Recall from Chapter 12.6 that chlorofluorocarbons (CFCs) catalyze the decomposition of ozone in the stratosphere. Infrared light is absorbed by CFCs while visible light is transmitted. Ultraviolet light can break the bonds in these molecules. Answer the following questions about CFCs. A) Draw the three-dimensional structure of trichlorofluoromethane, CC13F. B) How will CFCs interact with solar radiation in the troposphere? C) How will CFCs interact with solar radiation in the stratosphere?
Answers
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Consider the reaction below. Initially the concentration of SO2Cl2 is 0.1000 M. Solve for the equilibrium concentration of SO2Cl2((g). SO2Cl2(g) ←⎯⎯→ SO2(g) + Cl2(g) Kc = 2.99 x 10-7 at 227 °C
Answers
Answer:
[tex][SO_2Cl_2] = 0.09983 M[/tex]
Explanation:
Write the balance chemical equation ,
[tex]SO_2Cl_2((g) = SO_2(g) + Cl_2(g)[/tex]
initial concenration of [tex]SO_2Cl_2((g) =0.1M[/tex]
lets assume that degree of dissociation=[tex]\alpha[/tex]
concenration of each component at equilibrium:
[tex][SO_2Cl_2] = 0.1-0.1\alpha[/tex]
[tex][SO_2] = 0.1\alpha[/tex]
[tex][Cl_2] = 0.1\alpha[/tex]
[tex]Kc =\frac{0.1\alpha \times 0.1\alpha}{0.1-0.1\alpha}[/tex]
[tex]Kc =\frac{0.1\alpha \times \alpha}{1-\alpha}[/tex]
as [tex]\alpha[/tex] is very small then we can neglect [tex]1-\alpha[/tex]
therefore ,
[tex]Kc ={0.1\alpha \times \alpha}[/tex]
[tex]\alpha =\sqrt{\frac{Kc}{0.1}}[/tex]
[tex]\alpha = 1.73 \times 10^{-3}[/tex]
Eqilibrium concenration of [tex][SO_2Cl_2] = 0.1-0.1\alpha = 0.1-0.1\times 0.00173[/tex]
[tex][SO_2Cl_2] = 0.09983 M[/tex]
Consider the solubilities of a particular solute at two different temperatures. Temperature ( ∘ C ) Solubility ( g / 100 g H 2 O ) 20.0 32.2 30.0 70.2 Suppose a saturated solution of this solute was made using 56.0 g H 2 O at 20.0 °C. How much more solute can be added if the temperature is increased to 30.0 ∘ C? mass:
Answers
Answer:
21.28 grams solute can be added if the temperature is increased to 30.0°C.
Explanation:
Solubility of solute at 20°C = 32.2 g/100 grams of water
Solute soluble in 1 gram of water = [tex]\frac{32.2}{100}g=0.322 g[/tex]
Mass of solute in soluble in 56.0 grams of water:
[tex]0.322\times 56.0=18.032 g[/tex]
Solubility of solute at 30°C = 70.2g/100 grams of water
Solute soluble in 1 gram of water = [tex]\frac{70.2}{100}g=0.702 g[/tex]
Mass of solute in soluble in 56.0 grams of water:
[tex]0.702 \times 56.0=39.312 g[/tex]
If the temperature of saturated solution of this solute using 56.0 g of water at 20.0 °C raised to 30.0°C
Mass of solute in soluble in 56.0 grams of water 20.0°C = 18.032 g
Mass of solute in soluble in 56.0 grams of water at 30.0°C = 39.312 g
Mass of of solute added If the temperature of the saturated solution increased to 30.0°C:
39.312 g - 18.032 g = 21.28 g
21.28 grams solute can be added if the temperature is increased to 30.0°C.
Final answer:
By comparing the solubility at two different temperatures, an additional 21.280 g of solute can be dissolved in 56.0 g of water when the temperature increases from 20.0 °C to 30.0 °C.
Explanation:
Understanding how temperature influences solubility is crucial when trying to determine the concentration of a solute that can be dissolved in a solvent at different temperatures. According to the data provided, the solubility of a solute in water at 20.0 °C is 32.2 g per 100 g of water and at 30.0 °C it increases to 70.2 g per 100 g of water.
To calculate the additional amount of solute that can be dissolved when the solution temperature is raised from 20.0 °C to 30.0 °C, we should first calculate the solubility in 56.0 g of water at both temperatures:
At 20.0 °C: Proportionally,(32.2 g solute / 100 g water) = (x g solute / 56.0 g water) x = (32.2 g × 56.0 g) / 100 = 18.032 g
So, 18.032 g of solute can be dissolved in 56.0 g of water at 20.0 °C.
At 30.0 °C: Similarly,(70.2 g solute / 100 g water) = (y g solute / 56.0 g water) y = (70.2 g × 56.0 g) / 100 = 39.312 g
So, 39.312 g of solute can be dissolved in 56.0 g of water at 30.0 °C.
The difference between these two amounts (39.312 g - 18.032 g) gives us the additional amount of solute that can be added at the higher temperature:
Additional solute = 39.312 g - 18.032 g = 21.280 g
Therefore, an additional 21.280 g of solute can be added to the solution when the temperature is raised from 20.0 °C to 30.0 °C without reaching saturation.
