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Home-RamPort x M (no subject)-jmartinez 1 ? / D 2018-04-01 18:14 Page 1 × ?? Mr. Kevin A. Boudreaux × e chegg Study Guided S? C | (D file:///C:/users/Jordan%20Martinez/Downloads/2018-04-01%2018-14%20page%2019620(1).pdf 2018-04-01 18:14 page 1 1. A sample of oxygen gas is confined in a 5.00 L container at 964 torr and 25.0 C. How many (10 pts) grams of gas are present in the sample? (in mL) of a 1.00 gram sample of CCla gas (MM-i 53.82 g/mol) (10 pts) 2. What is the volume having a pressure of 842 torr and a temperature of 100.°C? Partial Pressures 3. A mixture of 50.0 g O and 150.0 g N2 is placed in a 1.00 L container at 27°C. Calculate the partial pressure of each gas, and the total pressure. (10 pts) 02 2 2018-04-01 18-14...pdf 2018-04-01 18-14 pdf ^ Show all | × ID 641 PM ^q )??4/1/2018

Explanation / Answer

1)

Given:

P = 964.0 torr

= (964.0/760) atm

= 1.2684 atm

V = 5.0 L

T = 25.0 oC

= (25.0+273) K

= 298 K

find number of moles using:

P * V = n*R*T

1.2684 atm * 5 L = n * 0.08206 atm.L/mol.K * 298 K

n = 0.2593 mol

Molar mass of O2 = 32 g/mol

use:

mass of O2,

m = number of mol * molar mass

= 0.2593 mol * 32 g/mol

= 8.30 g

Answer: 8.30 g

2)

Molar mass of CCl4,

MM = 1*MM(C) + 4*MM(Cl)

= 1*12.01 + 4*35.45

= 153.81 g/mol

mass(CCl4)= 1.00 g

use:

number of mol of CCl4,

n = mass of CCl4/molar mass of CCl4

=(1.0 g)/(153.81 g/mol)

= 6.502*10^-3 mol

Given:

P = 842.0 torr

= (842.0/760) atm

= 1.1079 atm

n = 0.0065 mol

T = 100.0 oC

= (100.0+273) K

= 373 K

use:

P * V = n*R*T

1.1079 atm * V = 0.0065 mol* 0.08206 atm.L/mol.K * 373 K

V = 0.180 L

Answer: 0.180 L

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