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Schach [20]
2 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]2 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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5 0
2 years ago
The solubility of o2 at 20c is 1.38 x10^-3. the partial presure of o2 in the air at sea level is 0.27 atm. using henery;s law, c
netineya [11]

<u>Answer:</u> The solubility of oxygen at 682 torr is 4.58\times 10^{-3}M

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{A}=K_H\times p_{A}

Or,

\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}

where,

C_1\text{ and }p_1 are the initial concentration and partial pressure of oxygen gas

C_2\text{ and }p_2 are the final concentration and partial pressure of oxygen gas

We are given:

Conversion factor used:  1 atm = 760 torr

C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm

Putting values in above equation, we get:

\frac{1.38\times 10^{-3}}{C_2}=\frac{0.27atm}{0.897atm}\\\\C_2=\frac{1.38\times 10^{-3}\times 0.897atm}{0.27atm}=4.58\times 10^{-3}M

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4 0
2 years ago
Which of the following elements matches the criteria below?
Tems11 [23]

Answer:

The answer to your question is below

Explanation:

1. Found in period 2. All the elements in the list are found in period 2.

a. F   This option is correct

b. Be  Beryllium is located in period two.

c. O  also oxygen is found in period 2.

d. C Carbon is found in period 2.

2.- Can gain lose 4 electrons to become its nearest stable noble gas. Only Carbon.

a. F    This option is wrong, F becomes stable when it gains 1 electron.

b. Be  Beryllium becomes stable when it loses 2 electrons.

c. O  Become stable when it gains 2 electrons.

<u>d. C </u><u>Become stable when it gains or loses 4 electrons.</u>

4 0
3 years ago
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