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Vitek1552 [10]
2 years ago
8

A large flask is evacuated and weighed, filled with argon gas, and then reweighed. When reweighed, the flask is found to have ga

ined 3.223 g. It is again evacuated and then filled with a gas of unknown molar mass. When reweighed, the flask is found to have gained 8.105 g.
Chemistry
1 answer:
Elden [556K]2 years ago
8 0

Answer:

100.52

Explanation:

from the ideal gas equation PV=nRT

for a given container filled with any ideal gas P and V remains constant.So T is also constant.R is as such a constant.

So n i.e no of moles will also be constant.

no of moles of Ar=3.224/40=0.0806

no of moles of unknown gas=0.0806

molecular wt of unknown gas=8.102/0.0806=100.52

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Why does iron not fit the pattern in column 7
son4ous [18]
Iron doesn't fit because it doesn't have enough atoms or protons in its nucleus so there for it belongs in column 2. <span />
3 0
2 years ago
Turbo the snail moves across the ground at a pace of 12 feet per day. If the garden is 48 feet away, how many days will it take
katen-ka-za [31]

Answer:4 days will it take for the snail to get to the garden.

Explanation:

Speed of the snail = 12 feet per day:

Distance between the garden and snail = 48 feet

Speed=\frac{Distance}{Time}

Time=\frac{48 feet}{12 \text{feet per day}}=4 days

4 days will it take for the snail to get to the garden.

8 0
2 years ago
A 0.271 g sample of an unknown vapor occupies 294 ml at 140.°c and 847 mmhg. the empirical formula of the compound is ch2. what
sdas [7]
When the molar mass M = mass (g)/ no.of moles (Mol)

∴ moles= 0.271 g / M

By using the gas equation:

PV = n RT

when P is the pressure = 847 mmHg / 760 = 1.11 atm

V is the volume = 0.294 L

n = 0.271 / M

R is constant = 0.0821 

T= 140+273 = 413 K

so by substitution:

when n = PV/RT

∴ 0.271/ M = 1.11 atm *0.294 L/ 0.0821 *413

∴ M = 28 


when the empirical formula of CH2 = 12+2 = 14 

∴ the exact no.of moles = 28/14 = 2

∴the molecular formula = 2(CH2) = C2H4

6 0
2 years ago
A chemist combined chloroform (CHCl3) and acetone (C3H6O) to create a solution where the mole fraction of chloroform is 0.187. T
ladessa [460]

Answer:

\large \boxed{\text{c = 2.50 mol/L; b = 3.96 mol/kg }}

Explanation:

1. Molar concentration

Let's call chloroform C and acetone A.

Molar concentration of C = Moles of C/Litres of solution

(a) Moles of C

Assume 0.187 mol of C.

That takes care of that.

(b) Litres of solution

Then we have 0.813 mol of A.

(i) Mass of each component

\text{Mass of C} = \text{0.187 mol C} \times \dfrac{\text{119.38 g C}}{\text{1 mol C}} = \text{22.32 g C}\\\\\text{Mass of A} = \text{0.813 mol A} \times \dfrac{\text{58.08 g A}}{\text{1 mol A}} = \text{47.22 g A}

(ii) Volume of each component

\text{Vol. of C} = \text{22.32 g C} \times \dfrac{\text{1 mL C}}{\text{1.48 g C}} = \text{15.08 mL C}\\\\\text{Vol. of A} = \text{47.22 g A} \times \dfrac{\text{1 mL A}}{\text{0.791 g A}} = \text{59.70 mL A}

(iii) Volume of solution

If there is no change of volume on mixing.

V = 15.08 mL + 59.70 mL = 74.78 mL

(c) Molar concentration of C

c = \dfrac{\text{0.187 mol}}{\text{0.07478 L}} = \textbf{2.50 mol/L }\\\\\text{ The molar concentration of chloroform is $\large \boxed{\textbf{2.50 mol/L}}$}

2. Molal concentration of C

Molal concentration = moles of solute/kilograms of solvent

Moles of C = 0.187 mol

Mass of A = 47.22 g = 0.047 22 kg

\text{b} = \dfrac{\text{0.187 mol}}{\text{0.047 22 kg}} = \textbf{3.96 mol/kg }\\\\\text{The molal concentration of chloroform is $\large \boxed{\textbf{3.96 mol/kg}}$}

4 0
2 years ago
An aqueous solution of licl is 34.0 % licl. what mass of water is present in 250.0 g of the solution? (density of water is 1.00
adell [148]
<span>If the aqueous solution is 34% Licl then it is 100 - 34% water = 66% From the calculation we've found out that it is 66% water. Then we need to find the weight from a 250 g solution. 66/100 * 250 = 165g Hence it is 165g</span>
4 0
2 years ago
Read 2 more answers
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