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svet-max [94.6K]
3 years ago
12

What is the mass in grams of 9.76 x 10 12 atoms of naturally occurring sodium?

Chemistry
1 answer:
Fiesta28 [93]3 years ago
4 0
The number of moles in a substance indicates the amount of the substance that contains the same number of particles as 12 g of the Carbon-12 isotope [or equivalent to 6.02 × 10²³] (which is used as a standard in the world of moles).

Now,

          if     6.02 × 10²³ atoms are found in 1 mole ofsodium
   then let  9.76 × 10¹² atoms are found in     x

⇒     x  =    (9.76 × 10¹² )  ÷  (6.02 × 10²³)

            =  1.619 × 10⁻¹¹ mol


Now, mass = moles × molar mass

∴ mass of Na = 1.619 × 10⁻¹¹ mol ×  23 g/mol
                      = 3.72 × 10⁻¹⁰ g


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3 years ago
Now that we have put a coefficient of 3 in front of caso4, what coefficient should go in front of cacl2 to balance calcium (ca)?
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3 years ago
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The atmosphere is about 21% oxygen. At a pressure of 749 torr, what is the partial pressure of O2?
Kitty [74]

Answer:

157.29 Torr

Explanation:

This question may be solved by the mole fraction.

Mole fraction → Moles of gas / Total moles = Partial pressure of gas / Total pressure

If the atmosphere is about 21% you assume that the mole fraction is 0.21 so:

0.21 = Partial pressure of gas / Total pressure

0.21 = Partial pressure of gas / 749 Torr

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3 years ago
Determine the mass of CO2 formed if 15 grams of C4H10
Svetlanka [38]

Answer:

m_{CO_2}=45gCO_2

Explanation:

Hello there!

In this case, since the combustion of butane is:

C_4H_{10}+\frac{13}{2} O_2\rightarrow 4CO_2+5H_2O

Thus, since the molar mass of butane is 58.12 g/mol and that of carbon dioxide is 44.01 g/mol, we obtain the following mass of CO2 product:

m_{CO_2}=15gC_4H_{10}*\frac{1molC_4H_{10}}{58.12gC_4H_{10}} *\frac{4molCO_2}{1molC_4H_{10}} *\frac{44.01gCO_2}{1molCO_2} \\\\m_{CO_2}=45gCO_2

Best regards!

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A). CH3OH I think !!!!!!!!
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3 years ago
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