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Schach [20]
3 years ago
13

A 3.06 gram sample of an unknown hydrocarbon with empirical formula CH2O was found to contain 0.0170 moles of the substance. Wha

t are the molecular mass and molecular formula, respectively, of the compound
Chemistry
1 answer:
Yanka [14]3 years ago
5 0

Answer:

180 amu

C₆H₁₂O₆

Explanation:

Step 1: Determine the molecular mass of the compound

The sample has a mass (m) of 3.06 g and it contains (n) 0.0170 moles. The molar mass M is:

M = m/n = 3.06/0.0170 mol = 180 g/mol

Then, the molecular mass is 180 amu.

Step 2: Determine the molar mass of the empirical formula.

M(CH₂O) = 1 × M(C) + 2 × M(H) + 1 × M(O)

M(CH₂O) = 1 × 12 g/mol + 2 × 1 g/mol + 1 × 16 g/mol = 30 g/mol

Step 3: Determine the molecular formula

First, we will determine "n" according to the following expression.

n = molar mass molecular formula / molar mass empirical formula

n = 180 g/mol / 30 g/mol = 6

The molecular formula is:

n × CH₂O = 6 × CH₂O = C₆H₁₂O₆

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hoa [83]

the reagents necessary to convert alcohol to ketoneNa_2Cr_2O_7 , H_2O which involves oxidation of alcohols.

<h3>What is oxidation of alcohols?</h3>
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To learn more about oxidation of alcohols with the given link

brainly.com/question/7207863

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<u>Question:</u>

Identify the reagents necessary to achieve each of the following transformations

a. O_3 , DMS

b.H_2SO_4 , H_2O , HgSO_4

c.Na_2Cr_2O_7 , H_2O

d.Fe ^ {2+}, NaOH

3 0
2 years ago
How many moles of nitrogen are present at STP if the volume is 846L
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Alternatively, you can go the long route and use the ideal gas law to solve for the number of moles of nitrogen given STP conditions (273 K and 1.00 atm). From PV = nRT, we can get n = PV/RT. Plugging in our values, and using 0.08206 L•atm/K•mol as our gas constant, R, we get n = (1.00)(846)/(0.08206)(273) = 37.8 moles, which confirms our answer.
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3 years ago
The 1995 Nobel Prize in Chemistry was shared by Paul Crutzen, F. Sherwood Rowland, and Mario Molina for their work concerning th
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Answer:

The enthalpy of reaction for the reaction of chlorine with ozone is -162.5 kJ.

Explanation:

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[2] - [1] = [3]

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The enthalpy of reaction for the reaction of chlorine with ozone is -162.5 kJ.

8 0
3 years ago
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Dafna11 [192]

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To know more about Effusion please click here : brainly.com/question/22359712

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