A carbon iota can bond with four other iotas and is just like the four-hole wheel, whereas an oxygen iota, which can bond only to two, is just like the two-hole wheel.
Here, we are required to determine the volume of the earth which is 1.08326 × 10¹² km³ in liters.
<em>The volume of the earth is approximately</em>,
, 1.08326 × 10²⁴ liters
By conversion factors;
- <em>1dm³ = 1liter</em>
- However; <em>1km = 10000dm = 10⁴ </em><em>dm</em>
- Therefore, 1km³ = (10⁴)³ dm³.
Consequently, 1km³ = 10¹²dm³ = 10¹²liters.
The conversion factor from 1km³ to liters is therefore, c.f = 10¹²liters/km³
Therefore, the volume of the earth which is approximately, 1.08326 × 10¹² km³ can be expressed in liters as;
<em>1.08326 × 10¹² km³ × 10¹²liters/km³ </em>
The volume of the earth is approximately,
1.08326 × 10²⁴ liters.
Read more:
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Answer:
150.1 mL
Explanation:
Step 1: Given data
- Density of benzene (ρ): 0.879 g/mL
- Mass of the sample of benzene (m): 131.9 g
- Volume of the sample of benzene (V): ?
Step 2: Calculate the volume of the sample of benzene
Density is an intrinsic property. It is equal to the quotient between the mass and the volume of the sample of benzene.
ρ = m/V
V = m/ρ
V = 131.9 g/(0.879 g/mL)
V = 150.1 mL
1. The molar mass of the unknown gas obtained is 0.096 g/mol
2. The pressure of the oxygen gas in the tank is 1.524 atm
<h3>Graham's law of diffusion </h3>
This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e
R ∝ 1/ √M
R₁/R₂ = √(M₂/M₁)
<h3>1. How to determine the molar mass of the gas </h3>
- Rate of unknown gas (R₁) = 11.1 mins
- Rate of H₂ (R₂) = 2.42 mins
- Molar mass of H₂ (M₂) = 2.02 g/mol
- Molar mass of unknown gas (M₁) =?
R₁/R₂ = √(M₂/M₁)
11.1 / 2.42 = √(2.02 / M₁)
Square both side
(11.1 / 2.42)² = 2.02 / M₁
Cross multiply
(11.1 / 2.42)² × M₁ = 2.02
Divide both side by (11.1 / 2.42)²
M₁ = 2.02 / (11.1 / 2.42)²
M₁ = 0.096 g/mol
<h3>2. How to determine the pressure of O₂</h3>
From the question given above, the following data were obtained:
- Volume (V) = 438 L
- Mass of O₂ = 0.885 kg = 885 g
- Molar mass of O₂ = 32 g/mol
- Mole of of O₂ (n) = 885 / 32 = 27.65625 moles
- Temperature (T) = 21 °C = 21 + 273 = 294 K
- Gas constant (R) = 0.0821 atm.L/Kmol
The pressure of the gas can be obtained by using the ideal gas equation as illustrated below:
PV = nRT
Divide both side by V
P = nRT / V
P = (27.65625 × 0.0821 × 294) / 438
P = 1.524 atm
Learn more about Graham's law of diffusion:
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Answer:
"1 M" will be the right solution.
Explanation:
The given values are:
Number of moles,
= 2 moles
Volume of solution,
= 2000 mL
or,
= 2 L
Now,
The molarity of the solution will be:
= 
On substituting the values, we get
= 
= 
