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3 years ago

5

The rate of effusion of a particular gas was measured and found to be 24.0 mL/min. Under the same conditions, the rate of effusi

on of pure methane (CH4) gas is 47.8 mL/min. What is the molar mass of the unknown gas?

1 answer:

Norma-Jean [14]3 years ago

3 0

Answer:

63.6g/mol

Explanation:

Use the equation of effusion:

47.8/24= $\sqrt{x/16.04}$

solve for x you get 63.6g/mol

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Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flas

hjlf

The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Suppose 1.20 mol $CS_2(g)$ of and 3.60 mol of $Cl_2(g)$ were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol of $CCl_4$ . Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Explanation:

Moles of $CS_2$ = 1.20 mole

Moles of $Cl_2$ = 3.60 mole

Volume of solution = 1.00 L

Initial concentration of $CS_2$ = $\frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M$

Initial concentration of $Cl_2$ = $\frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M$

The given balanced equilibrium reaction is,

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Initial conc. 1.20 M 3.60 M 0 0

At eqm. conc. (1.20-x) M (3.60-3x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}$

Now put all the given values in this expression, we get :

$K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}$

Given :Equilibrium concentration of $CCl_4$ , x = $\frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M$

$K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}$

$K_c=0.36$

Thus equilibrium constant at unknown temperature is 0.36

4 0

3 years ago

Which types of reactions would result in the formation of precipitate?

IRINA_888 [86]

Answer:

Double Displacement Reaction

Explanation:

A double displacement reaction is a type of chemical reaction in which the reactant ions exchange places to form new products. Usually, a double displacement reaction results in precipitate formation.

8 0

3 years ago

In a diagnostic test for leukemia, a person receives 4.1 mL of a solution containing selenium-75. If the activity of the seleniu

Tcecarenko [31]

To solve this problem, we can simply calculate for the dose by multiplying the volume of solution containing Selenium 75 and the activity of the Selenium 75. That is:

dose = 4.1 mL * (45 μCi/mL)

dose = 184.5 μCi

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3 years ago

Read 2 more answers

Methylpropane (complete and condensed structural formulas)

Tcecarenko [31]

Answer:

Here's what I get

Explanation:

1. Complete structural formula

Methylpropane consists of a chain of three carbons with another carbon atom attached to the middle carbon. Enough H atoms are added to give each C atom a total of four bonds.

The complete structural formula is shown below (There is a C atom at each intersection).

2. Condensed structural formula

A condensed structural formula is designed to be typed on one line.

The molecule has three CH₃ groups attached to a single carbon atom, so the condensed structural formula is

(CH₃)₃CH

The formula is also often written CH₃CH(CH₃)CH₃ and as (CH₃)₂CHCH₃.

5 0

3 years ago

A 254.5 g sample of a white solid is known to be a mixture of KNO3, BaCl2, and NaCl. When 116.5 g of this mixture is dissolved i

Margaret [11]

Answer:

Mass of KNO3 in the original mix is 146.954 g

Explanation:

mass of $KNO_3$ in original 254.5 mixture.

moles of $BaSO_4 = \frac{mass}{Molecular\ Weight}$

moles of $BaSO_4 = \frac{68.3}{233.38}$

= 0.2926 mol of BaSO4

Therefore,

0.2926 mol of BaCl2,

mass of $BaCl_2 = mol\times molecular weight$

$= 0.2926\times 208.23$

= 60.92 g

the AgCl moles $= \frac{mass}{Molecular\ Weight}$

$= \frac{199.1}{143.32}$

= 1.3891 mol of AgCl

note that, the Cl- derive from both, $BACl_2 and NaCl$

so

mole of Cl- f NaCl $= (1.3891) - (0.2926\times 2) = 0.8039$ mol of Cl-

mol of NaCl = 0.8039 moles

$mass = mol\times Molecular\ Weight = 0.8039 \times 58 = 46.626\ g \ of \ NaCl$

then

KNO3 mass = 254.5 - 60.92-46.626 = 146.954 g of KNO_3

Mass of KNO3 in the original mix is 146.954 g

8 0

3 years ago