Zinc metal is added to hydrochloric acid to generate hydrogen gas and is collected over a liquid whose vapor pressure is the sam

e as pure water at 20.0 degrees C (18 torr). The volume of the mixture is 1.7 L and its total pressure is 0.810 atm. Determine the number of moles of hydrogen gas present in the sample.
A. 0.272 mol
B. 0.04 mol
C. 0.997 mol
D. 0.139 mol
E. 0.0681 mol

Chemistry

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Explanation:

The rate at which gases effuse is inversely related to the square root of their molar masses.

In this case, half of the helium (1.5 L) passed through the membrane in 24 hours. Therefore, we can calculate the effusion rate of He gas as follows.

$\frac{1.5 L}{24 hr}$

= 0.0625 L/hr

Given that the molar mass of He is 4 g/mol and for $O_{2}$ it is 32 g/mol.

Now,

$\frac{\text{Rate of He}}{\text{Rate of Oxygen}} = \sqrt{\frac{32}{4}$

= 2.83

Thus, the effusion rate of $O_{2} = \frac{\text{Rate of He}}{2.83}$

= $\frac{0.0625 L/hr}{2.83}$

Rate of $O_{2}$ = 0.022 L/hr.

This implies that 0.022 L of $O_{2}$ gas will effuse in one hour.

Consequently, to find the duration needed for 1.5 L of $O_{2}$ gas to effuse, we calculate as follows.

$\frac{1.5 L}{0.022 L/hr}$

= 68.18 hours

Thus, we can conclude that it will require 68.18 hours for half of the oxygen to effuse through the membrane.

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A 0.784 g sample of magnesium is added to a 250 ml flask and dissolved in 150 ml of water. magnesium hydroxide obtained from the

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Although multiple values are given, our focus is on HCl.

We have 215 mL (0.215 L) of 0.300 M HCl fully consumed in the reaction. It's important to recall that the number of moles is found by multiplying volume by molarity:

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If a reaction occurs, what will be the products of the unbalanced reaction below?Cu(s) + Ni(NO3)2(aq)

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Does the result of the calculation in question 3 justify your original assumption that all of the SCN^- is in the form of FeNCS^

lorasvet [2795]

Fe 3+ + SCN- --> FeSCN 2+

.......Fe 3+.......SCN-.........FeSCN 2+ 
I.......0.04..........0.001.............. 
C........-x...............-x............. 
E.....0.04-x.....0.001-x...........x 

Keq = 203.4 = x / (0.04-x)(0.001-x) 
203.4 = x / (x^2 - 0.041x + 4x10^-5) 
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203.4x^2 - 9.34x + 0.00094 = 0 
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