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1.2 L sample of gas is determined to contain 0.5 moles of nitrogen. At the same temperature and pressure, how many moles of gas

hram777 [196]

A 20 L sample of the gas contains 8.3 mol N₂.

According to Avogadro’s Law, if p and T are constant

V₂/V₁ = n₂/n₁

n₂ = n₁ × V₂/V₁

___________

n₁ = 0.5 mol; V₁ = 1.2 L

n₂ = ?; V₂ = 20 L

∴n₂ = 0.5 mol × (20 L/1.2 L) = 8.3 mol

4 0

4 years ago

3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15

shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

$\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}$

(b) Mass of H

$\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}$

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

$\text{Moles of C = 1400 mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00$

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

$n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00 \approx 5$

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

8 0

4 years ago

3% Hydrogen Peroxide has an oral LD50 of 900 mg/kg. Acetic Acid has an oral LD50 of 3310 mg/kg. Which one is more hazardous to c

Lina20 [59]

Answer is: hydrogen peroxide is more hazardous to consume.
Median lethal dose, LD50 (lethal dose, 50%) is a measure of the lethal dose of a toxin. LD50 used as a general indicator of a substance's acute toxicity, lower LD50 is indicative of increased toxicity (in this case 900 mg/kg is lower than 3310 mg/kg).

4 0

4 years ago

Why do you require an acid catalyst to make an ester? Why not just mix acid and alcohol? Describe an alternate method of making

djverab [1.8K]

Answer:Acid catalyst is needed to increase the electrophilicity of Carbonyl group of Carboxylic acid as alcohol is a weak nucleophile.

Alternatively esters can be synthesised by converting carboxylic acid into acyl chloride using thionyl chloride(SOCl_{2} and then further treating acyl chloride with alcohol.

Carboxylic acid and esters can be easily distinguished on the basis of IR as carboxylic acid would contain a broad intense peak in 2500-3200cm_{-1} corresponding to OH stretching frequency whereas esters would not contain any such broad intense peak.

Alcohol and esters can also be distinguished using IR as alcohols would contain a broad intense peak at around 3200-3600cm_{-1}

Explanation: For the synthesis of esters using alcohol and carboxylic acid we need to add a little amount of acid in the reaction . The acid used here increases the electrophilicity of carbonyl carbon and hence makes it easier for a weaker nucleophile like alcohol to attack the carbonyl carbon of acid.

The oxygen of the carbonyl group is protonated using the acidic proton which leads to the generation of positive charge on the oxygen. The positive charge generated is delocalised over the whole acid molecule and hence the electrophilicity of carbonyl group is increased. Kindly refer attachment for the structures.

If we simply mix the acid and alcohol then no appreciable reaction would take place between them and ester formation would not take place because the carboxylic acid in that case is not a good electrophile whereas alcohol is also not a very strong nucleophile which can attack the carbonyl group.

Alternatively we can use thionyl chloride or any other reagent which can convert the carboxylic acid into acyl chloride. Acyl chloride is very elctrophilic and alcohol can very easily attack the acyl chloride and esters could be synthesized.

The carboxylic acid and ester can very easily be distinguished on the basis of broad intense OH stretching frequency peak at around 2500-3200cm_{-1} . The broad intense OH stretching frequency peak is present in carboxylic acids as they contain OH groups and absent in case of esters .

Likewise esters and alcohols can also be distinguished on the basis IR spectra as alcohols will have broad intense spectra at around 3200-3600cm_{-1}corresponding to OH stretching frequency whereas esters will not have any such peak. Rather esters would be having a Carbonyl stretching frequency at around 1720-1760

4 0

3 years ago

A mixture can be classified as _____ or ______

VikaD [51]

Mixtures can be classified as homogeneous or heterogeneous.

4 0

3 years ago

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