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2 years ago

10

2 grams of hydrogen diffuses from containers in 10 minutes. How many grams of oxygen would diffuse through same container in sam

e time under similar condition.

1 answer:

Leya [2.2K]2 years ago

4 0

Answer:20

Explanation:make sure you thank me

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Please answer, with explanation. Thanks!

nadya68 [22]

Answer:

Explanation:

a = 40.1 g of Ca

Number of moles = mass / molar mass

Number of moles = 40.1 g/ 40.1 g/mol

Number of moles = 1 mol

b = 11.5 g Na

Number of moles = mass / molar mass

Number of moles = 11.5 g/ 23 g/mol

Number of moles = 0.5 mol

c = 5.87 g Ni

Number of moles = mass / molar mass

Number of moles = 5.87 g/ 58.7 g/mol

Number of moles = 0.1 mol

d = 150 g of S

Number of moles = mass / molar mass

Number of moles = 150 g/ 32 g/mol

Number of moles = 4.7 mol

e = 2.65 g Fe

Number of moles = mass / molar mass

Number of moles = 2.65 g/ 55.85 g/mol

Number of moles = 0.05 mol

f = 0.00750 g Ag

Number of moles = mass / molar mass

Number of moles = 0.00750 g/ 107.9 g/mol

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g = 2.25 × 10²⁵ atoms Zn

1 mole = 6.022 × 10²³ atoms

1 mol / 6.022 × 10²³ atoms × 2.25 × 10²⁵ atoms

0.17 × 2.25 × 10²⁵ moles

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6 0

3 years ago

If the fugacity of a pure component at the conditions of an ideal solution is 40 bar and its mole fraction is 0.4, what is its f

yaroslaw [1]

Answer : The fugacity in the solution is, 16 bar.

Explanation : Given,

Fugacity of a pure component = 40 bar

Mole fraction of component = 0.4

Lewis-Randall rule : It states that in an ideal solution, the fugacity of a component is directly proportional to the mole fraction of the component in the solution.

Now we have to calculate the fugacity in the solution.

Formula used :

$f_i=X_i\times f_i^o$

where,

$f_i$ = fugacity in the solution

$f_i^o$ = fugacity of a pure component

$X_1$ = mole fraction of component

Now put all the give values in the above formula, we get:

$f_i=0.4\times 40\text{ bar}$

$f_i=16\text{ bar}$

Therefore, the fugacity in the solution is, 16 bar.

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3 years ago

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