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maria [59]
3 years ago
6

How many liters of ammonia are required to change 34.9 L of nitrogen monoxide to nitrogen gas? Assume 100% yield and that all ga

ses are measured at the same temperature and pressure.
Chemistry
1 answer:
frutty [35]3 years ago
4 0

Answer:

23.27 L

Explanation:

The balanced reaction of ammonia and nitrogen monoxide is shown below as:

4NH₃(g) + 6NO(g) ⇒ 5N₂(g) + 6H₂O(l)

Given that:

Volume of nitrogen monoxide = 34.9 L

Since, temperature and pressure are same, volume coefficients would be same as equation coefficients.

So,

6 L of nitrogen monoxide reacts with 4 L of ammonia.

1 L of nitrogen monoxide reacts with 4/6 L of ammonia.

34.9 L of nitrogen monoxide reacts with (4/6)*34.9 L of ammonia.

Amount of ammonia required = 23.27 L

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Photosynthesis in a plant is an example of which characteristic of living things?
lutik1710 [3]

Answer:

<h2><em><u>4. Ability to obtain and use energy</u></em></h2>
4 0
3 years ago
A 45 mL sample of nitrogen gas is cooled from 135ºC to 15C in a container that can contract or expand at constant pressure. Wha
Vanyuwa [196]

Answer:

V₂ =31.8 mL

Explanation:

Given data:

Initial  volume of gas = 45 mL

Initial temperature = 135°C (135+273 =408 K)

Final temperature = 15°C (15+273 =288 K)

Final volume of gas = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 45 mL × 288 K / 408 k

V₂ = 12960 mL.K / 408 K

V₂ =31.8 mL

8 0
3 years ago
An atom of hydrogen and an atom of carbon are
LenKa [72]

Answer: -

Organic compounds consisting of only carbon atoms and hydrogen atoms are known as hydrocarbons.

Hydrocarbons are used by us for mainly as a source of energy. Hydrocarbons make up gasoline which is used as fuel in automobiles.

Hydrocarbons depending on the type of unsaturation can be of three types - alkanes, alkenes and alkynes.

5 0
4 years ago
The natural abundances of elements in the human body, expressed as percent by mass, are oxygen (o), 65 percent; carbon (c), 18 p
Vlada [557]

It is given that the person weighs 62 kg = 62,000 g

Natural abundances in mass percent are:

O = 65%

C = 18%

H = 10%

N = 3.0%

Ca = 1.6%

P = 1.2%

Corresponding weights of the elements are:

O = 65/100 * 62000 g = 40.30 * 10^3 g

C = 18/100 * 62000 g = 11.16 * 10^3 g

H = 10/100 * 62000 g = 62.00 * 10^2 g

N = 3.0/100 * 62000 g = 18.60 * 10^2 g

Ca = 1.6/100 * 62000 g = 9.92 * 10^2 g

P = 1.2/100 * 62000 g = 7.44 * 10^2 g


3 0
3 years ago
Read 2 more answers
A sample of solid sodium hydroxide, weighing 13.20 grams is dissolved in deionized water to make a solution. What volume in mL o
Andreas93 [3]
<h3>Answer:</h3>

2.809 L of H₂SO₄

<h3>Explanation:</h3>

Concept tested: Moles and Molarity

In this case we are give;

Mass of solid sodium hydroxide as 13.20 g

Molarity of H₂SO₄ as 0.235 M

We are required to determine the volume of H₂SO₄ required

<h3>First: We need to write the balanced equation for the reaction.</h3>
  • The reaction between NaOH and H₂SO₄ is a neutralization reaction.
  • The balanced equation for the reaction is;

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

<h3>Second: We calculate the umber of moles of NaOH used </h3>
  • Number of moles = Mass ÷ Molar mass
  • Molar mass of NaOH is 40.0 g/mol
  • Therefore;

Moles of NaOH = 13.20 g ÷ 40.0 g/mol

                          = 0.33 moles

<h3>Third: Determine the number of moles of the acid, H₂SO₄</h3>
  • From the equation, 2 moles of NaOH reacts with 1 mole of H₂SO₄
  • Therefore, the mole ratio of NaOH: H₂SO₄ is 2 : 1.
  • Thus, Moles of H₂SO₄ = moles of NaOH × 2

                                    = 0.33 moles × 2

                                   = 0.66 moles of H₂SO₄

<h3>Fourth: Determine the Volume of the acid, H₂SO₄ used</h3>
  • When given the molarity of an acid and the number of moles we can calculate the volume of the acid.
  • That is; Volume = Number of moles ÷ Molarity

In this case;

Volume of the acid = 0.66 moles ÷ 0.235 M

                                = 2.809 L

Therefore, the volume of the acid required to neutralize the base,NaOH is 2.809 L.

7 0
4 years ago
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