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Zielflug [23.3K]
3 years ago
9

How much nano3 is needed to prepare 225 ml of a 1.55 m solution of nano3?

Chemistry
1 answer:
andrezito [222]3 years ago
5 0

Answer:

= 29.64 g  NaNO3

Explanation:

Molarity is given by the formula;

Molarity = Moles/Volume in liters

Therefore;

Number of moles = Molarity × Volume in liters

                             = 1.55 M × 0.225 L

                             = 0.34875 moles NaNO3

Thus; 0.34875 moles of NaNO3 is needed equivalent to;

   = 0.34875 moles × 84.99 g/mol

   = 29.64 g

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A sample of helium has a volume of 50.00 L at STP. How many helium atoms are in the sample? Show work
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Explanation:

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given volume=50.0 L

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Fritz Haber, a German chemist, discovered a way to synthesize ammonia gas (NH3) by combining hydrogen and nitrogen gases accordi
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1) Write the balanced equation to state the molar ratios:

<span>3H2(g) + N2(g) → 2NH3(g)

=> molar ratios = 3 mol H2 : 1 mol N2 : 2 mol NH3

What volume of nitrogen is needed to produce 250.0 L of ammonia gas at STP?

First, convert the 250.0 L of NH3 to number of moles at STP .

Use the fact that 1 mole of gas at STP occupies 22.4 L

=> 250.0 L * 1mol/22.4 L = 11.16 L

Second, use the molar ratio to find the number of moles of N2 that produces 11.16 L of NH3

=> 11.16 L NH3 * [1 mol N2 / 2 mol NH3] = 5.58 mol N2

Third, convert 5.58 mol N2 into liters at STP

=> 5.58 mol N2 * [22.4 L/mol] = 124.99 liters

Answer: 124,99 liters

What volume of hydrogen is needed to produce 2.50 mol NH3 at STP?
 

First, find the number of moles of H2 that produce 2.50 mol by using the molar ratios:

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Second, convert the number of moles to liters of gas at STP:

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Answer: 84 liters

 </span>



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3 years ago
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