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4 years ago

In terms of disorder and energy, systems in nature have a tendency to undergo changes toward

Chemistry

1 answer:

Simora [160]4 years ago

8 0

Answer:

Option C is correct

Explanation:

Entropy is defined as the randomness in a system. In nature, a system tends to have high entropy. The higher the entropy , the spontaneous the reactions would be. As a system loses energy, it becomes more ordered and less random. Hence, as the reaction proceeds in a system, the system has higher energy/entropy and less order.

Hence, option c is correct

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List 3 ways you interact with geomagnetism in your life.

nata0808 [166]

Answer:

anwer is c

Explanation:

7 0

3 years ago

Complete the dissociation reaction and the corresponding Ka equilibrium expression for each of the following acids in water. (Ty

anastassius [24]

Answer :

(A) The dissociation reaction of $HC_2H_3O_2$ will be:

$HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)$

The equilibrium expression :

$K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}$

(B) The dissociation reaction of $Co(H_2O)_6^{3+}$ will be:

$Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)$

The equilibrium expression :

$K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}$

(C) The dissociation reaction of $CH_3NH_3^+$ will be:

$CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)$

The equilibrium expression :

$K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}$

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

(A) The dissociation reaction of $HC_2H_3O_2$ will be:

$HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)$

The equilibrium expression of $HC_2H_3O_2$ will be:

$K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}$

(B) The dissociation reaction of $Co(H_2O)_6^{3+}$ will be:

$Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)$

The equilibrium expression of $Co(H_2O)_6^{3+}$ will be:

$K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}$

(C) The dissociation reaction of $CH_3NH_3^+$ will be:

$CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)$

The equilibrium expression of $CH_3NH_3^+$ will be:

$K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}$

3 0

3 years ago

Why is potassium less reactive than silver

Reika [66]

Because potassium is a gas that can evolve

3 0

3 years ago

The molecular weight of KMnO4 is 158 g/mol. If you wanted to make 500 mL of a 0.1M stock solution of KMnO4, how much solid KMnO4

labwork [276]

Solid KMnO₄ needed = 7.9 g

<h3>Further explanation</h3>

Given

MW KMnO₄ = 158 g/mol

500 mL(0.5 L) of a 0.1M stock solution of KMnO₄

Required

solid KMnO₄

Solution

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

$\large{\boxed {\bold {M ~ = ~ \frac {n} {V}}}$

Input the value :

n = M x V

n = 0.1 M x 0.5 L

n = 0.05 mol

Mass KMnO₄ :

= mol x MW

= 0.05 x 158 g/mol

= 7.9 g

5 0

3 years ago

A student was given the task of titrating a 20.mL sample of 0.10MHCl(aq) with 0.10MNaOH(aq) . The HCl(aq) was placed in an Erlen

gregori [183]

Answer:

Here's what I get

Step-by-step Explanation

(a) Effect of dilution

There will be no effect on the volume of NaOH needed.

The amount of HCl will be halved, so the amount of NaOH will be halved.

However, the concentration of NaOH is also halved, so you will need twice the volume.

You will be back to the same volume as before dilution.

(b) Net ionic equation

Molecular: HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)

Ionic: H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + OH⁻(aq) ⟶ Na⁺(aq) + Cl⁻(aq) + H₂O(l)

Net ionic: H⁺(aq) + OH⁻(aq) ⟶ H₂O(l)

(c) Proton acceptor

H⁺ is the proton. OH⁻ accepts the proton and forms water.

(d) Moles of HCl

$\text{Moles of HCl} = \text{20. mL HCl} \times \dfrac{\text{0.10 mmol HCl}}{\text{1 mL HCl}} = \text{2.0 mmol HCl} = \textbf{0.0020 mol HCl}$

(e) Equivalence point

The equivalence point is the point at which the titration curve intersects the pH 7 line.

(f) Schematic representation

Assume the box for 0.10 mol·L⁻¹ HCl contains four black dots (H⁺) and four open circles (Cl⁻).

The 0.20 mol·L⁻¹ solution is twice as concentrated.

It will contain eight black dots and eight open circles.

5 0

3 years ago