Need help with all 3 questions please! Home- RamPort M (no subject)- jmartine:1

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Need help with all 3 questions please!

Home- RamPort M (no subject)- jmartine:1 x 2018-04-01 18:14 page 2 x C Chegg Study | Guided Sc C | (D file:///C/Users/Jordan%20Martinez/Downloads/2018-04-01%2018-14%20page%202%20(1).pdf 2018-04-01 18:14 page 2 Gas Law How many liters of oxygen gas are produced at 125°C and 778 torr potassium chlorate (MM 122.55 g/mol) are decomposed? when 4.52 grams of (10 pts) 4. Gas Density S. Calculate the density of methane, CH4, at STP (5 pts) The density of a noble gas is 16.7 gL at 2.00 atm and 50.0°C. weight of the gas, and use the periodic table to identify the gns. Calculate the molecular (10 pts) 6. 2 2018-04-01 18-14.pdf ? 2018-04-01 18-14..pdf Show allX ^ 2018-04-01 18-14...pdf 828 PM ID ^q )??4/1/2018

Explanation / Answer

4)

Molar mass of KClO3 = 122.55 g/mol

mass of KClO3 = 4.52 g

mol of KClO3 = (mass)/(molar mass)

= 4.52/122.55

= 0.0369 mol

Balanced chemical equation is:

2KClO3 (s) ---> 2 KCl (s) + 3O2

According to balanced equation

mol of O2 formed = (3/2)* moles of KClO3

= (3/2)*0.0369

= 0.0553 mol

Given:

P = 778.0 torr

= (778.0/760) atm

= 1.0237 atm

n = 0.0553 mol

T = 125.0 oC

= (125.0+273) K

= 398 K

use:

P * V = n*R*T

1.0237 atm * V = 0.0553 mol* 0.08206 atm.L/mol.K * 398 K

V = 1.76 L

Answer: 1.76 L

5)

P = 1.0atm

T = 273 K

Molar mass of CH4,

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

= 1*12.01 + 4*1.008

= 16.042 g/mol

Lets derive the equation to be used

use:

p*V=n*R*T

p*V=(mass/molar mass)*R*T

p*molar mass=(mass/V)*R*T

p*molar mass=density*R*T

Put Values:

1.0 atm *16.042 g/mol = density * 0.08206 atm.L/mol.K *273.0 K

density = 0.716 g/L

Answer: 0.716 g/L

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