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3 years ago

What volume (in liters) does 3.91 moles of nitrogen gas at 5.35 atm and 323 K occupy

Chemistry

1 answer:

ivanzaharov [21]3 years ago

4 0

Data Given:
Moles = n = 3.91 mol

Pressure = P = 5.35 atm

Temperature = T = 323 K

Volume = V = ?

Formula used: Ideal Gas Equation is used,

P V = n R T
Solving for V,
V = n R T / P
Putting Values,
V = (3.91 mol × 0.0825 atm.L.mol⁻¹.K⁻¹ × 323 K) ÷ 5.35 atm

V = 19.36 L

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Distinguish between sugar and non-sugar with examples.

kap26 [50]

Answer:

Reducing sugars are sugars where the anomeric carbon has an OH group attached that can reduce other compounds. Non-reducing sugars do not have an OH group attached to the anomeric carbon so they cannot reduce other compounds. ... Maltose and lactose are reducing sugars, while sucrose is a non-reducing sugar

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3 years ago

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa

notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

$2NaOH+H_2SO_4-->Na_2SO_4+2H_2O$

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

$n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\$

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

$0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4$

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

$0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4$

Finally, the percent yield turns out into:

$Y=\frac{1.92g}{7.1g} *100$

$Y=27.0$ %

Best regards.

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3 years ago

What is the most common salt found in ocean water?

zhannawk [14.2K]

Hello.

The answer i believe would be sodium.

Have a great day.

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3 years ago

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