 234.25 K
Δ
S
fus = 9.776 J/mol*K
Since fusion is from solid to liquid. Freezing is the opposite process, so the entropy change of freezing is -9.776 J/mol*K
Step 3: Calculate moles Hg
Moles Hg = 5.590 grams / 200.59 g/mol
Moles Hg = 0.02787 moles
Step 4: Calculate the entropy change of the system:
Δ
S = -9.776 J/mol*K * 0.02787 moles
ΔS = -0.272 J/K
g Choose the best fit. Only choose one option once. A(n) ________ is one for which the rate law can be written from the molecularity, i.e., from coefficients in the balanced equation. Answer 1 The ________ is the number of species that must collide to produce the reaction represented by an elementary step in a reaction mechanism. Answer 2 The ________ of a reaction is the series of proposed elementary reactions that may occur to give the overall reaction. The sum of all the steps in the ________ gives the balanced chemical reaction.
Answer:
A)An elementary step reaction
B) Molecularity
C) Mechanism of reaction
D.)Balance chemical equation
Explanation:
Molecularity is the number of molecules that react in an elementary reaction which is equal to the sum of stoichiometric coefficients of all reactants that participate in the elementary reaction.The reaction can be unimolecular or bimolecular depending on the number of molecules.
Mechanism of reaction reffered to the step by step sequence of elementary reactions through which the overall chemical reaction change occurs. It gives detail of the reaction in each stage of an overall chemical reaction.
An elementary step reaction reffered to is one step of reaction among the series of simple reactions that show how reaction is progressing at the molecular level.