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3 years ago

7.5 moles of nitrogen gas (N2) is formed in the following reaction. How many grams

Chemistry

1 answer:

AfilCa [17]3 years ago

6 0

Answer:

Mass = 255 g

Explanation:

Given data:

Number of moles of nitrogen = 7.5 mol

Mass of ammonia formed = ?

Solution:

Chemical equation:

3H₂ + N₂ → 2NH₃

Now we will compare the moles of nitrogen and ammonia.

N₂ : NH₃

1 : 2

7.5 : 2/1×7.5 = 15

Mass of ammonia:

Mass = number of moles × molar mass

Mass = 15 mol × 17 g/mol

Mass = 255 g

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Answer:

32.07 g/mole.

Explanation:

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3 years ago

Write a balanced molecular reaction for the hydrogenation of ethyne to the saturated alkane.

vredina [299]

Explanation:

When ethyne undergoes a hydrogenation reaction with Ni catalyst and excess hydrogen,

C2H2 + 2H2 => C2H6.

Ethyne reacts with hydrogen gas to form ethane.

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3 years ago

A spirit burner used 1.00 g methanol to raise the temperature of 100.0 g water in a metal can from 28.00C to 58.0C. Calculate th

Andreas93 [3]

Complete question:

A spirit burner used 1.00 g methanol to raise the temperature of 100.0 g water in a metal can from 28.00C to 58.0C. Calculate the heat of combustion of methanol in kJ/mol.

Answer:

the heat of combustion of the methanol is 402.31 kJ/mol

Explanation:

Given;

mass of water, $m_w$ = 100 g

initial temperature of water, t₁ = 28 ⁰C

final temperature of water, t₂ = 58 ⁰C

specific heat capacity of water = 4.184 J/g⁰C

reacting mass of the methanol, m = 1.00 g

molecular mass of methanol = 32.04 g/mol

number of moles = 1 / 32.04

= 0.0312 mol

Apply the principle of conservation of energy;

$n\Delta H_{methanol} = Q_{water}\\\\n\Delta H_{methanol} = mc\Delta t\\\\n\Delta H_{methanol} = 100 \times 4.184\times (58-28)\\\\n\Delta H_{methanol} = 12,552 \ J\\\\n\Delta H_{methanol} = 12.552 \ kJ\\\\\Delta H_{methanol} = \frac{12.552}{n} \\\\H_{methanol} = \frac{12.552 \ kJ}{0.0312 \ mol} \\\\\Delta H_{methanol} = 402.31 \ kJ/mol$

Therefore, the heat of combustion of the methanol is 402.31 kJ/mol

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3 years ago

Afarmer applies 1705 kg of a fertilizer that contains 10.0% nitrogen to his fields each year. Fifteen percent (15.0%) of the fer

suter [353]

Answer:

The additional concentration of nitrogen in the river water due to the farmer's fertilizer is 0.0629 milligrams per liter

Explanation:

The mass of the fertilizer the farmer applies = 1705 kg

The percentage of nitrogen contained in the fertilizer = 10%

Therefore;

The mass of nitrogen contained in the fertilizer = 10% of 1705 kg = 10/100 × 1705 kg

The mass of nitrogen contained in the fertilizer = 10/100 × 1705 kg = 170.5 kg

The percentage of the fertilizer that washes into the river that runs through the farm = 15.0%

The mass of the fertilizer that washes into the river = 15/100 × 170.5 = 25.575 kg = 25575 g = 25575000 mg

The volume of water that flows through the farm = 0.455 cubic feet per second

The volume of water that flows through the farm in a year = 0.455 ft³ × 60 × 60 × 24 × 365 = 14348880 ft³ = 406,315.03 m³ = 406,315,034 l

The number of moles of nitrogen = Mass of nitrogen/(Molar mass of nitrogen)

Molar mass of nitrogen =14.0067 g/mol

Therefore;

The number of moles of nitrogen = 25575 kg/(14.0067 g/mol) = 1.825.912 moles

The additional concentration of the nitrogen in moles = 1.825.912 moles/(406,315,034 l) = 4.49 × 10⁻⁶ mol/liter

The additional concentration of the nitrogen in milligrams of nitrogen per liter of water = 25575000/406,315,034 = 0.0629 milligrams of nitrogen per liter.

Therefore, the additional concentration of nitrogen in the river water due to the farmer's fertilizer = 0.0629 milligrams per liter

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3 years ago

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lozanna [386]

Answer:

yes

Explanation:

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3 years ago