C(HClO) = 0,3 M.
<span>V(HClO) = 200 mL = 0,2 L.
n(HClO) = </span>c(HClO) · V(HClO).
n(HClO) = 0,06 mol.<span>
c(KClO</span>) =
0,2 M.
<span>V(KClO) = 0,3 L.
n(KClO) = 0,06 mol.
V(buffer solution) = 0,2 L + 0,3 L = 0,5 L.
ck</span>(HClO) = 0,06 mol ÷ 0,5 L = 0,12 M.
cs(KClO) = 0,06 mol ÷ 0,5 L = 0,12 M.<span>
Ka(HClO</span>) =
2,9·10⁻⁸.<span>
This is buffer solution, so use Henderson–Hasselbalch
equation:
pH = pKa + log(cs</span> ÷ ck).<span>
pH = -log(</span>2,9·10⁻⁸) + log(0,12 M ÷ 0,12 M).<span>
pH = 7,54 + 0.
pH = 7,54</span>
C.
centi- is essentially 10^2 of one meter.
If you had 100m, multiplying 100 by 10^2 (or 100) would give you 10000 cm.
Answer:a) 11.34 g of ethane 
can be formed
b) 
is the limiting reagent
c) 3.44 g of the excess reagent remains after the reaction is complete
Explanation:
To calculate the moles :
1. 
2. 
According to stoichiometry :
1 mole of require 1 mole of 
Thus 0.378 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
moles of left = (2.10-0.378) = 1.72 moles
mass of left=
According to stoichiometry :
As 1 mole of give = 1 mole of 
Thus 0.378 moles of give = of
Mass of 
Thus 11.34 g of ethane is formed.
1. From the balanced equation given above, the ratio of the number of moles of the hydrocarbon and oxygen is equal to 2/7. Given that there are 2 moles of oxygen,
moles hydrocarbon = (2 moles O2)(2 moles HC / 7 moles oxygen)
Simplifying,
moles HC = 4/7 or 0.57 moles HC
Answer: 0.57 moles HC
2. The calculation for the mass of water is shown below with the dimensional analysis and conversion factors,
(30 g C2H6)1 mole C2H6/30 g C2H6)(6 molesH2O/2 moles C2H6)(18 g H2O/1 mole C2H6)
Simplifying,
mass = 54 grams of water
