C. What is Cytoplasm? What does it contain?
1 answer:
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Q1)
we can use the ideal gas law equation to find the total pressure of the system ;
PV = nRT
where P - pressure
V - volume - 7 x 10⁻³ m³
n - number of moles
total number of moles - 0.477 + 0.265 + 0.115 = 0.857 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in K - 273 + 25 °C = 298 K
substituting the values in the equation
P x 7 x 10⁻³ m³ = 0.857 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
P = 303.33 kPa
1 atm = 101.325 kPa
Therefore total pressure - 303.33 kPa / 101.325 kPa/atm = 2.99 atm
Q2)
partial pressure is the pressure exerted by the individual gases in the mixture.
partial pressure for each gas can be calculated by multiplying the total pressure by mole fraction of the individual gas.
total number of moles - 0.477 + 0.265 + 0.115 = 0.857 mol
mole fraction of He -
mole fraction of Ne -
mole fraction of Ar -
partial pressure - total pressure x mole fraction
partial pressure of He - 2.99 atm x 0.557 = 1.67 atm
partial pressure of Ne - 2.99 atm x 0.309 = 0.924 atm
partial pressure of Ar - 2.99 atm x 0.134 = 0.401 atm
we can use the ideal gas law equation to find the total pressure of the system ;
PV = nRT
where P - pressure
V - volume - 7 x 10⁻³ m³
n - number of moles
total number of moles - 0.477 + 0.265 + 0.115 = 0.857 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in K - 273 + 25 °C = 298 K
substituting the values in the equation
P x 7 x 10⁻³ m³ = 0.857 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
P = 303.33 kPa
1 atm = 101.325 kPa
Therefore total pressure - 303.33 kPa / 101.325 kPa/atm = 2.99 atm
Q2)
partial pressure is the pressure exerted by the individual gases in the mixture.
partial pressure for each gas can be calculated by multiplying the total pressure by mole fraction of the individual gas.
total number of moles - 0.477 + 0.265 + 0.115 = 0.857 mol
mole fraction of He -
mole fraction of Ne -
mole fraction of Ar -
partial pressure - total pressure x mole fraction
partial pressure of He - 2.99 atm x 0.557 = 1.67 atm
partial pressure of Ne - 2.99 atm x 0.309 = 0.924 atm
partial pressure of Ar - 2.99 atm x 0.134 = 0.401 atm
The two solutions that have similar solubilities at 40°C will be Na₂HAsO₄ and Na₂SO₄.
<h3>What is solubity ?</h3>
It is the ability of a solute to be dissolved, especially in water.
As the temperature increases the solubility of the gas generally decreases"
According to given information and graph attached as reference ;
- Solubility of NaCl (sodium chloride) at 40°C is 36.3 grams.
- Solubility of Na₂SO₄ (sodium sulfate) at 40°C is 48.8 grams.
- Solubility of Na₂HAsO₄ (sodium arsenate dibasic) at 40°C is 48.9
grams. - Solubility of Ba(NO₃)₂ (barium nitrate) at 40°C is 14.1 grams.
- Solubility of Ce₂(SO₄)₃·9H₂O (ceasium sulfate nonahydrate) at 40°C is 6.2 grams.
Lear more about solubility here ;
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Answer:
Molar ratio of the compound is 1:1 and the type of hydrate is Mono hydrate.
Explanation:
From the given,
Mass of sodium carbonate = 8.85 g
Loss mass = 1.28 g
Actual weight of sodium carbonate = 8.85 g - 1.28 g = 7.57 g
Therefore, the compound has only one water molecule.
Molecular formula of the compound is an name of the compound is <u>sodium carbonate mono hydrate.</u>
Hence, the type of the compound is Mono hydrate.