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Olin [163]
3 years ago
13

Balance the following redox reaction occurring in an acidic solution. The coefficient of Mn2+(aq) is given. Enter the coefficien

ts (integers) into the cells before each substance below the equation. ___ AsO2−(aq) + 3 Mn2+(aq) + ___ H2O(l) \rightarrow→ ___ As(s) + ___ MnO4−(aq) + ___ H+(aq)
Chemistry
1 answer:
Vadim26 [7]3 years ago
7 0

Answer:

_5_ AsO2−(aq) + 3 Mn2+(aq) + _2_ H2O(l) → _5_ As(s) + _3_ MnO4−(aq) + _4_ H+(aq)

Explanation:

Step 1:

The unbalanced equation:

AsO2−(aq) + 3 Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)

Step 2:

Balancing the equation.

AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)

The above equation can be balanced as follow:

There are 3 atoms of Mn on the left side of the equation and 1 atom on the right side. It can be balance by putting 3 in front of MnO4− as shown below:

AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)

There are 12 atoms of O on the right side and a total of 3 atoms on the left side. It can be balance by putting 5 in front of AsO2− and 2 in front of H2O as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)

There are 4 atoms of H on the left side and 1 atom on the right side. It can be balance by putting 4 in front of H+ as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + 4H+(aq)

There are 5 atoms of As on the left side and 1 atom on the right side. It can be balance by putting 5 in front of As as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → 5As(s) + 3MnO4−(aq) + 4H+(aq)

Now the equation is balanced

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The following reaction occurs in aqueous solution: NH4 + (aq) + NO2 - → N2 (g) + 2H2O (l) The data below is obtained at 25°C.
GREYUIT [131]

Answer: The order with respect to NH_4^+ is 1.

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

NH_4^++NO_2^-\rightarrow N_2+2H_2O

Rate=k[NH_4^+]^x[NO_2^-]^y

k= rate constant

x = order with respect to NH_4^+

y = order with respect to ANO_2^-

n = x+y = Total order

From trial 1: 3.2\times 10^{-3}=k[0.0100]^x[0.200]^y    (1)

From trial 2: 6.4\times 10^{-3}=k[0.0200]^x[0.200]^y    (2)

Dividing 2 by 1 :\frac{6.4\times 10^{-3}}{3.2\times 10^{-3}}=\frac{k[0.0100]^x[0.2000]^y}{k[0.0200]^x[0.200]^y}

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8 0
3 years ago
In this experiment we will be using a 0.05 M solution of HCl to determine the concentration of hydroxide (OH-) in a saturated so
gulaghasi [49]

<u>Answer:</u> The moles of hydroxide ions present in the sample is 0.0008 moles

<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 HCl.

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is Ca(OH)_2

We are given:

n_1=1\\M_1=0.05M\\V_1=16mL\\n_2=2\\M_2=?M\\V_2=36.0mL

Putting values in above equation, we get:

1\times 0.05\times 16=2\times M_2\times 36\\\\M_2=\frac{1\times 0.05\times 16}{2\times 36}=0.011M

To calculate the moles of hydroxide ions, we use the equation used to calculate the molarity of solution:

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

Molarity of solution = 0.011 M

Volume of solution = 36.0 mL

Putting values in above equation, we get:

0.011=\frac{\text{Moles of }Ca(OH)_2\times 1000}{36}\\\\\text{Moles of }Ca(OH)_2=\frac{0.011\times 36}{1000}=0.0004mol

1 mole of calcium hydroxide produces 1 mole of calcium ions and 2 moles of hydroxide ions.

Moles of hydroxide ions = (0.0004 × 2) = 0.0008 moles

Hence, the moles of hydroxide ions present in the sample is 0.0008 moles

8 0
3 years ago
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