<span>B. S⁰(s) + 2H⁺ + 2e⁻ –--> H2S⁰(g)
by mass: 1 S and 2 H ----> 2 H and 1S True.
by charge : 0 +(2*(+1)) + 2*(-1) = 0, 0+2-2=0, 0 = 0 True.</span>
Molar mass of CO2 = 44.01 g/mol
B
a and c are unrelated
d: ph will increase
Answer: 0.0014 atm
Explanation:
Given that,
Original pressure of air (P1) = 1.08 atm
Original volume of air (T1) = 145mL
[Convert 145mL to liters
If 1000mL = 1l
145mL = 145/1000 = 0.145L]
New volume of air (V2) = 111L
New pressure of air (P2) = ?
Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law
P1V1 = P2V2
1.08 atm x 0.145L = P2 x 111L
0.1566 atm•L = 111L•P2
Divide both sides by 111L
0.1566 atm•L/111L = 111L•P2/111L
0.0014 atm = P2
Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm
For radioactive decay, we can relate current amount, initial amount, decay constant and time using:
N = No x exp(-λt)
Half-life = ln(2)/λ
λ = ln(2) / 5730
N/No = 80% = 0.8
0.8 = exp( -ln(2)/5730 x t)
t = 1844 years
