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larisa86 [58]
3 years ago
12

Suppose 20.8 g of sodium iodide is dissolved in 250. mL of a 0.70 M aqueous solution of silver nitrate. Calculate the final mola

rity of ALL IONS in the solution. You can assume the volume of the solution doesn't change when the sodium iodide is dissolved in it. Round your answer to 3 significant digits.
Chemistry
1 answer:
gtnhenbr [62]3 years ago
7 0

Answer:

[Ag+] = [NO3-] = 0.700M

0.555M = [Na+] = [I-]

Explanation:

To solve this question we need to find the moles of sodium iodide, NaI, using its molar mass -. With the moles and the volume we can find the molarity of Na+ and I-. The molarity of the ions of silver nitrate, AgNO3 doesn't change because we are assuming the volume doesn't change:

<em>Molarity Ag⁺ = Molarity NO₃⁻ = 0.700M</em>

<em>Moles NaI -Molar mass: 149.89g/mol-</em>

20.8g NaI * (1mol/149.89g) = 0.0139 moles NaI

<em>Molarity:</em>

0.0139 moles NaI / 0.250L = <em>0.555M = [Na+] = [I-]</em>

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Answer:

The mass percentage composition of nitrogen in the sample of the cleaning detergent is approximately 16.78%  

Explanation:

The given mass of the sample of the cleaning detergent, m₁ = 20.5 g

The mass of the ammonium hydroxide, NH₄OH in the detergent, m₂ = 8.61 g

The molar mass of NH₄OH = 35.04 g/mol

The molar mass of nitrogen, N = 14.01 g/mol

Therefore, the mass, m₃ of nitrogen, N, in 8.61 g of ammonium hydroxide, NH₄OH, is found as follows;

m₃ = (14.01/35.04) × 8.61 g = (402,087/118,800) g ≈ 3.44 g

The mass of nitrogen, N, in the ammonium hydroxide, NH₄OH, contained in the 20.5 g sample of the cleaning agent, m₃ ≈ 3.44 grams

The percentage composition of nitrogen in the sample of the cleaning detergent, %N is given as follows;

\% Composition = \dfrac{Mass \ of \ component}{Total \ mass \ of \ cleaning \ detergent} \times 100

Therefore;

%N ≈ ((3.44 g)/(20.5 g)) × 100 ≈ 16.78 %

The percentage composition of nitrogen, %N ≈ 16.78%.

6 0
3 years ago
The molar mass of V2O3 is<br> A. 66.94<br> B. 133.88<br> C. 149.88<br> D. 200.82
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prohojiy [21]

Answer:

0.924 g

Explanation:

The following data were obtained from the question:

Volume of CO2 at RTP = 0.50 dm³

Mass of CO2 =?

Next, we shall determine the number of mole of CO2 that occupied 0.50 dm³ at RTP (room temperature and pressure). This can be obtained as follow:

1 mole of gas = 24 dm³ at RTP

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1 mole of CO2 occupies 24 dm³ at RTP.

Therefore, Xmol of CO2 will occupy 0.50 dm³ at RTP i.e

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Thus, 0.021 mole of CO2 occupied 0.5 dm³ at RTP.

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Answer:

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