Answer:
2.5 g of platinum
Explanation:
Recall that a catalyst is a specie added to a reaction system to increase the rate of reaction. A catalyst does not participate in the chemical reaction hence it remains unchanged at the end of the chemical reaction. A catalyst merely provides an alternative reaction pathway by lowering the activation energy of the reaction system. Hence a catalysed reaction usually proceeds faster with less energy requirement than the uncatalysed reaction.
Since the catalyst does not participate in the reactions and remains unchanged at the end of the reaction, the mass of platinum will remain the same (2.5g). The mass can only change if a specie participates in the chemical reaction. Hence the answer.
<h3>Answer:</h3>
0.8133 mol
<h3>Solution:</h3>
Data Given:
Moles = n = ??
Temperature = T = 25 °C + 273.15 = 298.15 K
Pressure = P = 96.8 kPa = 0.955 atm
Volume = V = 20.0 L
Formula Used:
Let's assume that the Argon gas is acting as an Ideal gas, then according to Ideal Gas Equation,
P V = n R T
where; R = Universal Gas Constant = 0.082057 atm.L.mol⁻¹.K⁻¹
Solving Equation for n,
n = P V / R T
Putting Values,
n = (0.955 atm × 20.0 L) ÷ (0.082057 atm.L.mol⁻¹.K⁻¹ × 298.15 K)
n = 0.8133 mol
Answer:
The answer to your question is: 69.6 %
Explanation:
Freon -112 (C₂Cl₄F₂)
MW = (12 x 2) + (35.5 x 4) + (19 x 2)
= 24 + 142 + 38
= 204 g
204 g of C₂Cl₄F₂ ----------------- 100%
142 g ----------------- x
x = (142 x 100 ) / 204
x = 69.6 %
Answer:
= 913.84 mL
Explanation:
Using the combined gas laws
P1V1/T1 = P2V2/T2
At standard temperature and pressure. the pressure is 10 kPa, while the temperature is 273 K.
V1 = 80.0 mL
P1 = 109 kPa
T1 = -12.5 + 273 = 260.5 K
P2 = 10 kPa
V2 = ?
T2 = 273 K
Therefore;
V2 = P1V1T2/P2T1
= (109 kPa × 80 mL × 273 K)/(10 kPa× 260.5 K)
<u>= 913.84 mL</u>
