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Dima020 [189]
3 years ago
7

What is the mass percentage of iodine (I2) in a solution containing 4.0×10−2 mol I2 in 130 g of CCl4? Express your answer using

two significant figures. % Request Answer Part B Seawater contains 8.2×10−3 g Sr2+ per kilogram of water. What is the concentration of Sr2+ measured in ppm? Express your answer using two significant figures. ppm p p m Request Answer
Chemistry
1 answer:
musickatia [10]3 years ago
4 0

Answer:

A. 7.2%

B. 8.6 ppm

Explanation:

<em>Part A</em>

First let's <em>calculate the mass of I₂</em>, using the known value of moles and the molecular weight (253.8 g/mol)

  • 4.0x10⁻² mol I₂ * 253.8 g/mol = 10.152 g I₂

Mass percentage is calculated using the mass of I₂ and the total mass (mass of I₂ + mass of CCl₄)

  • Total mass = 130 + 10.152 = 140.152 g
  • Mass percentage I₂ = 10.152 / 140.152 * 100 = 7.2%

<em>Part B</em>

The concentration of Sr⁺² in ppm is calculated using the formula

  • mg Sr⁺² / L water

We're given the mass of Sr⁺² in grams, so now we <u>convert it into mg</u>:

  • 8.2x10⁻³g * \frac{1000mg}{1g} = 8.2 mg

Now to convert kg of water into L, we use the density of seawater (1050 kg/m³):

<em>Converting density</em>: 1050 \frac{kg}{m^{3}} *  \frac{1m^{3}}{1000L} = 1.05 kg/L

  • Volume of one kilogram of water = 1kgWater ÷ 1.05 kg/L = 0.95 L

Finally we<u> calculate the concentration of Sr⁺²</u>:

  • 8.2 mg / 0.95 L = 8.6 ppm

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3 years ago
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A sample of 23.2 g of nitrogen gas is reacted with
slavikrds [6]

Answer:

1.66 moles.

Explanation:

We'll begin by calculating the number of mole in 23.2 g of nitrogen gas, N2.

This is illustrated below:

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 = 23.2 g

Mole of N2 =.?

Mole = mass /Molar mass

Mole of N2 = 23.2/28

Mole of N2 = 0.83 mole

Next, we shall determine the number of mole in 23.2 g of Hydrogen gas, H2.

This is illustrated below:

Molar mass of H2 = 2x1 = 2 g/mol

Mass of H2 = 23.2 g

Mole of H2 =?

Mole = mass /Molar mass

Mole of H2 = 23.2/2

Mole of H2 = 11.6 moles

Next, the balanced equation for the reaction. This is given below:

N2 + 3H2 —> 2NH3

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2 to produce 2 moles of NH3.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2.

Therefore, 0.83 moles will react with = (0.83 x 3) = 2.49 moles of H2.

From the calculations made above, we can see that only 2.49 moles out of 11.6 moles of H2 is required to react completely with 0.83 mole of N2.

Therefore, N2 is the limiting reactant.

Finally, we shall determine the maximum amount of NH3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of NH3 since all of it is consumed in the reaction.

The limiting reactant is N2 and the maximum amount of NH3 produced can be obtained as follow:

From the balanced equation above,

1 mole of N2 reacted to produce 2 moles of NH3.

Therefore, 0.83 mole of N2 will react to produce = (0.83 x 2) = 1.66 moles of NH3.

Therefore, the maximum amount of NH3 produced from the reaction is 1.66 moles.

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