HELP An unknown gas effuses at a rate of 2.0 times the rate of CL2. What is the molar mass of the unknown gas?

Explanation:Explanation:

We can solve this equation using Graham's law of effusion:

r

1

r

2

=

√

M

M

2

M

M

1

We can substitute gas

X

and

oxygen

in for 1 and 2, respectively, to get

r

X

r

O

2

=

√

M

M

O

2

M

M

X

Since

O

2

effuses at a rate twice that of gas

X

, the ratio of the rate of effusion of

X

to

O

2

is

0.5

:

0.5

=

√

M

M

O

2

M

M

X

Since the molar mass of

O

2

is given as

32.00

g/mol

, we then have

0.5

=

√

32

l

g/mol

M

M

X

Using algebra to solve for

M

M

X

:

0.5

=

√

32

l

g/mol

√

M

M

X

√

M

M

X

=

√

32

l

g/mol

0.5

M

M

X

=

(

√

32

l

g/mol

0.5

)

2

=

128

g/mol

The molar mass of the unknown gas is thus

128

g/mol

A gas close to this value is

iodine

, with molar mass

126.90

g/mol

.

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.