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Paraphin [41]
2 years ago
13

What is the mass of 2.5 moles of hydrogen fluoride gas HF

Chemistry
1 answer:
Nady [450]2 years ago
7 0

Answer:

You will get 5.0 g of hydrogen.

Explanation:

As with any stoichiometry problem, we start with the balanced equation.

Sn

l

+

2HF

→

SnF

2

+

H

2

Moles of H

2

=

2.5

mol Sn

×

1 mol H

2

1

mol Sn

=

2.5 mol H

2

Mass of H

2

=

2.5

mol H

2

×

2.016 g H

2

1

mol H

2

=

5.0 g H

2

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Determine the number of neutrons in an atom of Rg-272.
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In order to find the number of neutrons in the atom,

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Radioactive elements decay at a know rate known as half-life. Choose all of the correct statements concerning half-life. Carbon-
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Explanation:

Half life is simply the amount of time it takes for half of a substance to decompose.

Options;

- Carbon-14 has a half-life of 5,730 years. A 30 gram sample will be 10 grams after 5,730 years. This is incorrect. After 5730 years, 15g of the sample ought to remain.

- Nickel-59 has a half-life of 76,000 years. A sample would go through 3 half-lives in 228,000 years. This is correct. 3 * 76000 = 228,000

- Hafnium-182 has a half-life of 9 million years. A 38 gram sample would be 4.75 grams in 27 million years. This is incorrect. Mass after 3 half lives (27/9) = 9.5 (38 / 2 / 2)

- Iron-60 has a half-life of 1.5 million years. In 6 million years a 40 gram sample would be reduced to 10 grams. This is incorrect. Mass after 4 half lives (6 / 1.5) = 2.5 gram (40 / 2 / 2 /2 / 2)

- Lead-202 has a half-life of 52,500 years. The original sample must have been 120 grams if you have a 60 gram sample after 105,000 years. This is incorrect. Original sampe = 240 gram. So after 2 half lives (105,000/52500), mass left = 60 (240 / 2 /2)

6 0
2 years ago
Read 2 more answers
A 15.0 mL sample of 0.013 M HNO3 is titrated with 0.017 M CH$NH2 which he Kb=3.9 X 10-10. Determine the pH at these points: At t
kramer

<u>Answer:</u> The pH of the solution in the beginning is 1.89 and the pH of the solution after the addition of base is

<u>Explanation:</u>

  • <u>For 1:</u> At the beginning

To calculate the pH of the solution, we use the equation:

pH=-\log[H^+]

We are given:

Nitric acid is a monoprotic acid and it dissociates 1 mole of hydrogen ions. So, the concentration of hydrogen ions is 0.013 M

[H^+]=0.013M

Putting values in above equation, we get:

pH=-\log(0.013)\\\\pH=1.89

  • <u>For 2:</u>

To calculate the number of moles, we use the equation:  

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}

  • <u>For nitric acid:</u>

Molarity of nitric acid solution = 0.013 M

Volume of solution = 15 mL

Putting values in above equation, we get:

0.013M=\frac{\text{Moles of }HNO_3\times 1000}{15}\\\\\text{Moles of }HNO_3=1.95\times 10^{-4}mol

  • <u>For methylamine:</u>

Molarity of methylamine solution = 0.017 M

Volume of solution = 10 mL

Putting values in above equation, we get:

0.017M=\frac{\text{Moles of }CH_3NH_2\times 1000}{10}\\\\\text{Moles of }CH_3NH_2=1.7\times 10^{-4}mol

  • The chemical equation for the reaction of nitric acid and methylamine follows:

                       HNO_3+CH_3NH_2\rightarrow CH_3NH_3^++NO_3^-

As, the mole ratio of nitric acid and methyl amine is 1 : 1. So, the limiting reagent will be the reactant whose number of moles are less, which is methyl amine.

By Stoichiometry of the reaction:

1 mole of methyl amine produces 1 mole of CH_3NH_3^+

So, 1.7\times 10^{-4}mol of methyl amine will produce = \frac{1}{1}\times 1.7\times 10^{-4}=1.7\times 10^{-4}\text{ moles of }CH_3NH_3^+

To calculate the pK_b of base, we use the equation:

pK_b=-\log(K_b)

where,

K_b = base dissociation constant = 3.9\times 10^{-10}

Putting values in above equation, we get:

pK_b=-\log(3.9\time 10^{-10})\\\\pK_b=9.41

  • To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:

pOH=pK_b+\log(\frac{[salt]}{[base]})

pOH=pK_b+\log(\frac{[CH_3NH_3^+]}{[CH_3NH_2]})

We are given:

pK_b=9.41

[CH_3NH_3^+]=\frac{1.7\times 10^{-4}}{10+15}=6.8\times 10^{-6}M

[CH_3NH_2]=\frac{1.7\times 10^{-4}}{10+15}=6.8\times 10^{-6}M

Putting values in above equation, we get:

pOH=9.41+\log(\frac{6.8\times 10^{-6}}{6.8\times 10^{-6}})\\\\pOH=9.41

To calculate pH of the solution, we use the equation:

pH+pOH=14\\pH=14-9.41=4.59

Hence, the pH of the solution is 4.59

4 0
2 years ago
If 29.4 mL of ethanol is dissolved in water to make 359 mL of solution, what is the concentration expressed in volume/volume % o
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Answer: 8.2\%

Explanation:- Volume percentage is the ratio of volume of solute to the volume of solution defined in terms of percentage.

{\text {volume percentage}}=\frac{\text {volume of solute}}{\text {volume of solution}}\times 100\%

Given: volume of solute = 29.4 ml

Volume of solution= 359 ml

{\text {volume percentage}=\frac{29.4}{359}\times 100\%=8.2\%


6 0
2 years ago
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