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juin [17]
2 years ago
8

A sample of glucose contains 1.250x10^21 carbon atoms, how many atoms of hydrogen does it contain?

Chemistry
1 answer:
dybincka [34]2 years ago
5 0

Answer:

Hydrogen = 2.5 * 10^21

Explanation:

Chemical Formula Glucose: C₆H₁₂O₆

One of the ways you could do  this is to notice that for every carbon atom there are two Hydrogen atoms. You can state this more formally by using the formula to set up a ratio: 12/6 = hydrogen to Carbon

So if there are 1.250 * 10^21 Carbon atoms in the Glucose sample, then there will be twice as many hydrogen atoms.

H = 2 * 1.25 * 10^21 = 2.5 * 10^21 atoms

You could do this more formally by setting up a proportion.

6 Carbon / 12 Hydrogen = 1.25*10^21 /  x         Cross Multiply

6*x = 12 * 1.25*10^21                                           Combine the right

6x = 1.5 * 10^22                                                  Divide by 6

x = 2.5 * 10^21

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3 years ago
Part 1. determine the molar mass of a 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm. show your work.
d1i1m1o1n [39]
  • The molar mass of 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm is 9.15g/mol.
  • If this sample was placed under extreme pressure, the volume of the sample will decrease.

<h3>How to calculate molar mass?</h3>

The molar mass of a substance can be calculated by first calculating the number of moles using ideal gas law equation:

PV = nRT

Where;

  • P = pressure
  • V = volume
  • T = temperature
  • R = gas law constant
  • n = no of moles

0.98 × 1.2 = n × 0.0821 × 287

1.18 = 23.56n

n = 1.18/23.56

n = 0.05moles

mole = mass/molar mass

0.05 = 0.458/mm

molar mass = 0.458/0.05

molar mass = 9.15g/mol

  • Therefore, the molar mass of 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm is 9.15g/mol
  • If this sample was placed under extreme pressure, the volume of the sample will decrease.

Learn more about gas law at: brainly.com/question/12667831

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<span>
Sodium Oxide= Na2O
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% Na= (46/62) x 100 = 74%
% O= (16/62) x 100 = 26%</span><span>
</span>
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