0.20 mol solute 0.300 kg solvent
calculate the molality of the solution
Answers
The molality of the solution is 0.67 m, which is calculated using the moles of solute and mass of the solvent. The molar mass of the solute would depend on additional data such as osmotic pressure and temperature, and requires using the van't Hoff equation.
Explanation:Calculating Molality and Molar Mass
To calculate the molality of a solution, which is the number of moles of solute per kilogram of solvent, you can use the formula:
Molality (m) = Moles of solute \/ Mass of solvent in kilograms
According to the given information, we have 0.20 moles of solute dissolved in 0.300 kg of solvent. Therefore, using the formula, we get:
Molality = 0.20 mol \/ 0.300 kg = 0.67 m (molal)
To determine the molar mass of a solute B from the given osmotic pressure, temperature, and the amount of solute (0.200 g) dissolved in a known molar volume (0.0180 L/mol) of solvent (1.00 mol), use the van't Hoff factor (i = 1 for non-electrolytes) and the equation:
y = i * (M/R * T)
Where,
n = number of moles of solute
R = gas constant (0.0821 L atm mol^-1 K^-1)
T = temperature in Kelvin
y = osmotic pressure of the solution
From the osmotic pressure formula, we can rearrange to solve for n, and then use the mass of the solute to find the molar mass:
n = y * V/(R * T)
and
Molar mass = mass of solute \/ n
The molar mass calculation will depend on more specific data such as the osmotic pressure and temperature data provided in your hypothetical variation of this exercise. To complete the calculation, one would need to work through the osmotic pressure formula with the given measurements (0.640 atm at 298 K).
A 14.3 g sample of HF is dissolved into 250 mL of solution. The concentration of the solution is *
A. 2.86 M
B. 0.14 M
C. 7.1 M
D. 3.6 M
Answers
Answer:
The concentration of the solution is 2.86 M
Explanation:
Molarity is a unit of concentration based on the volume of a solution. It is defined as the number of moles of solute that are dissolved in a given volume. In other words, molarity is defined as the number of moles of solute per liter of solution.
The Molarity of a solution is determined by the following expression:
[tex]Molarity (M)=\frac{number of moles of solute}{Volume}[/tex]
Molarity is expressed in units ([tex]\frac{moles}{liter}[/tex]).
In this case, you must then know the number of moles of HF, for which you must know the molar mass. Being:
H: 1 g/moleF: 19 g/molethe molar mass of HF is: HF= 1 g/mole + 19 g/mole= 20 g/mole
Then the following rule of three applies: if 20 g of HF are available in 1 mole, 14.3 g in how many moles will they be?
[tex]moles=\frac{14.3 g*1 mole}{20 g}[/tex]
moles= 0.715
So:
number of moles of solute: 0.715 molesVolume: 250 mL=0.250 L (being 1 L=1000 mL)Replacing:
[tex]Molarity=\frac{0.715 moles}{0.250 L}[/tex]
Solving:
Molarity= 2.86 [tex]\frac{moler}{liter}[/tex]=2.86 M
The concentration of the solution is 2.86 M
A chemist must prepare 550.0 mL of nitric acid solution with a pH of 2.00 at 25 °C. He will do this in three steps: Fill a 550.0 mL volumetric flask about halfway with distilled water. Measure out a small volume of concentrated (9.0M) stock nitric acid solution and add it to the flask Fill the flask to the mark with distilled water Calculate the volume of concentrated nitric acid that the chemist must measure out in the second step. Round your answer to 2 significant digits 0- mL.
Answers
Answer:
The correct answer is 0.61 ml
Explanation:
Nitric acid is a strong acid. That means that it dissociates completely in water as follows:
HNO₃ → H⁺ + NO₃⁻
As the dissociation is complete, the concentration of H⁺ ions is equal to the initial concentration of the acid (HNO₃). Thus, the pH can be calculated from the initial concentration of the acid:
pH= -log [H⁺] = -log [acid]
We want a nitric acid solution with a pH of 2.0. so we first calculate the concentration of acid we need:
2.0 = -log [acid]
10⁻²= [acid]
The chemist has a stock solution with C= 9.0 M and he/she wants a solution with C= 1 x 10⁻² M and V= 550 ml. We use the equation that relates the initial concentration and volume (Ci and Vi, respectively) of a solution with the final concentration and volume (Cf and Vf, respectively):
Ci x Vi = Cf x Vf
⇒ Vi= (Cf x Vf)/Ci = (1 x 10⁻² M x 550 ml)/9.0 M = 0.611 ml
Summarizing, the chemist must measure 0.611 ml of concentrated solution (9.0 M), add it to the flask and fill the flask to the mark until 550 ml in order to obtain a nitric acid solution with a pH of 2.0.
Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.02 g of ethane is mixed with 22. g of oxygen. Calculate the minimum mass of ethane that could be left over by the chemical reaction. Round your answer to significant digits.
