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lukranit [14]
3 years ago
9

A compound is found to contain 46.68 % nitrogen and 53.32 % oxygen by mass. To answer the question, enter the elements in the or

der presented above. QUESTION 1: The empirical formula for this compound is . QUESTION 2: The molar mass for this compound is 30.01 g/mol. The molecular formula for this compound is
Chemistry
1 answer:
dangina [55]3 years ago
8 0

Answer:

Q1: NO

Q2: NO

Explanation:

Step 1: Given data

Percentage by mass of nitrogen: 46.68%

Percentage by mass of oxygen: 53.32%

Step 2: Divide each percentage by the mass of the element

N: 46.68/14.01 = 3.332

O: 53.32/16.00 = 3.332

Step 3: Divide both numbers by the smallest one (in this case is the same)

N: 3.332/3.332 = 1

O: 3.332/3.332 = 1

The empirical formula is NO. Its molar mass is 30.01 g/mol.

Step 4: Calculate "n"

n = molar mass of the molecular formula / molar mass of the empirical formula

n = (30.01 g/mol) / (30.01 g/mol) = 1

Step 5: Determine the molecular formula

We do so by multiplying the empirical formula by "n".

NO × 1 = NO

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The iupac name of the compound will be hex-3-yne or 3-hexyne. By marking number of carbon in the given compound, it was found that triple bond comes at third position and there are total 6 carbon in the compound. So it will have Hex as a prefix and as it contains triple bond so it will have yne as a suffix and as the triple bond is at third position, so it will be hex-3-yne or 3-hexyne.

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3 years ago
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1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L
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Explanation : Given,

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Vapor presume of 2‑Propanol (P^o_2) = 45.2 torr

Mole fraction of 1‑Propanol (x_1) = 0.540

Mole fraction of 2‑Propanol (x_2) = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

p_1=x_1\times p^o_1

where,

p_1 = partial vapor pressure of 1‑Propanol

p^o_1 = vapor pressure of pure substance 1‑Propanol

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and,

p_2=x_2\times p^o_2

where,

p_2 = partial vapor pressure of 2‑Propanol

p^o_2 = vapor pressure of pure substance 2‑Propanol

x_2 = mole fraction of 2‑Propanol

p_2=(0.46)\times (45.2torr)=20.8torr

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352

and,

\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

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Answer:

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