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Mariana [72]
3 years ago
15

Now if your lab partner determined that there were 1.552 x 1022 atoms of H in an unknown sample of C2H4Cl2, how many milligrams

would the unknown sample weigh? 
There was 0.56 moles of H from the first question. 
Chemistry
1 answer:
STALIN [3.7K]3 years ago
3 0
To determine the mass of the sample in milligrams in this problem, we use the avogadro's number to convert from atoms to moles, relate the moles of element in the sample to the mole present and the molar mass of the sample. We do as follows:

1.552 x 10^22 atoms H ( 1 mol H / 6.022x10^23 atoms H ) ( 1 mol C2H4Cl2 / 4 mol H ) ( 98.96 g C2H4Cl2 / 1 mol C2H4Cl2 ) = 0.625 g C2H4Cl2 = 625 mg <span>C2H4Cl2</span>
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3 years ago
What is altered by a catalyst?
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7 0
3 years ago
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When 74.8g of alanine C3H7NO2 are dissolved in 1450.g of a certain mystery liquid X, the freezing point of the solution is 8.30°
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Explanation:

Depression in freezing point is given by:

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\Delta T_f=T_f^0-T_f=8.30^0C = Depression in freezing point

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K_f = freezing point constant =  

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7 0
2 years ago
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2 years ago
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