Answers
Answer:
3.13g
Explanation:
First, we'll begin by writing a balanced equation between the reaction of gaseous ethane with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. This is illustrated below:
2C2H6 + 7O2 —> 4CO2 + 6H2O
Next, let us calculate the mass of C2H6 and the mass of O2 that reacted from the balanced equation. This is illustrated below:
Molar Mass of C2H6 = (12x2) + (6x1) = 24 + 6 = 30g/mol
The mass of C2H6 that reacted from the balanced equation = 2 x 30 = 60g
Molar Mass of O2 = 16x2 = 32g/mol
Mass of O2 that reacted from the balanced equation = 7 x 32 = 224g
Now, to calculate the left over mass of ethane (C2H6), let us first calculate the mass of ethane (C2H6) that will react with 22g of oxygen(O2). This can be achieved by doing the following:
From the balanced equation above,
60g of C2H6 reacted with 224g of O2.
Therefore, Xg of C2H6 will react with 22g of O2 i.e
Xg of C2H6 = (60 x 22)/224
Xg of C2H6 = 5.89g
From the calculations above, 5.89g will react completely with 22g of O2.
The left over mass of C2H6 can be obtained as follow:
Mass of C2H6 from the question = 9.02g
Mass of C2H6 that reacted = 5.89g
The left over mass of C2H6 =?
The left over mass of C2H6 = Mass of C2H6 - mass of C2H6 that reacted
Left over mass of C2H6 = 9.02 - 5.89
Left over mass of C2H6 = 3.13g
Cu(s), CuCl2 (0.50 M) || Ag(s), AgNO3 (0.010 M) 1.
Draw the schematic of the electrochemical cell that you created including all the components (metals, solutions, salt bridges, voltmeters, etc.) in this portion of the experiment. Annotate on the schematic which side is the anode, which side is the cathode, the sign of each half cell, the composition of the metals and solutions, and the direction of the flow of the electrons through the cell.
Answers
Answer:
See explaination
Explanation:
An electrochemical cell is a device capable of either generating electrical energy from chemical reactions or using electrical energy to cause chemical reactions.
See attachment for the step by step solution of the given problem.
Which precaution should you take when you see this symbol?
A. do not look directly at the experiment
B. watch the teacher carefully and follow what he or she does
C.wear goggles if you are doing the experiment
D. wear goggles if you are performing or observing the experiment
Answers
Answer:
D
Explanation:
You should wear goggles whether you are the person performing the experiment or the person watching. Analogy: If someone is using fireworks and they say anyone around should wear ear protection would you wear it if you are next to them?
The researcher performed a follow-up experiment to measure the rate of oxygen consumption by muscle and brain cells. Predict the effect of the MT - ND5 mutation on the rate of oxygen consumption in muscle and brain cells. Justify your prediction. The researcher had hypothesized that the addition of the vitamin that is similar in structure to NADH would increase the activity of the mutated NADH dehydrogenase enzyme in individuals with the disorder. Explain how the vitamin most likely increased the activity of the enzyme.
Answers
Answer:
The rate of oxygen consumption by muscle and brain cells will decrease. The MT-ND5 mutation causes a buildup of lactic acid which occurs as a result of glycolysis not entering the Krebs cycle and instead entering fermentation. Because it is not entering the krebs cycle, there is no oxygen being consumed.
Explanation:
The MT-ND5 mutation potentially disrupts mitochondrial function, decreasing the rate of oxygen consumption in muscle and brain cells. Introducing a NADH-like vitamin may increase the activity of the affected NADH dehydrogenase enzyme, thereby restoring oxygen consumption and ATP production.
Explanation:The MT-ND5 mutation would likely affect the rate of oxygen consumption by disrupting mitochondrial function in both muscle and brain cells. Mitochondria are responsible for cellular respiration where oxygen is consumed to create ATP, a high-energy compound. Any mutation disrupting this process would likely decrease the rate of oxygen consumption.
As the MT-ND5 mutation affects the NADH dehydrogenase enzyme, it impedes the electron transport chain which is critical for ATP production. The researcher's hypothesis is that introducing a vitamin similar to NADH could potentially increase the enzyme's activity. As NADH plays a crucial role in transferring electrons in the electron transport chain, increasing NADH or a similar compound could restore or enhance the function of the enzyme, thereby potentially increasing ATP production and oxygen consumption.
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Bacteria are only found in decaying matter.
A. True
B. False
Answers
Answer:
False
Explanation:
You have bacterias inside of each one of us, some are good and some are bad.
They can be found in the air ou objects, and objects don't decay.
Assuming that the distances between the two ions are the same in all cases, which of the following ion pairs has the greatest electrostatic potential energy (i.e., largest in magnitude)? Please explain your answer.a.) Na+ - Cl- b.) Na+ - O-2. c.) Al+3 - O-2. d.) Mg+2-O-2 e.) Na- -Mg+2
Answers
Answer:
Correct option: C
Explanation:
As given in the question that distance between two ions are same in all cases hence r is same for all.
potential energy:
[tex]P.E =\frac{k\times q_{1} \times q_{2}}{r}[/tex]
therefore potential energy depend on the two charge muliplication
so higher the charge multiplication higer will be the potential energy.
magnitude of charge multiplication follow as:
a. 1
b. 2
c. 6
d. 4
e. 2
in option C it is higher
so correct option is C
Final answer:
The ion pair with the greatest electrostatic potential energy among the given pairs is Al+3 - O-2, as it has the highest product of charges according to Coulomb's Law. Option C is correct.
