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3 years ago

8

Which expression represents the percent by mass of nitrogen in NH4NO3?

2 answers:

guajiro [1.7K]3 years ago

8 0

Answer:

This question is incomplete. The completed question is below

Which expression represents the percent by mass of nitrogen in NH4NO3?

A) 14 g N/80 g NH₄ NO₃ ₓ 100%

B) 28 g N/80 g NH₄ NO₃ ₓ 100%

C) 80 g NH₄ NO₃ /14 g N ₓ 100%

D) 80 g NH₄ NO₃ /28 g N ₓ 100%

The correct option is B

Explanation:

To calculate the percentage by mass of nitrogen in the compound, the formula will be

total mass of nitrogen/total mass of compound × 100

Atomic mass of nitrogen (N) is 14 g/mol

There are 2 nitrogen atoms present in the compound as highlighted below

<u><em>Total mass of nitrogen</em></u> (N) present in NH₄ NO₃ = 14 × 2 = 28g

Atomic mass of hydrogen (H) is 1 g/mol. There are 4 hydrogen atoms present in the compound, NH₄NO₃.

Atomic mass of oxygen (O) is 16 g/mol. There are 3 oxygen atoms present in the compound, NH₄NO₃

<u><em>Total mass of the compound</em></u> =

(14 × 2) + (1 ×4) + (16 × 3) = 80g

Hence, the percentage by mass of nitrogen in the compound (as shown in the formula earlier) is

28g of N/80g of NH₄NO₃ × 100%

podryga [215]3 years ago

7 0

W(N)={2M(N)/M(NH₄NO₃)}*100%

w(N)={2*14.0 g/mol/80.0 g/mol}*100% = 35%

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A precipitate forms when mixing solutions of sodium fluoride (NaF) and lead II nitrate (Pb(NO3)2). Complete and balance the net

Margarita [4]

Answer:

Pb^2+(aq) + 2F-(aq) → PbF2(s)

Explanation:

Step 1: Data given

sodium fluoride = NaF

lead(II)nitrate Pb(NO3)2

Step 2: The unbalanced equation

NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + NaNO3(aq)

Step 3: Balancing the equation

NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + NaNO3(aq)

On the left side we have 2x NO3 (in Pb(NO3)2), on the right side we have 1x NO3 (in NaNO3). To balance the amount of NO3 we hvae to multiply NaNO3 on the right side by 2.

NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + 2NaNO3(aq)

On the left side we have 1x Na (in NaF), on the right side we have 2x Na (in 2NaNO3). To balance the amount of Na we have to multiply NaF on the left side by 2. Now the equation is balanced.

2NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + 2NaNO3(aq)

Step 4: Calculate net ionic equation

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

2Na+(aq) + 2F-(aq) + Pb^2+(aq) + 2NO3-(aq) → PbF2(s) + 2Na+(aq) + NO3-(aq)

Pb^2+(aq) + 2F-(aq) → PbF2(s)

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3 years ago

Which of the statements is correct?

Step2247 [10]

Answer:

<u>A. Magnet has a north pole and south pole.</u>

Explanation:

- A magnet has a north and south pole

- Magnets attract paper clips due to the metals present

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Only the first statement is correct, the answer is (A).

8 0

3 years ago

Read 2 more answers

Calculate the mass of aluminum that would have the same number of atoms as 6.35 g of cadmium

Sergeu [11.5K]

Answer:

1.62 g of Al contain the same number of atoms as 6.35 g of cadmium have.

Explanation:

Given data:

mass of cadmium = 6.35 g

Number of atoms of aluminum as 6.35 g cadmium contain = ?

Solution:

Number of moles of cadmium = 6.35 g/ 112.4 g/mol

Number of moles of cadmium = 0.06 mol

Number of atoms of cadmium:

1 mole = 6.022×10²³ atoms of cadmium

0.06 mol × 6.022×10²³ atoms of cadmium/ 1mol

0.36×10²³ atoms of cadmium

Number of atoms of Al:

Number of atoms of Al = 0.36×10²³ atoms

1 mole = 6.022×10²³ atoms

0.36×10²³ atoms × 1 mol /6.022×10²³ atoms

0.06 moles

Mass of aluminum:

Number of moles = mass/molar mass

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m = 0.06 mol ×27 g/mol

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3 years ago

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nata0808 [166]

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concentration of iodine, C(I) = 1.6 mol/1 = 1.6

<h3>Create an ICE table</h3><h3>What is ICE table?</h3>

An ICE table is a tabular system of keeping track of changing concentrations in an equilibrium reaction.

ICE is an abbreviation that stands for initial, change, equilibrium.

Create ICE table for the reactants and products formed;

Br2(g) + I2(g) ↔ 2IBr(g)

I 0.6 1.6 0

C -1.19 -1.19 1.19

E 0.6 - 1.19 1.6 - 1.19 1.19

E = -0.59 0.41 1.19

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The equilibrium constant is calculated as follows;

KC = [IBr]²/[Br][I]

KC = (1.19²) / (0.59 x 0.41)

KC = 5.85

Thus, the equilibrium constant for this reaction at 350°C is determined as 5.85.

Learn more about equilibrium constant here: brainly.com/question/19340344

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2 years ago