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2 answers:
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60.7 ml is the volume of a sample of CO2 at STP that has a volume of 75.0mL at 30.0°C and 91kPa.
Explanation:
Data given:
V1 = 75 ml
T1 = 30 Degrees or 273.15 + 30 = 303.15 K
P1 = 91 KPa
V2 =?
P2 = 1 atm or 101.3 KPa
T2 = 273.15 K
At STP the pressure is 1 atm and the temperature is 273.15 K
applying Gas Law:
=
putting the values in the equation of Gas Law:
V2 =
V2 =
V2 = 60.7 ml
at STP the volume of carbon dioxide gas is 60.7 ml.
Answer:
Rate = 116m⁻¹s⁻¹[lactose][H]⁺
Explanation:
the formula for rate of reaction is given as
Rate = k[lactose]∧α[H]⁺∧β
we solve for the value of α and β
([lactose]₁/[lactose]₂)∧α
α =
when we divide this equation
α =
α = 1
we find β
R₁/R₂ = 0.01/0.02(0.001/0.001)∧β
0.00116/0.00232 = 0.5(1)∧β
β = 1
Rate = k[lactose]∧α[H]⁺∧β
we have to find the value for k
k = 0.00116/0.01(0.001)
k = 0.00116/0.00001
= 116m⁻¹s⁻¹
<u>Rate = 116m⁻¹s⁻¹[lactose][H]⁺</u>
