Answer:
1.654 atm.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and V are constant, and have different values of P and T:
<em>(P₁T₂) = (P₂T₁)</em>
<em></em>
P₁ = 1.0 atm, T₁ = 25°C + 273 = 298 K,
P₂ = ??? atm, T₂ = 220°C + 273 = 493 K,
- Applying in the above equation
<em>(P₁T₂) = (P₂T₁)</em>
<em></em>
<em>∴ P₂ = (P₁T₂)/(T₁) </em>= (1.0 atm)(493 K)/(298 K) = <em>1.654 atm.</em>
<span>a) 7.9x10^9
b) 1.5x10^9
c) 3.9x10^4
To determine what percentage of an isotope remains after a given length of time, you can use the formula
p = 2^(-x)
where
p = percentage remaining
x = number of half lives expired.
The number of half lives expired is simply
x = t/h
where
x = number of half lives expired
t = time spent
h = length of half life.
So the overall formula becomes
p = 2^(-t/h)
And since we're starting with 1.1x10^10 atoms, we can simply multiply that by the percentage. So, the answers rounding to 2 significant figures are:
a) 1.1x10^10 * 2^(-5/10.5) = 1.1x10^10 * 0.718873349 = 7.9x10^9
b) 1.1x10^10 * 2^(-30/10.5) = 1.1x10^10 * 0.138011189 = 1.5x10^9
c) 1.1x10^10 * 2^(-190/10.5) = 1.1x10^10 * 3.57101x10^-6 = 3.9x10^4</span>