Explanation:
The electrostatic potential energy between two ions is determined by Coulomb's Law, which states that the energy is proportional to the product of the charges of the ions divided by the distance between them. With all distances being equal, the pair with the greatest product of charges would have the highest electrostatic potential energy. Thus, we can compare the products of charges for each pair:
Na+ - Cl- gives us 1(+1)×1(-1) = 1Na+ - O-2 gives us 1(+1)×2(-1) = 2Al+3 - O-2 gives us 3(+1)×2(-1) = 6Mg+2 - O-2 gives us 2(+1)×2(-1) = 4Na- - Mg+2 gives us 1(-1)×2(+1) = 2Therefore, the ion pair Al+3 - O-2 has the greatest electrostatic potential energy because the product of their charges (6) is the largest.
Alkali halides commonly have either the sodium chloride structure or the cesium chloride structure. The molar mass of CsCl is 2.88 times the molar mass of NaCl, and the edge length of the unit cell for NaCl is 1.37 times the edge length of the CsCl unit cell. Determine the ratio of the density of CsCl to the density of NaCl.
Answers
Answer:
[tex]\large \boxed{1.85:1}[/tex]
Explanation:
The density of a substance is directly proportional to the molar mass and the number of atoms per unit cell, and inversely proportional to the volume of the unit cell.
The BCC unit cell of CsCl contains one K⁺ and one Cl⁻ ion, while the SC unit cell contains four Na⁺ and four Cl⁻ ions.
[tex]\dfrac{\text{CsCl density}}{\text{NaCl density}} = \dfrac{\text{Atoms of Cs}}{\text{Atoms of Ca}} \times \dfrac{\text{MM of CsCl}}{\text{MM of NaCl}} \times \dfrac{\text{Vol. of NaCl unit cell}}{\text{Vol. of CsCl unit cell}}\\\\= \dfrac{1}{4} \times \dfrac{\text{2.88}}{\text{1}} \times \dfrac{1.37^{3}}{1^{3}} = \dfrac{1.85}{1}\\\\\text{The ratio of the density of CsCl to that of NaCl is $\large \boxed{\mathbf{1.85:1}}$}[/tex]
Final answer:
To determine the ratio of the density of CsCl to NaCl, one must consider their molar masses, unit cell structures, and edge lengths. CsCl's density is calculated based on its simple cubic unit cell structure, while NaCl's is based on its FCC unit cell.
Explanation:
The ratio of the density of CsCl to the density of NaCl can be calculated by considering their molar masses, unit cell structures, and the edge lengths of their unit cells.
Sodium chloride (NaCl) crystallizes in a face-centered cubic (FCC) lattice, whereas cesium chloride (CsCl) forms a simple cubic unit cell. Each FCC cell for NaCl contains four formula units, while each simple cubic cell for CsCl contains one formula unit.
Given that the molar mass of CsCl is 2.88 times that of NaCl, and the edge length of NaCl's unit cell is 1.37 times the edge length of CsCl's unit cell, we can calculate the densities by using the formula: density = mass/volume.
The volumes of the unit cells can be found by cubing the respective edge lengths.To find the mass of the unit cells, multiply the molar masses by Avogadro's number and divide by the number of formula units per unit cell.
The final step is to compare the calculated densities by taking the ratio of CsCl's density to NaCl's density. This will yield the required density ratio.
In an experiment to determine the enthalpy change for this reaction, you combine 0.158 g of Mg metal with enough HCl to make 100.0 mL of solution in a coffee-cup calorimeter. The HCl is sufficiently concentrated so that the Mg completely reacts. The temperature of the solution rises from 25.6 °C to 32.8 °C as a result of the reaction. Find ΔHrxn for the reaction as written. Use 1.00 g/mL as the dens
Answers
Answer:
The value of Δ [tex]H_{rxn}[/tex] for the reaction = 463 [tex]\frac{KJ}{mol}[/tex]
Explanation:
[tex]q_{rxn} = - q_{sol}[/tex]
[tex]q_{rxn} = H_{rxn}[/tex]
Δ [tex]H_{rxn}[/tex] = - [tex]q_{sol}[/tex] ------ (1)
We know that
[tex]q_{sol} = m c ( T_{2} - T_{1} )[/tex]
[tex]q_{sol} =[/tex] (1)(100) × 4.18 × (32.8 - 25.6)
[tex]q_{sol} =[/tex] 3010 J = 3.01 Kilo Joule
From equation (1)
Δ [tex]H_{rxn}[/tex] = 3.01 Kilo Joule
No. of moles
[tex]N = \frac{m}{M}[/tex]
m = 0.158 gm & M = 24.31 gm Mg
No. of moles
[tex]N = \frac{0.158}{24.31}[/tex]
N = 0.0065
Therefore
Δ [tex]H_{rxn}[/tex] = [tex]\frac{3.01}{0.0065}[/tex]
Δ [tex]H_{rxn}[/tex] = 463 [tex]\frac{KJ}{mol}[/tex]
This is the value of Δ [tex]H_{rxn}[/tex] for the reaction.
A student collects 629ml of oxygen at 0.500at, the student collected 0.0337 moles. At what temperature did the student collect the oxygen?
Answers
Answer:
114 K
Explanation:
Given data
Volume of oxygen (V): 629 mL = 0.629 LPressure of oxygen (P): 0.500 atmMoles of oxygen (n): 0.0337 molTemperature (T): ?We can calculate the temperature at which the student collected the oxygen using the ideal gas equation.
