Answer:
a)23.2 L
b)68.3kPa
c)7.5 atm
d)60.5L
e)1.67 atm
Explanation:
From Boyle's law:
P1V1=P2V2
P1= 748mmHg
P2=725mmHg
V1= 22.5L
V2??
V2= P1V1/P2= 748×22.5/725= 23.2 L
b)
V1=4.0L
P1= 205×10^3Pa
V2= 12.0L
P2=???
P2= P1V1/V2= 205×10^3×4/12
P2= 68.3×10^3 Pa or 68.3kPa
c)
P1= 1 atm
V1= 196.0L
P2= ??
V2= 26.0L
P2= P1V1/V2=1×196.0/26.0
P2= 7.5 atm
d)
V1= 40.0L
P1= 12.7×10^3Pa
V2=???
P2= 8.4×103Pa
V2= P1V1/P2= 12.7×10^3×40.0/8.4×103
V2=60.5L
e)
V1= 100mL
P1= 1atm
V2= 60mL
P2=???
P2= P1V1/V2= 1×100/60
P2= 1.67 atm
Answer:
I'm sure but send thru this picture for the question so I can help.
Answer:
6.53g of K₂SO₄
Explanation:
Formula of the compound is K₂SO₄
Given parameters:
Volume of K₂SO₄ = 250mL = 250 x 10⁻³L
= 0.25L
Concentration of K₂SO₄ = 0.15M or 0. 15mol/L
Unknown:
Mass of K₂SO₄ =?
Methods:
We use the mole concept to solve this kind of problem.
>>First, we find the number of moles using the expression below:
Number of moles= concentration x volume
Solving for number of moles:
Number of moles = 0.25 x 01.5
= 0.0375mole
>>Secondly, we use the number of moles to find the mass of K₂SO₄ needed. This can be obtained using the expression below:
Mass(g) = number of moles x molar mass
Solving:
To find the molar mass of K₂SO₄, we must know the atomic mass of each element in the compound. This can be obtained using the periodic table.
For:
K = 39g
S = 32g
O = 16g
Molar mass of K₂SO₄ = (39x2) + 32 + (16x4)
= 78 +32 + 64
= 174g/mol
Using the expression:
Mass(g) = number of moles x molar mass
Mass of K₂SO₄ = 0.0375 x 174 = 6.53g
Answer:
HI.
Explanation:
- Thomas Graham found that, at a constant temperature and pressure the rates of effusion of various gases are inversely proportional to the square root of their masses.
Rate of effusion ∝ 1/√molar mass.
- <em>(Rate of effusion of O₂) / (Rate of effusion of unknown gas) = (√molar mass of unknown gas) / (√molar mass of O₂).</em>
- An unknown gas effuses at one half the speed of that of oxygen.
∵ Rate of effusion of unknown gas = 1/2 (Rate of effusion of O₂)
∴ (Rate of effusion of O₂) / (Rate of effusion of unknown gas) = 2.
Molar mass of O₂ = 32.0 g/mol.
∵ (Rate of effusion of O₂) / (Rate of effusion of unknown gas) = (√molar mass of unknown gas) / (√molar mass of O₂).
∴ 2.0 = (√molar mass of unknown gas) / √32.0.
(
√molar mass of unknown gas) = 2.0 x √32.0
By squaring the both sides:
∴ molar mass of unknown gas = (2.0 x √32.0)² = 128 g/mol.
∴ The molar mass of sulfur dioxide = 80.91 g/mol and the molar mass of HI = 127.911 g/mol.
<em>So, the unknown gas is HI.</em>
<em></em>
An atom is the basic unit of matter.
