The 3rd diagram is a mixture because it has 2 substances
Answer:
P = 2.145kPa
Explanation:
Mass = 22.1g
Molar mass of CO2 = 44g/mol
Vol = 165mL = 0.165L
T = -188°C = (-188 + 273.15)K = 85.15K
R = 8.314J/mol.K
From ideal gas equation,
PV = nRT
P = pressure of the ideal gas
V = volume the gas occupies
n = number of moles if the gas
R = ideal gas constant
T = temperature of the gas
n = number of moles
n = mass / molar mass
n = 22.1 / 44 = 0.50moles
PV = nRT
P = nRT/ V
P = (0.5 × 8.314 × 85.15) / 0.165
P = 2145.26Pa = 2.145kPa
Pressure of the gas is 2.145kPa
The half-life equation is written as:
An = Aoe^-kt
We use this equation for the solution. We do as follows:
5.5 = 176e^-k(165)
k = 0.02
<span>What is the half-life of the goo in minutes?
</span>
0.5 = e^-0.02t
t = 34.66 minutes <----HALF-LIFE
Find a formula for G(t) , the amount of goo remaining at time t.G(t)=?
G(t) = 176e^-0.02t
How many grams of goo will remain after 50 minutes?
G(t) = 176e^-0.02(50) = 64.75 g
D. The energy released or absorbed during the reaction