[tex]P \times V = n \times R \times T\\T = \frac{P \times V}{n \times R} = \frac{0.500atm \times 0.629L}{0.0337mol \times 0.0821atm.L/mol.K} = 114 K[/tex]
The oxygen gas was collected at 114 K.
Answer:
The temperature is 114 K or -159ºC
Explanation:
We must use the ideal gas equation to calculate the temperature in K:
P x V = n x R x T
⇒ T = (P x V)/(n x R)
We have to introduce the data expressed in the adequate units:
V = 629 ml x 1 L/1000 ml= 0.629 L
n = 0.0337 moles
P = 0.500 atm
R = 0.082 L.atm/K.mol (it is the gas constant)
T = (P x V)/(n x R) = (0.500 atm x 0.629 L)/(0.0337 mol x 0.082 L.atm/K.mol)
T = 113.8 K ≅ 114 K
Thus, the temperature is 114 K or -159ºC.
g Suppose you are titrating vinegar, which is an acetic acid solution of unknown strength, with a sodium hydroxide solution according to the equation H C 2 H 3 O 2 + N a O H ⟶ H 2 O + N a C 2 H 3 O 2 If you require 31.26 mL of 0.1871 M N a O H solution to titrate 10.0 mL of H C 2 H 3 O 2 solution, what is the concentration of acetic acid in the vinegar?
Answers
Answer: The concentration of acetic acid in the vinegar is 0.585 M
Explanation:
To calculate the volume of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]CH_3COOH[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=1\\M_1=?\\V_1=10.0mL\\n_2=1\\M_2=0.1871M\\V_2=31.26mL[/tex]
Putting values in above equation, we get:
[tex]1\times M_1\times 10.0=1\times 0.1871\times 31.26\\\\M_1=0.585M[/tex]
Thus the concentration of acetic acid in the vinegar is 0.585 M
The concentration of acetic acid in the vinegar solution is 0.585 M. This was determined by calculating the moles of acetic acid using the volume and molarity of the NaOH solution necessary to neutralize the acetic acid and then applying the definition of molarity.
Explanation:In a titration of acetic acid (found in vinegar) with sodium hydroxide, the point at which all the acetic acid has been neutralized by the sodium hydroxide is called the equivalence point. The number of moles of sodium hydroxide used will be equal to the number of moles of acetic acid in the solution.
From the given data of the sodium hydroxide solution volume (31.26 mL) and molarity (0.1871 M), the moles of NaOH can be calculated as follows: Moles of NaOH = Volume (L) x Molarity = 0.03126 L x 0.1871 mol/L = 0.00585 mol
This is equal to the moles of acetic acid in the 10 mL sample of vinegar. Hence the molarity (and therefore the concentration) of acetic acid in vinegar is given by: Molarity = Moles / Volume (L) = 0.00585 mol / 0.01 L = 0.585 M
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Ammonia gas combines with hydrogen chloride gas, forming solid ammonium chloride.a.Write a balanced chemical equation for this reaction.b.In a reaction mixture of 3.0 g ammonia and 5.0 g hydrogen chloride, which of the two is the limiting reagent?c.How many grams of ammonium chloride could form from the reaction mixture in part b?d.How much of the reactant is left over in the reaction mixture o
Answers
Answer:
a. NH₃(g) + HCl(g) → NH₄Cl(s)
b. HCl is the limiting reagent
c. 7.35 g of NH₄Cl
d. 0.039 moles of NH₃ remains after the reaction is complete (0.663 grams of ammonia)
Explanation:
First of all, we define the reaction where the reactants are the ammonia and the hydrogen chloride. The product is the ammonium chloride.
NH₃(g) + HCl(g) → NH₄Cl(s)
First of all, we need to determine the limiting reactant. As the ratio is 1:1, the compound that has the lowest mol, will be the limiting.
3 g / 17g/mol = 0.176 moles NH₃
5 g / 36.45 g/mol = 0.137 moles HCl (limiting reactant)
For 0.176 moles of ammonia, we need the same of HCl, but we have 0.137 moles. We convert the moles to mass, in order to know how many grams of NH₄Cl are produced → 0.137 mol . 53.45 g /1mol = 7.32 g
As ratio is 1:1, and the limiting reagent is the HCl, after the reaction goes complete, some ammonia remains. This is because, the ammonia is the excess reagent.
0.176 mol - 0.137 moles = 0.039 moles of NH₃ remains after the reaction is complete; we convert the moles to mass → 0.039 mol . 17 g /mol = 0.663 g are left over in the reaction mixture.
Final answer:
The balanced chemical equation for the reaction between ammonia and hydrogen chloride to form ammonium chloride is NH3(g) + HCl(g) → NH4Cl(s). Ammonia is the limiting reagent in the given mixture, and the theoretical yield of ammonium chloride is 9.42 g. No excess reactant will be left over after the reaction.
Explanation:
Chemical Reaction Between Ammonia and Hydrogen Chloride
When ammonia gas (NH3) combines with hydrogen chloride gas (HCl), solid ammonium chloride (NH4Cl) is formed. The balanced chemical equation for this reaction is:
NH3(g) + HCl(g) → NH4Cl(s).
