Answer:
Explanation:
The expected product is MgO, so the 1-to-1 mole ratio Mg to O in the product is all that is required.
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
Answer:
You need 375 mL of BaCl2 solution.
Explanation:
M1V1=M2V2
Dilution formula. Substitute known values and solve for V1.
M1 = 2.0 M
M2 = 1.50 M
V2 = 500 mL
(2.0 M)(V1) = (1.50 M)(500 mL)
V1 = (1.50 M)(500 mL) / (2.0 M)
V1 = 375 mL
Answer:
The mass of the precipitate that AgCl is 3.5803 g.
Explanation:
a) To calculate the molarity of solution, we use the equation:
We are given:
Mass of solute (NaCl) = 1.46 g
Molar mass of sulfuric acid = 58.5 g/mol
Volume of solution = 
Putting values in above equation, we get:
0.09982 M is the concentration of the sodium chloride solution.
b) 
Moles of NaCl = 
according to reaction 1 mol of NaCl gives 1 mol of AgCl.
Then 0.02495 moles of NaCl will give:
of AgCl
Mass of 0.02495 moles of AgCl:
The mass of the precipitate that AgCl is 3.5803 g.
The isotopes are elements with the same number of protons but different number of neutrons.
Boron atomic number is 5. That is the number of protons.
The natural occurring isotopes of boron contain 5 or 6 neutrons. Other, non stable isotopes, contain from 1 to 16 neutrons.
