How many moles of sulfur dioxide are in 2.26 x 10^33 sulfur dioxide molecules?
1 answer:
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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>
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 =
∴
∴,
∴, Ct = 85.35%
Thus, <span>85.35 % of the initial amount of SO2Cl2 will remain after 2.00 hours.</span>