For the 1st order reactions,rate constant (k) is mathematically expressed as
k =
where, t = time
Co = initial conc. of reactant
Ct = conc. of reactant after time 't'
Given: k = <span>2.20 × 10^-5 s-1, t = 2 hours = 7200 s
Therefore, we have
</span>2.20 × 10^-5 =
∴
= 0.06877
∴,
= 1.1716
∴, Ct = 85.35%
Thus, <span>
85.35 % of the initial amount of SO2Cl2 will remain after 2.00 hours.</span>
Answer:
7.07
Explanation:
HA = weak acid = 0.053
A+ = conjugate base = 0.045
Ka = 7.2x10^-8
Ka = [H+][A-]/HA
7 2x10^-8 = [H+][0.045]/0.053
[H+] = 7.2x10^-8 x 0.053/0.045
= 8.48x10^-8
PH = -log[H+]
= -log[8.48x10^-8]
PH = -[login.48 + log10^-8]
PH = -0.928 - (-8)log10
= 7.07
To get the answer you use the Law of Raoult.
Raoult's law states that the decrease of the vapor pressure of a liquid is proportional to the molar fraction of the solute.
ΔP = Pa * Xa
Here Pa = 0.038 atm
And Xa = N a / (Na + Nb), where Na is number of moles of A and Nb is number of moles of b
Na = mass of urea / molar mass of urea = 60 g / (molar mass of CH4N2O)
molar mass of CH4N2O = 12 g/mol + 4*1g/mol + 2*14 g/mol + 16 g/mol = 60 g/mol
Na = 60 g / 60 g/mol = 1 mol
Nb = mass of water / molar mass of water = 180g / 18g/mol = 10 mol
Xa = 1 mol / (10 mol + 1 mol) = 1/11 =0.09091
ΔP = Pb * Xa = 0.038 atm * 0.09091 = 0.0035 atm
Then, the final vapor pressure of water is Pb - ΔP = 0.038atm - 0.0035atm = 0.035 atm.
Answer: 0.035 atm
Answer:
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