The pressure exerted by 0.57 moles of CO2 at a temperature of 25°C and a volume of 500 ml is 28 atm.
<u>Explanation:</u>
According to ideal gas law,
PV = nRT
where P represents the pressure of a gas,
V represents the volume of a gas,
n represents the number of moles,
R represents the gas constant = 0.0821 L atm / mol K.
T represents the temperature of a gas.
Given V = 500 ml = 0.5 l, T = 25°C = 298 K, n = 0.57 mol
PV = nRT
P = nRT / V
= (0.57 0.0821 298) / 0.5
P = 28 atm.
The pressure of a gas is 28 atm.
Answer:
Alcohol has greater value of temperature coefficient of expansion than mercury.
The answer is: "Twelve (12) moles of Fe (iron) would be required."
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or:
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The answer is: "12" .
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Note: The balanced chemical equation is given:
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4 Fe + 3 O₂ → 2 Fe₂O₃
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The product is: 2 moles of Fe₂O₃ ; that is, "2 moles of iron (III) oxide" .
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→ If SIX (6) moles of iron oxide were produced, that would be "3 (three) times the number of moles of iron oxide produced (3 TIMES 2) ;
→ since: "6 ÷ 2 = 3" ; and since the equation is balanced.
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→ So we want to find the number of moles of "iron (Fe)" needed in such a situation. In the balanced chemical, we have "4 Fe" (that is, "FOUR (4) moles of iron") ; so we multiply that number by "3", proportionally:
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→ 4 (moles Fe) * (3) = 12 (moles Fe) .
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Answer:
Final number of moles = 0.675 mol
Mass = 2.7 g
Explanation:
Given data:
Initial volume of gas = 2.00 L
Final volume of gas = 2.70 L
Initial number of moles = 0.500 mol
Final number of moles = ?
Solution:
Formula:
V₁/n₁ = V₂/n₂
V₁ = Initial volume
n₁ = Initial number of moles
V₂ = Final volume of gas
n₂ = Final number of moles
Now we will put the values in formula.
2.00 L /0.500 mol = 2.70 L / n₂
n₂ = 2.70 L× 0.500 mol /2.00 L
n₂ = 1.35 L.mol / 2.00 L
n₂ = 0.675 mol
B)
Grams of helium added = ?
Solution:
Mass = number of moles × molar mass
Mass = 0.675 mol× 4 g/mol
Mass = 2.7 g
A 10.0 g piece of metal is placed in an insulated calorimeter containing 250.0 g of water initially at 20.0 °C. If the final temperature of the mixture is 25.0 °C, the heat change of water will be 5230 J.
Weight of metal = 10.0 g
Weight of water = 250.0 g
Initial temperature of water (Ti) = 20.0 °C
Final temperature of water (T2) = 25.0 °C
The specific heat capacity of water is 4.184 J/g.°C
We can calculate the heat change of water by the following equation;
<em>Q = mwater × Cwater × (T2-Ti)</em>
Q = 250.0 g × 4.184 J/g.°C × (25.0 - 20.0) °C
Q = 5230 J
You can also learn about heat change from the following question:
brainly.com/question/18912282
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