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Maslowich
3 years ago
12

Manganese(II) oxide, lead(IV) oxide, and nitric acid react to produce permanganic acid, lead(II) nitrate, and water according to

the reaction above. How many moles of each product could be produced by the reaction of 8.90 moles of nitric acid with excess manganese(II) oxide and excess lead(IV) oxide
Chemistry
1 answer:
Leokris [45]3 years ago
7 0

Answer:

There will be produced:

2.97 moles HMnO4

4.45 moles Pb(NO3)2

2.97 moles H2O

Explanation:

Step 1: Data given

Manganese(II) oxide = MnO2

lead(IV) oxide = PbO2

nitric acid = HNO3

Moles of HNO3 = 8.90 moles

Step 2: The balanced equation

2MnO2 + 3PbO2 + 6HNO3 → 2HMnO4 + 3Pb(NO3)2 + 2H2O

Step 3: Calculate moles of reactants and products

For 2 moles MnO2 we need 3 moles PbO2 and 6 moles HNO3 to produce 2 moles HMnO4, 3 moles Pb(NO3)2 and 2 moles of water

For 8.90 moles of HNO3, there will react:

8.90 / 3 = 2.97 moles MnO2

8.90 / 2 = 4.45 moles PbO2

There will be produced:

8.90/3 = 2.97 moles HMnO4

8.90/2 = 4.45 moles Pb(NO3)2

8.90 / 3 = 2.97 moles H2O

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Aneli [31]

Answer:

0.0008 m

Explanation:

We are given that 0.080 cm

We have to convert 0.080 cm into meter

To find the value of 0.080 cm in meter we are using unitary method

We know that

100 cm =1 m

1 cm =\frac{1}{100} m

Therefore, 0.080 cm =\frac{1}{100}\times 0.080 m

0.080 cm =\frac{1}{100}\times \frac{80}{1000}m

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7 0
3 years ago
When the equation Sn + HNO₃ → SnO₂ + NO₂ + H₂O is balanced in acidic solution, what is the smallest whole-number coefficient for
balu736 [363]

Answer:

1.

Explanation:

Hello,

In this case, for the given reaction we first assign the oxidation state for each species:

Sn^0 + H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 + N^{4+}O^{2-}_2 + H^+_2O^-

Whereas the half reactions are:

Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow  N^{4+}O^{2-}_2+H_2O

Next, we exchange the transferred electrons:

1\times(Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-)\\\\4\times (H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow  N^{4+}O^{2-}_2+H_2O)\\\\\\Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\4H^++4H^+N^{5+}O^{-2}_3 +4e^-\rightarrow  4N^{4+}O^{2-}_2+4H_2O

Afterwards, we add them to obtain:

Sn^0+2H_2O+4H^++4H^+N^{5+}O^{-2}_3  \rightarrow Sn^{4+}O_2 +4H^++4N^{4+}O^{2-}_2+4H_2O

By adding and subtracting common terms we obtain:

Sn^0+4H^+N^{5+}O^{-2}_3  \rightarrow Sn^{4+}O_2 +4N^{4+}O^{2-}_2+2H_2O

Finally, by removing the oxidation states we have:

Sn + 4HNO_3 \rightarrow SnO_2 + 4NO_2 + 2H_2O

Therefore, the smallest whole-number coefficient for Sn is 1.

Regards.

6 0
3 years ago
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Hydrogen 3 has a half life of 12.32 years a sample of h-3 weighing 3.02 grams is left for 15.0 years what will the final weight
yawa3891 [41]

Answer:

The final mass of sample is 1.3 g.

Explanation:

Given data:

Half life of H-3 = 12.32 years

Amount left for 15.0 years = 3.02 g

Final amount = ?

Solution:

First all we will calculate the decay constant.

t₁/₂ = ln² /k

t₁/₂ =12.32 years

12.32 y =  ln² /k

k = ln²/12.32 y

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Now we will find the original amount:

ln (A°/A) = Kt

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ln (3.02 g/ A) = 0.8439

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A = 3.02 g/ 2.33

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8 0
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