Question 1 :
V1/T1 = V2/T2
3.0L/273K = V2/373K
To get the value of Z, cross multiply
3.0L x 373K = 273K x V2
1119 = 273V2
Divide both sides by 273
1119/273 = 273V2/273
4.10L = V2
The new volume is 4.10 liters
Question 2 :
P1/T1 = P2 /T2
P1 = 880 kPA= 880 *10^3 Pa
T1 = 250 K
T2 = 303 K
P2 =?
Substituting for P2
P2 = P1 T2/ T1
P2 = 880 kPa * 303 / 250
P2 = 266,640 kPa/ 250
P2 = 1066.56 kPa.
The new pressure of the gas is 1066.56 kPa
Question 3 :
Given that:
Volume of gas V = 4.80L
(since 1 liter = 1dm3
4.80L = 4.80dm3)
Temperature T = 62°C
Convert Celsius to Kelvin
(62°C + 273 = 335K)
Pressure P = 2.9 atm
Number of moles of gas N = ?
Apply ideal gas equation
pV = nRT
2.9atm x 4.8dm3 = n x (0.0082 atm dm3 K-1 mol-1 x 335K)
13.92 atm dm3 = nx 2.747 atm dm3 mol-1
n = 13.92/2.747
n = 5.08 moles
There are 5.08 moles of gas contained in the sample
Question 4 :
Volume of gas V = 3.47L
(since 1 liter = 1dm3
3.47L = 3.47dm3)
Temperature T = 85.0°C
Convert Celsius to Kelvin
(85.0°C + 273 = 358K)
Pressure P = ?
Number of moles of gas N = 0.100 mole
Apply ideal gas equation
pV = nRT
p x 3.47dm3 = 0.10 x (0.0082 atm dm3 K-1 mol-1 x 358K)
p x 3.47dm3 = 0.29 atm dm3
p = (0.29 atm dm3 / 3.47 dm3)
p = 0.085 atm
If 1 atm = 760 mm Hg
0.085atm = 0.085 x 760
= 64.6 mm Hg
The pressure of the gas is 64.6 mm hg
Flammmability is a chemical change because it changes the composition of the object. the product is very different from the rectant.
4 NH₃ + 3O₂ --> 2N₂ + 6H₂O
First, make sure that this is a balanced equation.
There are 4 moles of nitrogen on the left side, and 4 moles of nitrogen on the right side.
There are 12 moles of hydrogen on the left side, and 12 moles of hydrogen on the right side.
There are 6 moles of oxygen on the left side, and 6 moles of oxygen on the right side.
The equation is therefore balanced, and we may proceed.
a) the mole ratio for NH₃ to N₂ is 4 to 2, which can be simplified to 2:1 or 2/1.
b) the mole ratio for H₂O to O₂ is 6 to 3, which can be simplified to 2:1 or 2/1.
Answer:
copy and paste this equation into the website
Explanation:
https://en.intl.chemicalaid.com/tools/equationbalancer.php
Explanation:
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