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2 years ago

What’s the answer??????

Chemistry

1 answer:

atroni [7]2 years ago

7 0

Answer:

A evaperation

Explanation:

humidity would not change the water level condensation is the collection of water on objects and that would cause more not less water and precipitation is rain fall A would be the correct answer.

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What is the reduction half-reaction for the following unbalanced redox equation?

bazaltina [42]

First find the oxidation states of the various atoms:
in Cr2O2 2- Cr @ +1; In NH3 N @ +3; in CrO3 Cr @ +3, N2 N @ 0 
Note that N gained electrons, ie, was reduced; Cr was oxidized 
Now there is a problem, because B has NH4+ which the problem did not, and is not balanced, showing e- in/out 
B.NH4+ → N2 

Which of the following is an oxidation half-reaction? 
A.Sn 2+ →Sn 4+ + 2e- 
Sn lost electrons so it got oxidized

5 0

3 years ago

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1. A student collected the following data for a fixed volume of gas: Temperature (⁰C) Pressure (mm of Hg) 10 726 20 750 40 800 7

zimovet [89]

Answer is: the missing pressure is 1088.66 mmHg.

Gay-Lussac's Law states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.

p₁/T₁ = p₂/T₂.

p₁ = 960 mmHg; pressure of the gas.

T₁ = 100°C + 273.15.

T₁ = 373.15 K; temperature of the gas.

T₂ = 150°C + 273.15.

T₂ = 423.15 K.

p₂ = p₁T₂/T₁.

p₂ = 960 mmHg · 423.15 K / 373.15 K.

p₂ = 1088.66 mmHg.

6 0

3 years ago

What is the range for the following set of scores? Scores:6, 12, 9, 17, 11, 4, 14

MrRissso [65]

This is the answer is 17

5 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:

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

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3 years ago

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alisha [4.7K]

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3 years ago

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