the mass if an element is .007502 x 10^-26 g. find the Number of significant figures when the mass is converted to mg. (both mas
ses values have same order of magnitude)
1 answer:
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Answer:
The number of hydrogen atoms is 4.96x10²⁴.
Explanation:
The number of atoms can be found with the following equation:
Where:
N: is the Avogadro's number = 6.022x10²³ atoms/mol
η: is the number of moles of hydrogen
n: is the number of hydrogen atoms
First, we need to find the number of hydrogen moles. The number of moles of CH₄ is:
Where:
m: is the mass of methane = 33 g
M: is the molar mass of methane = 16.04 g/mol
Now, since we have 4 hydrogen atoms in 1 mol of methane, the number of moles of hydrogen is:
Hence, the number of hydrogen atoms is:
Therefore, the number of hydrogen atoms is 4.96x10²⁴.
I hope it helps you!
<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
- <u>For NaBr:</u>
Given mass of NaBr = 11.1 g
Molar mass of NaBr = 103 g/mol
Putting values in equation 1, we get:
The chemical equation for the reaction of 1-butanol and NaBr is:
By Stoichiometry of the reaction
1 mole of NaBr produces 1 mole of 1-bromobutane
So, 0.108 moles of NaBr will produce = moles of 1-bromobutane
- Now, calculating the mass of 1-bromobutane from equation 1, we get:
Molar mass of 1-bromobutane = 137 g/mol
Moles of 1-bromobutane = 0.108 moles
Putting values in equation 1, we get:
- To calculate the percentage yield of 1-bromobutane, we use the equation:
Experimental yield of 1-bromobutane = 7.2 g
Theoretical yield of 1-bromobutane = 14.80 g
Putting values in above equation, we get:
Hence, the percent yield of the 1-bromobutane is 48.65 %