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Vedmedyk [2.9K]
3 years ago
13

How many grams of CO2 are in 10 mol of the compound?

Chemistry
2 answers:
Rama09 [41]3 years ago
5 0
One mole of CO2 is the sum of all the compound's elements' molar masses
(C = 12.01g/mol; O = 16g/mol). That being said, one mole of CO2 is 12.01g + 16g + 16g which is equal to 44.01g. That being said, 10 mol of CO2 is 44.01g • 10 which is
440.1 grams
IrinaK [193]3 years ago
3 0
The answer is 60 grams.
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20.9%

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Which is an "enumerated" power of Congress?
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A scientist experiments ima unknown solution. She notices that when a base, sodium hydroxide, is added to the solution, it forms
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The unknown solution is acidic
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3 years ago
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Open the Balancing Chemical Equations interactive and select Introduction mode. Then choose Separate Water. Adjust the coefficie
aleksklad [387]
<h2>H_2O  + H_2 + O_2</h2>

Explanation:

1. Water decomposition

  • Decomposition reactions are represented by-

       The general equation: AB → A + B.

  • Various methods used in the decomposition of water are -
  1. Electrolysis
  2. Photoelectrochemical water splitting
  3. Thermal decomposition of water
  4. Photocatalytic water splitting
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  • The chemical equation will be -

        H_2O  + H_2 + O_2

Hence, balancing the equation we need to add a coefficient of 2 in front of H_2O on right-hand-side of the equation and  2 in front of H_2 on left-hand-side of the equation.

     ∴The balanced equation is -

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2. Formation of ammonia

  • The formation of ammonia is by reacting nitrogen gas and hydrogen gas.

      N_2 + H → NH_3

Hence, for balancing equation we need to add a coefficient of 3 in front of hydrogen and 2 in front of ammonia.

   ∴The balanced chemical equation for the formation of ammonia gas is as  follows -

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5 0
2 years ago
Boron has two naturally-ocurring isotopes. Boron-10 has an abundance of 19.8% and actual mass of 10.013 amu, and boron-11 has an
aniked [119]

Answer:

Average atomic mass = 10.812 amu

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

<u>For first isotope, Boron-10: </u>

% = 19.8 %

Mass = 10.013 amu

<u>For second isotope, Boron-11: </u>

% = 80.2 %

Mass = 11.009 amu

Thus,  

Average\ atomic\ mass=\frac{19.8}{100}\times {10.013}+\frac{80.2}{100}\times {11.009}=1.982574+8.829218=10.811792

<u>Average atomic mass = 10.812 amu</u>

6 0
2 years ago
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