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miss Akunina [59]
3 years ago
7

G. Amount of charge required to reduce

Chemistry
2 answers:
kirill [66]3 years ago
7 0
The answer would be c 4f
Nady [450]3 years ago
4 0

Answer:

\boxed{\text{c) 4 F}}

Explanation:

1. Write the skeleton equation for the half-reaction

NO₃⁻ ⟶ N₂O

2. Balance all atoms other than H and O

2NO₃⁻ ⟶ N₂O

3. Balance O by adding H₂O molecules to the deficient side.

2NO₃⁻ ⟶ N₂O + 5H₂O

4. Balance H by adding H⁺ ions to the deficient side.

2NO₃⁻ + 10H⁺ ⟶ N₂O + 5H₂O

5. Balance charge by adding electrons to the deficient side.

2NO₃⁻ + 10H⁺ + 8e⁻ ⟶ N₂O + 5H₂O

The amount of charge required to reduce 2 mol of NO₃⁻ is 8 F

\text{The amount of charge required to reduce 1 mol of NO$_{3}^{-}$ is \boxed{\textbf{4 F}}}

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___ replacement involves one element replacing another element in a compound.
Solnce55 [7]

Answer:

single

Explanation:

4 0
3 years ago
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For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi
natka813 [3]

<u>Answer:</u> The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       ....(1)

  • <u>For Iron:</u>

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:  

\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol

For the given chemical reaction:

2Fe(s)+O_2(g)\rightarrow 2FeO(s)

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = \frac{2}{2}\times 0.0771=0.0771mol of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:  

0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g

To calculate the percentage yield of iron (ii) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0
3 years ago
Please help me with this homework
ella [17]

Answer:

Density is the correct answer choice.

5 0
3 years ago
Read 2 more answers
6.00 moles of sodium perchlorate contains the same number of atoms as __________. 6.00 moles of sodium perchlorate contains the
Ksju [112]

Answer:

6.00 moles of sodium permanganate

Explanation:

Let us attempt to count the number of atoms present in sodium perchlorate. The formula of the compound is NaClO4

Sodium atoms - 1

Chlorine atoms - 1

Oxygen atoms -4

Total number of atoms = 6

Among the options, only KMnO4 also has six atoms. One atom of potassium, one atom of manganese and four atoms of oxygen.

3 0
3 years ago
Imagine that you have a 5.00 L gas tank and a 3.50 L gas tank. You need to fill one tank with oxygen and the other with acetylen
Lorico [155]

Answer:

77.14 atm of pressure should be of an acetylene in the tank.

Explanation:

2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O

According to reaction, 2 moles of acetylene reacts with 5 moles of oxygen.

Moles of oxygen=n_1

Moles of acetylene =n_2

\frac{n_1}{n_2}=\frac{5 mol}{2 mol}=\frac{5}{2}

Volume of large tank with oxygen gas, V_1 = 5.00 L

Pressure of the oxygen gas inside the tank = P_1=135 atm

RT=\frac{P_1V_1}{n_1} ..[1]

Volume of small tank with acetylene gas ,V_2= 3.50 L

Pressure of the acetylene gas inside the tank = P_2=?

RT=\frac{P_2V_1}{n_2} ..[2]

Considering both the gases having same temperature T, [1]=[2]

\frac{P_1V_1}{n_1}=\frac{P_2V_2}{n_2}

P_2=\frac{P_1V_1\times n_2}{V_2\times n_1}

=\frac{135 atm\times 5.00 L\times 2}{3.50 L\times 5}=77.14 atm

77.14 atm of pressure should be of an acetylene in the tank.

8 0
3 years ago
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