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Katyanochek1 [597]
3 years ago
9

At some temperature, the reaction: 3 clo- ? clo3- 2 cl- has an equilibrium constant kc = 3.2 x 103. if the components of this re

action are mixed such that their initial concentrations are [cl-] = 0.05 m; [clo3-] = 0.32; and [clo-] = 0.74, is the mixture at equilibrium, yes or no? if the mixture is not at equilibrium in which direction, left to right or right to left, will reaction occur so that the mixture can reach equilibrium?
Chemistry
1 answer:
Scilla [17]3 years ago
4 0
Gagagagaagagry rftdadadad
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Gravity is the force that keeps the planets in orbit.
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Differences in air pressure drive the global wind system on earth. Which term describes polar winds that blow from east to west?
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Answer: Polar Easterlies

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For the reaction: 3 H2(g) + N2(g) &lt;--&gt; 2 NH3(g), the concentrations at equilbrium were [H2] = 0.10 M, [N2] = 0.10 M, and [
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The equilibrium constant, k of the reaction in which case, the concentrations of the given reactants and products are as indicated is; Choice A; K = 3.1 x 10⁵

<h3>What is the equilibrium constant , k of the reaction as described in the task content?</h3>

It follows from above that the concentrations of the reactants and products are as follows; [H2] = 0.10 M, [N2] = 0.10 M, and [NH3] = 5.6 M at equilibrium.

Hence, the equilibrium constant of the reaction in discuss is;

K = [5.6]²/[0.10]³[0.10]

k = 5.6² × 10⁴

k = 3.136 × 10⁵

K = 3.1 × 10⁵.

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5 0
2 years ago
Which two periods have the same number of elements ​
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In a single displacement reaction between sodium phosphate and barium, how much of each product (in grams) will be formed from 1
weeeeeb [17]

Answer:

A. 3.36g of Na.

B. 14.62g of Ba3(PO4)2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

3Ba + 2Na3PO4 → 6Na + Ba3(PO4)2

Next, we shall determine the mass of Ba that reacted and the mass of Na and Ba3(PO4)2 produced from the equation.

This is illustrated below:

Molar Mass of Ba = 137g/mol

Mass of Ba from the balanced equation = 3 x 137 = 411g

Molar mass of Na = 23g/mol

Mass of Na from the balanced equation = 6 x 23 = 138g

Molar mass of Ba3(PO4)2 = (3 x 137) + 2[31 + (4x16)] = 411 + 2[31 + 64] = 601g/mol

Mass of Ba3(PO4)2 from the balanced equation = 1 x 601 = 601g

Summary:

From the balanced equation above,

411g of Ba reacted to produce 138g of Na and 601g of Ba3(PO4)2.

A. Determination of the mass of Na produced by reacting 10g of Ba.

From the balanced equation above,

411g of Ba reacted to produce 138g of Na.

Therefore, 10g of Ba will react to produce = (10 x 138)/411 = 3.36g of Na.

Therefore, 3.36g of Na is produced.

B. Determination of the mass of Ba3(PO4)2 produced by reacting 10g of Ba.

From the balanced equation above,

411g of Ba reacted to produce 601g of Ba3(PO4)2.

Therefore, 10g of Ba will react to produce = (10 x 601)/411 = 14.62g of Ba3(PO4)2.

Therefore, 14.62g of Ba3(PO4)2 is produced.

7 0
3 years ago
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