What is the mole fraction of calcium chloride in 3.35 m CaCl2(aq)? The molar mass of CaCl2 is 111.0 g/mol and the molar mass of water is 18.02 g/mol.

Answers

Answer 1
Answer:

Answer: The mole fraction of calcium chloride and water in the solution is 0.057 and 0.943 respectively

Explanation:

We are given:

Molality of calcium chloride = 3.35 m

This means that 3.35 moles of calcium chloride are present in 1 kg or 1000 g of water

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of water = 1000 g

Molar mass of water = 18.02 g/mol

Putting values in above equation, we get:

\text{Moles of water}=(1000g)/(18.02g/mol)=55.49mol

Total moles of solution = [3.35 + 55.49] = 58.84 moles

Mole fraction of a substance is given by:

\chi_A=(n_A)/(n_A+n_B)

  • For calcium chloride:

\chi_(CaCl_2)=(n_(CaCl_2))/(n_(CaCl_2)+n_(H_2O))\n\n\chi_(CaCl_2)=(3.35)/(58.84)=0.057

  • For water:

\chi_(H_2O)=(n_(H_2O))/(n_(CaCl_2)+n_(H_2O))\n\n\chi_(H_2O)=(55.49)/(58.84)=0.943

Hence, the mole fraction of calcium chloride and water in the solution is 0.057 and 0.943 respectively

Answer 2
Answer:

Answer:

49.3% water

Explanation:


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Write the balanced chemical equation for each of these reactions. Include phases.1) When aqueous sodium hydroxide is added to a solution containing lead(II) nitrate, a solid precipitate forms.

2) However, when additional aqueous hydroxide is added the precipitate redissolves forming a soluble Pb(OH)42-(aq) complex ion

Answers

Answer: The chemical equations are given below.

Explanation:

  • For 1:

The chemical equation for the reaction of lead nitrate and sodium hydroxide follows:

Pb(NO_3)_2(aq.)+2NaOH(aq.)\rightarrow Pb(OH)_2(s)+2NaNO_3(aq.)

By Stoichiometry of the reaction:

1 mole of aqueous solution of lead nitrate reacts with 2 moles of aqueous solution of sodium hydroxide to produce 1 mole of solid lead hydroxide and 2 moles of aqueous solution of sodium nitrate.

  • For 2:

The chemical equation for the reaction of lead hydroxide and hydroxide ions follows:

Pb(OH)_2(s)+2OH^-(aq.)\rightarrow [Pb(OH)_4]^(2-)(aq.)

By Stoichiometry of the reaction:

1 mole of lead hydroxide reacts with 2 moles of aqueous solution of hydroxide ions to produce 1 mole of aqueous solution of tetra hydroxy lead (II) complex

Hence, the chemical equations are given above.

The balanced chemical equation for each of the reactions is as follows:

  • Pb(NO3)2(aq) + 2NaOH(aq) → 2 NaNO3(aq) + Pb(OH)2(s)
  • Pb(OH)2 + 2OH- → Pb(OH)42-

How to balance chemical equation?

A chemical equation is said to be balanced when the number of atoms of each element on both sides of the equation are the same.

According to this question, when aqueous sodium hydroxide is added to a solution containing lead(II) nitrate, a solid precipitate forms. The balanced equation are as follows:

Pb(NO3)2(aq) + 2NaOH(aq) → 2 NaNO3(aq) + Pb(OH)2(s)

Also, when additional aqueous hydroxide is added the precipitate redissolves forming a soluble Pb(OH)42-(aq) complex ion. The balanced equation is as follows:

Pb(OH)2 + 2OH- → Pb(OH)42-

Learn more about balanced equation at: brainly.com/question/1301642

The rate constant for the second-order reaction 2NOBr(g) ¡ 2NO(g) 1 Br2(g) is 0.80/M ? s at 108C. (a) Starting with a concentration of 0.086 M, calculate the concentration of NOBr after 22 s. (b) Calculate the half-lives when [NOBr]0 5 0.072 M and [NOBr]0 5 0.054 M.

Answers

Explanation:

2NOBr(g) --> 2NO(g) 1 Br2(g)

Rate constant, k = 0.80

a) Initial concentration, Ao = 0.086 M

Final Concentration, A = ?

time = 22s

These parameters are connected with the equation given below;

1 / [A] = kt + 1 / [A]o

1 / [A] = 1 / 0.086 + (0.8 * 22)

1 / [A] = 11.628 + 17.6

1 / [A] = 29.228

[A] = 0.0342M

b) t1/2 = 1 / ([A]o * k)

when [NOBr]0 5 0.072 M

t1/2 = 1 / (0.072 * 0.80)

t1/2 = 1 / 0.0576 = 17.36 s

when [NOBr]0 5 0.054 M

t1/2 = 1 / (0.054 * 0.80)

t1/2 = 1 / 0.0432 = 23.15 s

Answer:

(a)

0.0342M

(b)

t_(1/2)=17.36s\nt_(1/2)=23.15s

Explanation:

Hello,

(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:

(1)/([NOBr])=kt +(1)/([NOBr]_0)\n(1)/([NOBr])=(0.8)/(M*s)*22s+(1)/(0.086M)=(29.3)/(M)\n

[NOBr]=(1)/(29.2/M)=0.0342M

(b) Now, for a second-order reaction, the half-life is computed as shown below:

t_(1/2)=(1)/(k[NOBr]_0)

Therefore, for the given initial concentrations one obtains:

t_(1/2)=(1)/((0.80)/(M*s)*0.072M)=17.36s\nt_(1/2)=(1)/((0.80)/(M*s)*0.054M)=23.15s

Best regards.

Which group is the most reactive?*
Alkaline Earth Metals
Alkali metals
Noble Gases
Lanthanides

Answers

Answer:

alkali metals- Group 1

Explanation:

they have less valence electrons and therefore are more reactive

Mr. Hall was conducting an experiment. He dissolved an unknownubstance in water. He performed multiple trials while varying the
emperature. What is the independent variable in his experiment?
a. The unknown substance, because it's the only thing he changed
b.The temperature, because it's the only thing he changed

Answers

Answer:

a

Explanation:

What are four properties of water?

Answers

Answer:

Polarity

Cohesion  

Adhesion

High Specific Heat

Explanation:

Please answer my question. #6-10: Consider the energy diagram below:
6. What is the activation energy for this reaction? _____
7. What is the change in free energy for the reaction? _____
8. How many intermediates are involved in this reaction? _____
9. How many transition states are involved in the reaction mechanism? _____
10. Is the reaction endergonic or exergonic overall?

Answers

6. Activation energy is the difference in energy between the reactant and the transition state. The energy of the reactant and the transition state are 5 kJ/mol and 15 kJ/mol respectively. The activation energy for this reaction is 10 kJ/mol.

7. The change in free energy for the reaction is the difference in energy between the reactant and the product. The energy of the reactant and the product are 5 kJ/mol and 10 kJ/mol respectively. The change in free energy for the reaction is 5 kJ/mol.

8. There are no intermediates involved in this reaction.

9.  There is only one transition state involved in the reaction mechanism. The transition state is indicated by the highest point of the graph.

10. The reaction is endergonic overall. The energy of the product is higher than the energy of the reactants, which is only possible if energy is absorbed by the reaction.