Answer:
3.60 ml
Explanation:
First of all we must put down the equation of the reaction. This will serve as a guide to our solution;
KOH(aq) + HBr(aq) -----> KBr(aq) + H2O(l)
The following were given in the question;
Concentration of acid CA= 2M
Volume of acid VA= 18ml
Concentration of base CB= 0.01 M
Volume of base VB= ????
Number of moles of acid NA= 1
Number of moles of base NB= 1
From;
CAVA/CBVB = NA/NB
CAVANB= CBVBNA
Therefore;
VB= CAVANB/CBNA
Substituting values;
VB= 2 × 18 ×1 / 0.01×1
VB= 3.60 ml
Therefore; 3.60 ml of base was used.
Organic matter is a variety of living and dead plant and animal material.
You can use any form of the Ideal Gas Law, but usually one form is more convenient, depending on the information given.
There are three forms of the Ideal Gas Law.
Form 1
P
V
=
n
R
T
If the question gives you the number of moles then
, this is the most convenient one to use.
Form 2
P
V
=
m
M
R
T
, where m is the mass of the gas and
M is its molar mass.
If the question gives you the mass
m
, this is the most convenient one to use.
Form 3
P =
ρ
M
R
T
, where ρ is the density of the gas.
If the question gives you the density
ρ
, this is the most convenient one to use.
The final temperature is 28 °C.
There are two heat transfers involved.
heat from combustion of propane + heat gained by water = 0
<em>q</em>_1 + <em>q</em>_2 = 0
<em>q</em>_1+ <em>C</em>_calΔ<em>T</em> = 0
<em>q</em>_1 = -50 kJ
<em>q</em>_2 = 6.00 kJ·°C^(-1) × Δ<em>T</em> = 6.00 Δ<em>T</em> kJ·°C^(-1)
<em>q</em>_1 +<em> q</em>_2 = -50 kJ + 6.00 Δ<em>T</em> kJ·°C^(-1) = 0
Δ<em>T</em> = 50/[6.00 °C^(-1)] = 8.33 °C
<em>T</em>_f = <em>T</em>_i + Δ<em>T</em> = 20 °C + 8.33 °C = 28 °C
