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2 years ago

5

At a certain temperature, the reaction 2NO + Cl2 ⇌ 2NOCl has an equilibrium constant Kc of 45.0. A chemist creates a mixture wit

h the following initial concentrations: [NO] = 0.10 M, [Cl2] = 0.20 M, and [NOCl] = 0.30 M. What will happen as the reaction begins?
A. [NOCl] will increase.
B. [NOCl] will decrease.
C. [NOCl] will stay the same.
D. There is not enough information to answer the question.

1 answer:

Dmitriy789 [7]2 years ago

5 0

Answer:

C. [NOCl] will stay the same

Explanation:

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In a hexagonal-close-pack (hcp) unit cell, the ratio of lattice points to octahedral holes to tetrahedral holes is 1:1:2. What i

Alex777 [14]

Explanation:

In a hexagonal-close-pack (HCP) unit cell, the ratio of lattice points to octahedral holes to tetrahedral holes = 1 : 1 : 2

let the :

Number of lattice point = 1x.

Number of octahedral points = 1x

Number of tetrahedral points = 2x

If anions occupy the HCP lattice points and cations occupy half of the tetrahedral holes.

Number of anions occupying the HCP lattice points, A= 1x

Number of cations occupying the tetrahedral points, B = $\frac{2x}{2}=x$

The formula of the compound will be = $A_{1x}B_{1x}=AB$

If anions occupy the HCP lattice points and cations occupy all of the tetrahedral holes.

Number of anions occupying the HCP lattice points, A= 1x

Number of cations occupying the tetrahedral points, B = 2x

The formula of the compound will be = $A_{1x}B_{2x}=AB_2$

6 0

2 years ago

A 0.530 M Ca(OH)2 solution was prepared by dissolving 36.0 grams of Ca(OH)2 in enough water. What is the total volume of the sol

Paladinen [302]

The concentration of a solution is the number of moles of solute per fixed volume of solution.

Concentration (C) = number of moles of solute (n) / volume of the solution (v)

we have to find the volume of the solution when 36.0 g of Ca(OH)₂ is added to water to make a solution of concentration 0.530 M

mass of Ca(OH)₂ added - 36.0 g

number of moles of Ca(OH)₂ - 36.0 g / 74.1 g/mol = 0.486 mol

we know the concentration of the solution prepared and the number of moles of Ca(OH)₂ added, substituting these values in the above equation, we can find the volume of the solution

C = n/v

0.530 mol/L = 0.486 mol / V

V = 0.917 L

answer is 0.917 L

6 0

2 years ago

Read 2 more answers

Which of the following is a valid conversion factor?

VMariaS [17]

Answer:

100 cg/1g

Step-by-step explanation:

1 cg = 0.01 g Multiply by 100

100 cg = 1 g

(a) is <em>wrong</em>. The correct conversion factor is 1000 cm³/1 L.

(b) is <em>wrong</em>. The correct conversion factor is 1000 mL/1 L.

(c) is <em>wrong</em>. The correct conversion factor is 1 m/10 dm.

7 0

2 years ago

Read 2 more answers

Calculate the oxidation number of s in S2O8^2-

mixer [17]

Given problem:

S₂O₈²⁻

Find the oxidation number of S;

Oxidation number presents the extent of oxidation of each atom of elements a molecular formular or formula unit or an ionic radical.

For radicals:

"the algebraic sum of all the oxidation numbers of all atoms in an ion containing more than one kind of atom is equal to the charge on the ion "

S₂O₈²⁻; oxidation number of O is usually -2

2(S) + 8(-2) = -2

2S - 16 = -2

2S = -2 + 16

2S = 14

S = +7

The oxidation state of S in the radical is +7

3 0

2 years ago

An unknown salt is either NaF, NaCl, or NaOCl. When 0.050 mol of the salt is dissolved in water to form 0.500 L of solution, the

victus00 [196]

<u>Answer:</u> The unknown salt is NaF

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of salt = 0.050 moles

Volume of solution = 0.500 L

Putting values in above equation, we get:

$\text{Molarity of salt}=\frac{0.050mol}{0.500L}\\\\\text{Molarity of salt}=0.1M$

To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:

pH + pOH = 14

We are given:

pH = 8.08

$pOH=14-8.08=5.92$

To calculate pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

Putting values in above equation, we get:

$5.92=-\log[OH^-]$

$[OH^-]=10^{-5.92}=1.202\times 10^{-6}M$

The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH

The chemical equation for the hydrolysis of $X^-$ ions follows:

$X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b$

<u>Initial:</u> 0.1

<u>At eqllm:</u> 0.1-x x x

Concentration of $OH^-=x=1.202\times 10^{-6}M$

The expression of $K_b$ for above equation follows:

$K_b=\frac{[OH^-][HX]}{[X^-]}$

Putting values in above expression, we get:

$K_b=\frac{(1.202\times 10^{-6})\times (1.202\times 10^{-6})}{(1-(1.202\times 10^{-6}))}\\\\K_b=1.445\times 10^{-11}M$

To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant

$K_b$ = Base dissociation constant = $1.445\times 10^{-11}$

Putting values in above equation, we get:

$10^{-14}=1.445\times 10^{-11}\times K_a\\\\K_a=\frac{10^{-14}}{1.445\times 10^{-11}}=6.92\times 10^{-4}$

We know that:

$K_a\text{ for HF}=6.8\times 10^{-6}$

$K_a\text{ for HCl}=1.3\times 10^{6}$

$K_a\text{ for HClO}=3.0\times 10^{-8}$

So, the calculated $K_a$ is approximately equal to the $K_a$ of HF

Hence, the unknown salt is NaF

6 0

2 years ago