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Anastasy [175]
3 years ago
15

The volume of a gas at 2.0 atm is 3.0 L. What is the volume of the gas at 1.5 atm at the same temperature?

Chemistry
1 answer:
Amanda [17]3 years ago
6 0

Answer:4.0L

Explanation:

You might be interested in
How many molecules of N204 are in 85.0 g of N2O4?
alukav5142 [94]

Answer:

5.56 × 10^23 molecules

Explanation:

The number of molecules in a molecule can be calculated by multiplying the number of moles in that molecule by Avagadro's number (6.02 × 10^23)

Using mole = mass/molar mass

Molar mass of N2O4 = 14(2) + 16(4)

= 28 + 64

= 92g/mol

mole = 85.0/92

= 0.9239

= 0.924mol

number of molecules of N2O4 (nA) = 0.924 × 6.02 × 10^23

= 5.56 × 10^23 molecules

4 0
3 years ago
A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.
makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4'  to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

              %     =>         grams                 =>                 moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H =>  7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio,  divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0
3 years ago
A sample of gas contains 0.1500 mol of HCl(g) and 7.500×10-2 mol of Br2(g) and occupies a volume of 9.63 L. The following reacti
Furkat [3]

Answer:

9.63 L.

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2HCl(g) + Br_2(g)\rightarrow 2HBr(g) + Cl_2(g)

So the consumed amounts of hydrochloric acid and bromine are the same to the beginning based on:

n_{Br_2}^{consumed}=0.1500molHCl*\frac{1molBr_2}{2molHCl}=0.075molBr_2

In such a way, the yielded moles of hydrobromic acid and chlorine are:

n_{HBr}=0.1500molHCl*\frac{2molHBr}{2molHCl}=0.1500molHBr \\n_{Cl_2}=0.1500molHCl*\frac{1molCl_2}{2molHCl}=0.075molCl_2

Thus, the volume of the sample, after the reaction is the same as no change in the total moles is evidenced, that is 9.63L.

Best regards.

7 0
3 years ago
What<br>was the initial volume of the hydrogen in cm3?​
svetlana [45]

Answer:

255.51cm3

Explanation:

Data obtained from the question include:

V1 (initial volume) =?

T1 (initial temperature) = 50°C = 50 + 273 = 323K

T2 (final temperature) = - 5°C = - 5 + 237 = 268K

V2 (final volume) = 212cm3

Using the Charles' law equation V1/T1 = V2/T2, the initial volume of the gas can be obtained as follow:

V1/T1 = V2/T2

V1/323 = 212/268

Cross multiply to express in linear form

V1 x 268 = 323 x 212

Divide both side by 268

V1 = (323 x 212)/268

V1 = 255.51cm3

Therefore, the initial volume of the gas is 255.51cm3

5 0
3 years ago
Read 2 more answers
Consider a closed containing a solid in equilibrium with its vapor. The volume of the solid is much less than that of the contai
Furkat [3]

Answer:

Explanation:

check the attachment below

5 0
3 years ago
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