Enter the molecular formula for butane, C4H10

What stress would shift the equilibrium position of the following system to the right?

yanalaym [24]

Answer:

Decreasing the concentration of N2O3

Explanation:

This is because the products on the right of the reaction occupy more space. One (1) mole of NO and another mole of NO2 will occupy more space than the one (1) mole of N2O3. Therefore decreasing the concentration of N2O3 will shift the reaction to the right because the products will have more space to occupy – hence favoring equilibrium.

4 0

3 years ago

Read 2 more answers

What's the balanced equation for Iron and copper (II) sulfate solution.

Georgia [21]

Answer:

2Fe + 3CuSO₄ → Fe₂(SO₄)₃ + 3Cu

Explanation:

The reactants are:

Iron = Fe

Copper (II) sulfate solution = CuSO₄

The reaction is given as:

Fe + CuSO₄ → Fe₂(SO₄)₃ + Cu

This is a single displacement reaction;

Now let us balance the equation;

aFe + bCuSO₄ → cFe₂(SO₄)₃ + dCu

Conserving Fe;

a = 2c

Conserving Cu;

b = d

Conserving S;

b = 3c

Conserving O;

4b = 12c

let b = 1;

c = $\frac{1}{3}$

d = 1

a = $\frac{2}{3}$

Multiply through by 3 to get whole numbers;

a = 2, b = 3, c = 1 and d = 3

2Fe + 3CuSO₄ → Fe₂(SO₄)₃ + 3Cu

5 0

3 years ago

What kind of reaction is this ?

AlexFokin [52]

This reaction is double replacement because two ionic compounds are exchanged, making two new ionic compounds.

6 0

3 years ago

The oxidation of ammonia produces nitrogen and water via the following reaction: 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l) Suppose the

Sonbull [250]

Answer:

The rate of consumption of $NH_{3}$ is 2.0 mol/L.s

Explanation:

Applying law of mass action to this reaction-

$-\frac{1}{4}\frac{\Delta [NH_{3}]}{\Delta t}=-\frac{1}{3}\frac{\Delta [O_{2}]}{\Delta t}=\frac{1}{2}\frac{\Delta [N_{2}]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_{2}O]}{\Delta t}$

where $-\frac{\Delta [NH_{3}]}{\Delta t}$ represents rate of consumption of $NH_{3}$ , $-\frac{\Delta [O_{2}]}{\Delta t}$ represents rate of consumption of $O_{2}$ , $\frac{\Delta [N_{2}]}{\Delta t}$ represents rate of formation of $N_{2}$ and $\frac{\Delta [H_{2}O]}{\Delta t}$ represents rate of formation of $H_{2}O$ .