Answer:
2.09 atm
Explanation:
We can solve this problem by using the equation of state for an ideal gas, which relates the pressure, the volume and the temperature of an ideal gas:
where
p is the pressure of the gas
V is its volume
n is the number of moles
R is the gas constant
T is the absolute temperature
In this problem we have:
n = 0.65 mol is the number of moles of the gas
V = 8.0 L is the final volume of the gas
is the temperature of the gas
is the gas constant
Solving for p, we find the final pressure of the gas:
Answer:
Explanation:
Hello!
In this case, we can divide the problem in two steps:
1. Dilution to 278 mL: here, the initial concentration and volume are 1.20 M and 52.0 mL respectively, and a final volume of 278 mL, it means that the moles remain the same so we can write:
So we solve for C2:
2. Now, since 111 mL of water is added, we compute the final volume, V3:
So, the final concentration of the 139 mL portion is:
Best regards!
Answer:2.27 g H2O
Explanation:
Step 1. Get the limiting reagent
1.24 atm of H2 would react completely with 1.24 x (1/2) = 0.62 atm of O2, but there is more O2 present than that (0.98). O2 is in excess and H2 is the limiting reactant.
If the reaction goes to completion:
(1.24 atm H2) x (2 mol H2O / 2 mol H2) = 1.24 atm H2O vapor
n = PV / RT = (1.24 atm) x (2.5 L) / ((0.08205746 L atm/K mol) x (28 + 273)K) = 0.126 mol H2O
(0.126 mol H2O) x (18.01532 g H2O/mol) = 2.27 g H2O
Answer:
4.99 x 10⁻⁷ meters or 499 nanometers
Explanation:
Use the formula:
λ = c/ν, where λ = wavelength, c = the speed of light (it's constant, 2.998 x 10⁸ m/s), and ν = frequency
λ = (2.998 x 10⁸ m/s)/(6.01 x 10¹⁴ 1/s)
λ = 4.98835 x 10⁻⁷
Round to nearest hundreth and you get 4.99 x 10⁻⁷ meters, or 499 nanometers.
