From the chemical equation given:
H2SO4+2KOH--->K2SO4+2H2O
the two reactants, H2SO4 and KOH, are in 1:2 stoichiometric ratio.
No. of moles of KOH = 2* no. of moles of H2SO4
=2*0.1*0.033
The concentration of KOH = no. of moles / volume
=2*0.1*0.033/0.05
=0.132M
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Answer:
Methyl pentanoate.
Explanation:
Hello there!
In this case, according to the given information, we can see the correct structure will be:
O
||
CH3CH2CH2CH2COCH3
Which matches with the structure of an ester due to the -COO- functional group. In such a case, the first part of the name is in function of the right side of the ester, in this case, methyl, followed by the left side, pentanoate, as it has five carbon atoms and is an ester (similar to an inorganic salt, but organic) and therefore, the name will be methyl pentanoate.
Regards!
Answer is: it takes 116,8 seconds to fall to one-sixteenth of its initial value
<span>
The half-life for the chemical reaction is 29,2 s and is
independent of initial concentration.
c</span>₀
- initial concentration the reactant.
c - concentration of the reactant remaining
at time.
t = 29,2 s.<span>
First calculate the rate constant k:
k = 0,693 ÷ t = 0,693 ÷ 29,2 s</span> = 0,0237 1/s.<span>
ln(c/c</span>₀) = -k·t₁.<span>
ln(1/16 </span>÷ 1) = -0,0237 1/s ·
t₁.
t₁ = 116,8 s.
