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ANEK [815]
2 years ago
7

What is the mass-action expression Qc for the following chemical reaction:Zn(s)+2Ag+(aq) ⇌ Zn+2(aq)+2Ag(s) a. [Zn-2]/[Ag-]b. [

Zn-2][Ag(s)]^2/[Zn(s)][Ag-]^2c. [Ag-]^2/[Zn2-]d. [Zn(s)][Ag-]^2/[Zn2-][Ag(s)]^2e. [Zn2-]/[Ag-]^2
Chemistry
1 answer:
sattari [20]2 years ago
7 0

Answer : The expression for reaction quotient will be :

Q_c=\frac{[Zn^{2+}]}{[Ag^{+}]^2}

Explanation :

Reaction quotient (Qc) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given redox reaction is :

2Ag^{+}(aq)+Zn(s)\rightarrow 2Ag(s)+Zn^{2+}(aq)

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

The expression for reaction quotient will be :

Q_c=\frac{[Zn^{2+}]}{[Ag^{+}]^2}

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5 0
3 years ago
Which of the following is true about a system at equilibrium? a The concentration(s) of the reactant(s) is equal to the concentr
anyanavicka [17]

Answer:

c The concentration(s) of reactant(s) is constant over time.  

Step-by-step explanation:

When the reaction A ⇌ B reaches equilibrium, the concentrations of reactants and products are constant over time.

a is <em>wrong</em>, because the concentrations of reactants and products are usually quite different.

b is <em>wrong</em>, because both product and reactant molecules are being formed at equilibrium.

d is <em>wrong</em>. The rates of the forward and reverse reactions are equal, but they are not zero.

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3 years ago
How many moles is 3.85 x 1025 atoms of gold?
stich3 [128]

Answer:

3.45 x 10^24 x 1 / 6.022 x 10^23 = 573 

4 0
2 years ago
An element has an average atomic mass of 1.008 amu. It consists of two isotopes , one having a mass of 1.007 amu, and one having
dimulka [17.4K]

Answer:

The most abundant isotope is 1.007 amu.

Explanation:

Given data:

Average atomic mass = 1.008 amu

Mass of first isotope = 1.007 amu

Mass of 2nd isotope = 2.014 amu

Most abundant isotope = ?

Solution:

First of all we will set the fraction for both isotopes

X for the isotopes having mass  2.014 amu

1-x for isotopes having mass 1.007 amu

The average atomic mass is 1.008 amu

we will use the following equation,

2.014x + 1.007  (1-x) = 1.008  

2.014x + 1.007  - 1.007 x = 1.008  

2.014x - 1.007x  =  1.008  -  1.007

1.007 x = 0.001

x= 0.001/ 1.007

x= 0.0009

0.0009 × 100 = 0.09 %

0.09 % is abundance of isotope having mass  2.014 amu because we solve the fraction x.

now we will calculate the abundance of second isotope.

(1-x)

1-0.0009 = 0.9991

0.9991 × 100= 99.91%

6 0
3 years ago
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Zielflug [23.3K]
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8 0
3 years ago
Read 2 more answers
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