Answer:
Explanation:
moles of acetic acid = 500 x 10⁻³ x .1 M
= 5 X 10⁻³ M
.005 M
Moles of NaOH = .1 M
Moles of sodium acetate formed = .005 M
Moles of NaOH left = .095 M
pOH = 4.8 + log .005 / .095
= 4.8 -1.27875
= 3.52125
pH = 14 - 3.52125
= 10.48
Answer:
iron, graphite,NaCl solution etc are some examples of good conductor of electricity
The molarity of formic acid is 100 mM or
. The dissociation reaction of formic acid is as follows:

The expression for dissociation constant of the reaction will be:
![K_{a}=\frac{[HCOO^{-}][H^{+}]}{[HCOOH]}](https://tex.z-dn.net/?f=K_%7Ba%7D%3D%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%5BH%5E%7B%2B%7D%5D%7D%7B%5BHCOOH%5D%7D)
Rearranging,
![[HCOO^{-}]=\frac{K_{a}[HCOOH]}{[H^{+}]}](https://tex.z-dn.net/?f=%5BHCOO%5E%7B-%7D%5D%3D%5Cfrac%7BK_%7Ba%7D%5BHCOOH%5D%7D%7B%5BH%5E%7B%2B%7D%5D%7D)
Here, pH of solution is 4.15 thus, concentration of hydrogen ion will be:
![[H^{+}]=10^{-pH}=10^{-4.15}=7.08\times 10^{-5}M](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%3D10%5E%7B-pH%7D%3D10%5E%7B-4.15%7D%3D7.08%5Ctimes%2010%5E%7B-5%7DM)
Similarly,
thus,

Putting the values,
![[HCOO^{-}]=\frac{(1.78\times 10^{-4}M)(100\times 10^{-3}M)}{(7.08\times 10^{-5}M}=0.2511 M](https://tex.z-dn.net/?f=%5BHCOO%5E%7B-%7D%5D%3D%5Cfrac%7B%281.78%5Ctimes%2010%5E%7B-4%7DM%29%28100%5Ctimes%2010%5E%7B-3%7DM%29%7D%7B%287.08%5Ctimes%2010%5E%7B-5%7DM%7D%3D0.2511%20M)
Therefore, the concentration of formate will be 0.2511 M.
B. Hydrogen is the answer
Answer:
If 1,079.75 Joules of heat are added to 77.75 grams of water, by 3.32 degrees Celsius the temperature of water will increase

Explanation:

Here , q = heat added / removed from the substance
m = mass of the substance taken
= Change in temperature
C = specific heat capacity of the substance
In liquid state the value of C for water is :

Given values :
q = 1079.75 J
m = 77.75 gram
Insert the value of C, m , q in the given equation
on transposing ,


