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EastWind [94]
3 years ago
7

How many moles of electrons are required to reduce one mole of bromine gas (Br2) to two moles of bromide ions (Br-)? 1 e- 2 e- 3

e- 4 e-
Chemistry
1 answer:
rosijanka [135]3 years ago
8 0
The correct answer would be the second option. It would would need 2 moles of electrons to reduce one mole of bromine gas into two moles of the bromide ions. This is a reduction reaction. It would be written as:

Br2 = 2Br- + 2e-
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You have a 2.0 mL sample of acetic acid (molar mass 60.05 g/mol) of unknown concentration. You titrate it to its endpoint with 2
Katyanochek1 [597]

<u>Answer:</u> The mass of acetic acid used is 0.12 grams

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is CH_3COOH

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=?M\\V_1=2.0mL\\n_2=1\\M_2=0.1M\\V_2=20.0mL

Putting values in above equation, we get:

1\times M_1\times 2.0=1\times 0.1\times 20.0\\\\M_1=\frac{1\times 0.1\times 20.0}{1\times 2.0}=1M

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Given mass of acetic acid = ? g

Molar mass of acetic acid = 60.05 g/mol

Molarity of solution = 1 M

Volume of the solution = 2.0 mL

Putting values in above equation, we get:

1mol/L=\frac{\text{Mass of acetic acid}\times 1000}{60.05g/mol\times 2.0}\\\\\text{Mass of acetic acid}=\frac{1\times 60.05\times 2}{1000}=0.12g

Hence, the mass of acetic acid used is 0.12 grams

5 0
4 years ago
How can the prevailing winds be different during an El Nino year?
ad-work [718]

Answer: During an El Nino year, weakening winds along the equator lead to warming water surface temperatures that lead to further weakening of the winds

3 0
3 years ago
Read 2 more answers
N2 + 3H2, -&gt; 2NH3<br> If I have 50.6 L of N2 and excess H2, how many liters of NH3 can I produce?
Anastaziya [24]

Answer:

C. 101.2 L

Explanation:

N2 + H2= NH3

Balancing it,

N2 + 3 H2 = 2.NH3

(1 mol) (3 mol) (2 mol)

which means

1 molecule of nitrogen reacts with 3 molecule of hydrogen to form ammonia.

Likewise,

50.6 l of nitrogen reacts with 50.6 × 3= 151.8 l of hydrogrn to form 50.6 × 2= 101.2 l of ammonia.

3 0
3 years ago
An ellipse is composed of which two of the following?
Oduvanchick [21]
C i think is the ansower hope it helped
4 0
3 years ago
A mixture initially contains A, B, and C in the following concentrations: [A] = 0.650 M, [B] = 1.35 M, and [C] = 0.300 M. The fo
dimaraw [331]

<u>Answer:</u> The value of K_c for given reaction is 0.465

<u>Explanation:</u>

We are given:

Initial concentration of A = 0.650 M

Initial concentration of B = 1.35 M

Initial concentration of C = 0.300 M

Equilibrium concentration of A = 0.550 M

Equilibrium concentration of B = 0.400 M

For the given chemical equation:

                           A+2B\rightarrow C

<u>Initial:</u>                0.65     1.35     0.30

<u>At eqllm:</u>        0.65-x   1.35-2x   0.30+x

Evaluating the value of 'x'

0.650-x=0.550\\\\x=0.100

So, equilibrium concentration of B = 1.35 - 2x = [1.35 - 2(0.100)] = 1.15 M

Equilibrium concentration of C = (0.30 + x) = (0.300 + 0.100) = 0.400 M

The expression of K_c for above equation follows:

K_c=\frac{[C]}{[A][B]^2}

Putting values in above equation, we get:

K_c=\frac{0.400}{0.650\times (1.15)^2}\\\\K_c=0.465

Hence, the value of K_c for given reaction is 0.465

7 0
3 years ago
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