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MA_775_DIABLO [31]
2 years ago
6

According to the following balanced reaction, how many moles of HNO3 are formed from 8.44 moles of NO2 if there is plenty of wat

er present? 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)
Chemistry
1 answer:
kotegsom [21]2 years ago
3 0

Answer:

5.63 mol.

Explanation:

  • The balanced chemical equation between NO₂ and H₂O is:

<em>3NO₂(s) + H₂O(l) → 2HNO₃(aq) + NO(g), </em>

It is clear that 3 mol of NO₂ reacts with 1 mol of H₂O to produce 2 mol of HNO₃ and 1 mol of NO.

<em>Water is present as an excess reactant and NO₂ is limiting reactant.</em>

<em></em>

  • To find the no. of moles of HNO₃ produced:

3 mol of NO₂ produces → 2 mol of HNO₃, from stichiometry.

8.44 mol of NO₂ produces → ??? mol of HNO₃.

∴ The no. of moles of HNO₃ are formed = (8.44 mol)(2 mol)/(3 mol) = 5.63 mol.

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In November 1987, a massive iceberg broke loose from the antartic ice mass and floated free in the ocean. The chunk of ice was e
salantis [7]
<h2>Answer:</h2>

1.58  × 10∧16 pools.

<h3>Explanation:</h3>

Given:

Length of ice berg= 98 miles = 1557716 meters

Width of iceberg = 25 miles = 40233.6 meters

Thickness of iceberg = 750 ft = 230 meters

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The volume of the ice berg:

Volume = Length . width . thickness

Volume = 1557716 . 40233.6 . 230 = 1,441, 468, 016, 5248 m3 =  1,441, 468, 016, 5248 × 10 ∧3 L.

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1,441, 468, 016, 5248 × 10 ∧3 liters contains = 1/90850 . 1,441, 468, 016, 524.8 × 10 ∧3 .

= 1.100 .  1,441, 468, 016, 5248 × 10 ∧3 L.

= 1.58  × 10∧16 pools.

Hence 1.6 × 10∧16 pools will be filled with that chunk of ice.


4 0
1 year ago
Read 2 more answers
The standard reduction potentials for two half-cells involving iron are: given below. Fe2+ (aq) + 2e– ® Fe (s) Eο = –0.44 V Fe3+
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Answer:

Explanation:

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Fe⁺³ (aq) + e⁻ =  ® Fe²⁺ (aq) ;   E⁰ = + .77 V

Reduction potential of second reaction is more , so it will take place , ie Fe⁺³ will be reduced and Fe will be oxidised .

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In this case, we apply the Gay-Lussac's law which allows us to understand the pressure-temperature behavior as a directly proportional relationship:

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Thus, we solve for the final pressure P2 to obtain it as shown below:

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