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3 years ago

17. A balloon contains 4.50 X 1022 atoms of helium gas. Calculate the mass of helium in grams.

Chemistry

2 answers:

Rashid [163]3 years ago

4 0

Answer:

1.08 grams

Explanation:

first, we need to find the number of moles

we divide 12.1/22.4=0.54 moles

we multiply the number of moles with the molecular mass 0.54x2=1.08g

castortr0y [4]3 years ago

3 0

I agree with the person on top have a wonderful day :)

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The half life of radium-226 is 1600 years. If you have 200 grams of radium today how many grams would be present in 8000 years?

Sveta_85 [38]

Answer:

Half life is the time taken by a radio active isotope to reduce by half of its original amount. Radium-226 has a half life of 1602 years meaning that it would take 1602 years for a mass of radium to reduce by half.

Number of half lives in 9612 years = 9612/1602 = 6 half lives

New mass = Original mass x (1/2)n where n is the number of half lives.

Therefore, New mass= 500 x (1/2)∧6

= 500 x 0.015625

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Explanation:

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Calculate the number of milliliters of 0.710 M Ba(OH)2 required to precipitate all of the Mn2+ ions in 161 mL of 0.796 M KMnO4 s

USPshnik [31]

<u>Answer:</u> The volume of barium hydroxide is 183 mL.

<u>Explanation:</u>

To calculate the moles of cadmium nitrate, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of $MnSO_4$ = 0.796 M

Volume of $MnSO_4$ = 161 mL = 0.161 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.796mol/L=\frac{\text{Moles of }MnSO_4}{0.161L}\\\\\text{Moles of }MnSO_4=0.13mol$

The chemical equation for the reaction of manganese sulfate and barium hydroxide follows:

$MnSO_4(aq.)+Ba(OH)_2(aq.)\rightarrow Mn(OH)_2(s)+BaSO_4(aq.)$

By Stoichiometry of the reaction:

1 mole of manganese sulfate reacts with 1 mole of barium hydroxide.

So, 0.13 moles of manganese sulfate will react with = $\frac{1}{1}\times 0.13=0.13mol$ of barium hydroxide

Now, calculating the volume of barium hydroxide by using equation 1, we get:

Moles of barium hydroxide = 0.13 moles

Molarity of barium hydroxide = 0.710 M

Putting values in equation 1, we get:

$0.710mol/L=\frac{0.13mol}{\text{Volume of barium hydroxide}}\\\\\text{Volume of barium hydroxide}=0.183L$

Converting this into milliliters, we use the conversion factor:

1 L = 1000 mL

So, $0.183L=0.183\times 1000=183mL$

Hence, the volume of barium hydroxide is 183 mL.

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