Limiting Reagent and Yield
In a reaction mixture with 3.0 g of ammonia and 5.0 g of hydrogen chloride, to find the limiting reagent we must compare the mole ratio of the reactants to their stoichiometric ratio in the balanced equation. Using molar masses (NH3: 17.03 g/mol, HCl: 36.46 g/mol), we find that:
NH3: 3.0 g × (1 mol / 17.03 g) ≈ 0.176 moles
HCl: 5.0 g × (1 mol / 36.46 g) ≈ 0.137 moles
The stoichiometry of the reaction is 1:1, so ammonia is the
limiting reagent
because we have fewer moles of it relative to HCl.
The mass of ammonium chloride formed is based on the limiting reagent. Since all of the ammonia will react, the theoretical yield of NH4Cl is the moles of NH3 times the molar mass of NH4Cl (53.49 g/mol):
0.176 moles × 53.49 g/mol ≈ 9.42 g of NH4Cl.
Excess Reactant Left Over
To determine the amount of excess reactant left over, we subtract the number of moles of HCl that reacted (equivalent to the moles of NH3) from the initial moles of HCl:
0.137 moles HCl - 0.176 moles NH3 = -0.039 moles HCl (not possible, indicating no excess HCl remains).
A student is asked to standardize a solution of barium hydroxide. He weighs out 0.978 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid). It requires 35.8 mL of barium hydroxide to reach the endpoint. A. What is the molarity of the barium hydroxide solution? M This barium hydroxide solution is then used to titrate an unknown solution of hydrochloric acid. B. If 17.1 mL of the barium hydroxide solution is required to neutralize 18.6 mL of hydrochloric acid, what is the molarity of the hydrochloric acid solution? M
Answers
Answer:
(A) 0.129 M
(B) 0.237 M
Explanation:
(A) The reaction between potassium hydrogen phthalate and barium hydroxide is:
2HA + Ba(OH)₂ → BaA₂ + 2H₂OWhere A⁻ is the respective anion of the monoprotic acid (KC₈H₄O₄⁻).
We convert mass of phthalate to moles, using its molar mass:
0.978 g ÷ 156 g/mol = 9.27x10⁻³ mol = 9.27 mmolNow we convert mmol of HA to mmol of Ba(OH)₂:
9.27 mmol HA * [tex]\frac{1mmolBa(OH)_{2}}{2mmolHA}[/tex] = 6.64 mmol Ba(OH)₂Finally we calculate the molarity of the Ba(OH)₂ solution:
6.64 mmol / 35.8 mL = 0.129 M(B) The reaction between Ba(OH)₂ and HCl is:
2HCl + Ba(OH)₂ → BaCl₂ + 2H₂OSo the moles of HCl that reacted are:
17.1 mL * 0.129 M * [tex]\frac{2mmolHCl}{1mmolBa(OH)_2}[/tex] = 4.41 mmol HClAnd the molarity of the HCl solution is:
4.41 mmol / 18.6 mL = 0.237 MUsing the name of the ionic compound, select the
appropriate chemical formula. Please use the
periodic table that has been provided for your use.
Calcium bicarbonate
o CaHCO3
O Ca(CO3)2
o Ca(HCO3)2
DONE
Answers
Answer:
option C. Ca(HCO3)2
Explanation:
did it on edg
Answer:
answer c
Explanation:
A mixture of gaseous reactants is put into a cylinder, where a chemical reaction turns them into gaseous products. The cylinder has a piston that moves in or out, as necessary, to keep a constant pressure on the mixture of . The cylinder is also submerged in a large insulated water bath. (See sketch at right.) From previous experiments, this chemical reaction is known to absorb of energy. The temperature of the water bath is monitored, and it is determined from this data that of heat flows out of the system during the reaction. Is the reaction exothermic or endothermic
Answers
Answer:
See explaination
Explanation:
Exothermic reactions are reactions or processes that release energy, usually in the form of heat or light.
Hence when heat is sent out or released we have an Exothermic reaction.
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A geochemist in the field takes a small sample of the crystals of mineral compound X from a rock pool lined with more crystals of X. He notes the temperature of the pool, 26.° C, and caps the sample carefully. Back in the lab, the geochemist dissolves the crystals in 3.00 L of distilled water. He then filters this solution and evaporates all the water under vacuum. Crystals of X are left behind. The researcher washes, dries and weighs the crystals. They weigh 0.36 kg1) Using only the information above can you calculate the solubility of X in water at 26 degrees Celsius? yes or no2) If yes calculate the solubility. Round answer to 2 signifacnt digits
Answers
Yes, you can use the given information to calculate the solubility of substance X in water at 26 degrees Celsius. You would do this by dividing the mass of X (0.36 kg) by the volume of the water (3.00 L), resulting in a solubility of approximately 120 g/L.
Explanation:1. Yes, you can calculate the solubility of X in water at 26 degrees Celsius with the information provided. The reason being solubility is defined as the maximum quantity of a substance that can be dissolved in a given amount of solvent at a certain temperature.
2. To calculate the solubility, you would take the mass of the compound dissolved (0.36 kg or 360 g) and divide it by the volume of water in which it was dissolved (3.00 L).
Solubility = (360 g / 3.00 L) ≈ 120 g/L. Hence, the solubility of X at 26° C is approximately 120 g/L, to two significant figures.
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Suppose that you want to find the molarity of a solution that contains 15.0 g of KCI in 150.0 mL of solution.
