Freezing, condensation, Deposition.
(35/18.02)(6.02kJ)=11.6926 kiloJoules
Answer:
22.9 Liters CO(g) needed
Explanation:
2CO(g) + O₂(g) => 2CO₂(g)
? Liters 32.65g
= 32.65g/32g/mol
= 1.02 moles O₂
Rxn ratio for CO to O₂ = 2 mole CO(g) to 1 mole O₂(g)
∴moles CO(g) needed = 2 x 1.02 moles CO(g) = 2.04 moles CO(g)
Conditions of standard equation* is STP (0°C & 1atm) => 1 mole any gas occupies 22.4 Liters.
∴Volume of CO(g) = 1.02mole x 22.4Liters/mole = 22.9 Liters CO(g) needed
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*Standard Equation => molecular rxn balanced to smallest whole number ratio coefficients is assumed to be at STP conditions (0°C & 1atm).
Answer:
4,38%
small molecular volumes
Decrease
Explanation:
The percent difference between the ideal and real gas is:
(47,8atm - 45,7 atm) / 47,8 atm × 100 = 4,39% ≈ <em>4,38%</em>
This difference is considered significant, and is best explained because argon atoms have relatively <em>small molecular volumes. </em>That produce an increasing in intermolecular forces deviating the system of ideal gas behavior.
Therefore, an increasing in volume will produce an ideal gas behavior. Thus:
If the volume of the container were increased to 2.00 L, you would expect the percent difference between the ideal and real gas to <em>decrease</em>
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I hope it helps!
Given parameters:
Volume of CuSO₄ = 250mL
Concentration of CuSO₄ = 2.01M
Unknown:
Mass of CuSO₄.5H₂O = ?
To solve this problem, we must write the chemical relationship between both species.;
CuSO₄.5H₂O → CuSO₄ + 5H₂O
Now that we know the expression, it is possible to solve for the unknown mass.
First find the number of moles of CuSO₄;
Number of moles = Concentration x Volume
Take 250mL to L so as to ensure uniformity of units;
Volume = 250 x 10⁻³L
Input the parameters and solve for number of moles;
Number of moles = 250 x 10⁻³ x 2.01 = 0.5mol
From the equation;
1 mole of CuSO₄ is produced from 1 mole of CuSO₄.5H₂O
So 0.5 moles of CuSO₄ will be produced from 0.5 moles of CuSO₄.5H₂O
Now let us find the molar mass of CuSO₄.5H₂O = 63.6 + 32 + 4(16) + 5(2x1 + 16) = 249.6g/mole
Mass of CuSO₄.5H₂O = number of moles x molar mass
= 0.5 x 249.6
= 124.8g
The mass of CuSO₄.5H₂O is 124.8g
