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Akimi4 [234]
3 years ago
5

How many grams are 5.82 x 10^24 molecules of H2S

Chemistry
1 answer:
kotykmax [81]3 years ago
0 0

As we calculate the the no. of grams of any molecules then we should know about every method or every conversion method how we concert gram into moles and moles into gram so in this conversion method we know that 1 mole of any atoms = 6.022*1023(avogadro's#)

So 5.82*1023 / 6.022*1023 = 9.66 gram of H2S

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victus00 [196]

Answer:

200 mL

Explanation:

You simply add the volumes :) If this was a case that involved titration, you would use the formula M1V1 = M2V2. I do not know if that is what you were referring to but based on the information you gave, you simply add the two volumes.

4 0
3 years ago
Silver (Ag) has two stable isotopes: 107Ag, 106.90 amu, and 109Ag, 108.90 amu. If the average atomic mass of silver is 107.87 am
inysia [295]

Answer:

  • The abundance of 107Ag is 51.5%.
  • The abundance of 109Ag is 48.5%.

Explanation:

The <em>average atomic mass</em> of silver can be expressed as:

107.87 = 106.90 * A1 + 108.90 * A2

Where A1 is the abundance of 107Ag and A2 of 109Ag.

Assuming those two isotopes are the only one stables, we can use the equation:

A1 + A2 = 1.0

So now we have a system of two equations with two unknowns, and what's left is algebra.

First we<u> use the second equation to express A1 in terms of A2</u>:

A1 = 1.0 - A2

We <u>replace A1 in the first equation</u>:

107.87 = 106.90 * A1 + 108.90 * A2

107.87 = 106.90 * (1.0-A2) + 108.90 * A2

107.87 = 106.90 - 106.90*A2 + 108.90*A2

107.87 = 106.90 + 2*A2

2*A2 = 0.97

A2 = 0.485

So the abundance of 109Ag is (0.485*100%) 48.5%.

We <u>use the value of A2 to calculate A1 in the second equation</u>:

A1 + A2 = 1.0

A1 + 0.485 = 1.0

A1 = 0.515

So the abundance of 107Ag is 51.5%.

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bit.^{}ly/3a8Nt8n

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nekit [7.7K]

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