(The molar mass of KCl is 74.45 g/mol.)
Calculate the number of moles of KCI:
Answers
Answer :
Moles = 0.2mol
Explanation:
n= m/M = 15/74.45 = 0.2mol
Answer:
A is 15.0 g
B is 74.45 g/mol
C is 0.201 mol
D is 150.0 mL
E is 1.34 M
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A mixture of krypton and argon gas is expanded from a volume of 33.0L to a volume of 61.0L , while the pressure is held constant at 58.0atm . Calculate the work done on the gas mixture. Be sure your answer has the correct sign (positive or negative) and the correct number of significant digits.
Answers
Answer:
W = +1624J
Explanation:
V1 = 33.0L
V2 = 61.0L
P = 58 atm
Work = force * distance (∇x)
Work = F.∇x
But force = pressure * area
F = P.A
Work = P.∇x.A
Work = P∇v
Work = P (V₂ - V₁)
Work = 58 * (61 - 33)
Work = 58 * 28
Work = 1624J
The work done is +1624J
For the reaction shown, calculate how many moles of NH3 form when 16.72 moles of reactant completely reacts:
3 N2H4 (l)
⟶
4 NH3(g) + N2(g)
Answers
Answer : The moles of [tex]NH_3[/tex] formed are, 22.3 moles.
Explanation : Given,
Moles of [tex]N_2H_4[/tex] = 16.72 mol
The given chemical reaction is:
[tex]3N_2H_4(l)\rightarrow 4NH_3(g)+2N_2(g)[/tex]
From the balanced chemical reaction, we conclude that:
As, 3 moles of [tex]N_2H_4[/tex] react to give 4 moles of [tex]NH_3[/tex]
So, 16.72 moles of [tex]N_2H_4[/tex] react to give [tex]\frac{4}{3}\times 16.72=22.3[/tex] moles of [tex]NH_3[/tex]
Therefore, the moles of [tex]NH_3[/tex] formed are, 22.3 moles.
Consider again the thermite reaction. If 0.0257 g Al react completely, what mass of Fe forms? Use the periodic table to find molar masses.
Answers
Answer:
C) 0.0532
Explanation:
Stoichiometry is an important concept in chemistry which helps us to use the balanced chemical equations to calculate the amount of reactants and products. Here the mass of 'Fe' produced from 0.0257 g Al is 0.0532 g.
What is stoichiometry?Chemical Stoichiometry refers to the quantitative study of the reactants and products involved in a chemical reaction. Here we make use of the ratios from the balanced equation to calculate the amount of reactants and products.
The number of moles of 'Al' = Given mass / Molar mass
Molar mass of 'Al' = 27 g/mol
n = 0.0257 / 27 = 9.5 × 10⁻⁴ moles
Here the balanced reaction is:
Fe₂O₃ + 2 Al → Al₂O₃ + 2 Fe
using the mole ratio to determine the moles of Fe
The mole ratio of Al:Fe is 2:2 = 1:1
So the moles of Fe is also = 9.5 × 10⁻⁴ moles
The mass of iron is:
Mass = moles x molar mass of Fe
Molar mass of Fe= 56 g/mol
Mass= 9.5 × 10⁻⁴ x 56 =0.0532 grams
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How many c atoms are there in 5K2CO3
Answers
There is 1 carbon atom in 5K2CO3.
Explanation:To determine the number of carbon (C) atoms in 5K2CO3, we need to consider the molecular formula of potassium carbonate (K2CO3). The formula indicates that there are 2 potassium (K) atoms, 1 carbon (C) atom, and 3 oxygen (O) atoms per molecule.
Since there is only 1 carbon atom in the formula, regardless of the coefficient 5 in front, there will always be 1 carbon atom in 5K2CO3.
Therefore, there is 1 carbon atom in 5K2CO3.
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Taylor, a MIC 206 student needs to determine the OCD of a sample of Escherichia coli. She performed dilutions using four 9 ml dilution blanks and plated 1.0 ml from the final dilution tube. 137 colonies grew on the plate. Calculate the number of CFU/ml that should be present in the original E. Coli sample.
Answers
Answer:
1.37 x [tex]10^6[/tex] CFU/mL
Explanation:
First, the dilution factor needs to be calculated.
Since four 9 ml dilution blanks were prepared, the dilution factor that yielded 137 colonies is of [tex]10^{-4}[/tex].
Next is to divide the colony forming unit from the dilution by the dilution factor:
137/[tex]10^{-4}[/tex] = 137 x [tex]10^4[/tex]
In order to get the CFU/ml, divide the CFU from the dilution by the plated volume (1 mL) from the final dilution tube.
137 x [tex]10^4[/tex]/1 = 1.37 x [tex]10^6[/tex]
Hence, the CFU/ml present in the original E. coli sample is 1.37 x [tex]10^6[/tex].
cfu/ml = (no. of colonies x dilution factor) / volume of culture plate
Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction:
2 S(s) + 3 O2(g) → 2 SO3(g) ΔH°rxn = ?
Given:
SO2(g) → S(s) + O2(g) ΔH°rxn = +296.8 kJ
2 SO2(g) + O2(g) → 2 SO3(g) ΔH°rxn = -197.8 kJ
a. -494.6 kJ
b. -692.4 kJ
c. -791.4 kJ
d. 1583 kJ
Answers
Answer:
The answer is -791.5 kJ (option c)
Explanation:
You know:
SO₂ (g) → S (s) + O₂ (g) ΔH°rxn = +296.8 kJ
2 SO₂ (g) + O₂ (g) → 2 SO₃ (g) ΔH°rxn = -197.8 kJ
You must add them to obtain the desired equation:
2 S(s) + 3 O₂(g) → 2 SO₃(g) ΔH°rxn = ?
You want to calculate the ∆H (heat of reaction) of the combustion reaction, that is, the heat that accompanies the entire reaction. The calculation is made using Hess's law. This law states: when the reactants are converted to products, the enthalpy change is the same, regardless of whether the reaction is carried out in one step or in a series of steps.
In Hess's law he explains that the enthalpy changes are additive, ΔHneta = ΣΔHr and contains three rules:
If the chemical equation is inverted, the ΔH symbol is inverted as well. If the coefficients are multiplied, multiply ΔH by the same factor. If the coefficients are divided, divide ΔH by the same divisor.The sum of the fitted equations should give the problem equation. In this case:
2*( S(s) + O₂(g) → SO₂ ) To obtain the desired reaction, this equation must be inverted, so the enthalpy value is also inverted. It must also be multiplied by 2, then the whole equation is multiplied, both reactants and products and the value of the enthalpy. So ΔH°rxn = (-296.8 kJ)*2= -593.6 kJ
2 SO₂(g) + O₂(g) → 2 SO₃ (g) ΔH°rxn = -197.8 kJ
____________________________________________________
2 S(s) + 3 O₂(g) → 2 SO₃(g) ΔH°rxn = -791.4 kJ ( Enthalpies are added algebraically)
The answer is -791.5 kJ (option c)
The value of ΔH°rxn for the 2S(s) + 3O₂(g) → 2SO₃(g) is -791.4 kJ.
What is enthalpy?Enthalpy of the reaction tells about the amount of heat released or absorbed during any chemical reaction.
Given that,
2SO₂(g) + O₂(g) → 2SO₃(g) ΔH°rxn = -197.8 kJ
SO₂(g) → S(s) + O₂(g) ΔH°rxn = 296.8 kJ
To obtain the required reaction first we have to invert the second equation and multiply that by 2 and then add with the left hand side as well as the right hand side of the reaction, we get:
2S(s) + 3O₂(g) → 2SO₃(g)
ΔH°rxn of the reaction will be calculated as-
ΔH°rxn = -2(296.8 kJ) + (-197.8 kJ)
ΔH°rxn = -791.4 kJ
Hence, option (C) is correct i.e. -791.4 kJ.
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The pH of water is about 7. The pH of lemon juice is about 2. Which liquid is an acid?
Answers
Answer:
lemon juice
Explanation:
lemons are acidic.
Answer:
the answer is lemon.
Explanation:
Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 3.97 g of magnesium ribbon burns with 8.05 g of oxygen, a bright, white light and a white, powdery product are formed. Enter the balanced chemical equation for this reaction. Be sure to include all physical states.
Answers
Answer:
The balanced chemical equation is given as:
[tex]2Mg(s)+O_2(g)\rightarrow 2MgO(s)[/tex]
Explanation:
When magnesium metal burns in presence of oxygen it gives white color powdered compound called magnesium oxide.
The balanced chemical equation is given as:
[tex]2Mg(s)+O_2(g)\rightarrow 2MgO(s)[/tex]
According to reaction, 2 moles of magnesium metal when reacts with 1 mole of oxygen gas it gives 2 moles of solid magnesium oxide.
When water cools from 10 °C and then freezes to become ice, which of the following best describes describe the heat flow between the system and surroundings? A) The energy of the universe is decreasing because the water is cold. B) The energy of the universe is increasing because the energy had to be removed from the water C) The water is absorbing energy as its temperature decreases, and that energy was released by the surroundings. D) The water is releasing energy as its temperature decreases, and that energy is absorbed by the surroundings.
Answers
Answer:
D) The water is releasing energy as its temperature decreases, and that energy is absorbed by the surroundings.
Explanation:
When the temperature decreases it loses energy to the surroundings. When water cools and then freezes to ice it becomes a solid which has a lower energy compared to liquid and gas. Solid is more compact and little or no collision occurs between its particles.
This means energy was lost during the process of freezing to ice. The energy lost is normally absorbed by the surrounding environment.
Final answer:
The water releases energy as it cools and turns to ice, which is absorbed by the surroundings. This describes an exothermic process where the heat flow involves energy leaving the system and increasing the entropy of the surroundings. Thus, option D is correct.
Explanation:
When water cools from 10 °C and freezes to become ice, the correct description of the heat flow between the system and the surroundings is that the water is releasing energy as its temperature decreases, and that energy is absorbed by the surroundings. So the best answer is D) The water is releasing energy as its temperature decreases, and that energy is absorbed by the surroundings.
This process of cooling and freezing is exothermic, which means heat is exiting the system. During the phase change from liquid to solid, the temperature remains constant at the melting/freezing point, and the heat energy released due to the formation of a more structured lattice of ice molecules is absorbed by the surroundings. This results in an increased entropy in the surroundings which compensates for the decreased entropy of the system (water turning into ice).