Answer:
255.6
Explanation:
If you have 12 gallons and get 21.3mpg,
-Multiply 21.3 by 12
-you can travel 255.6 miles before running out of gas.
-If you need to estimate, round up to 256 miles.
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
Answer:
<u>Mass concentration (g/L) </u><u><em>= 2.49g/L.</em></u>
Explanation:
No. of moles = 
= 
= 0.001245 moles
Concentration of KHP (C1) in litres = n/v
= 
= 0.062 mol/L
We know that:
= 
where c1v1 and c2v2 are the products of concentration and volumes of KHP and NaOH respectively.
Since mole ratio is 1 : 1.
1 mole of NaOH - 40g
0.001245 mole of NaOH = 40 × 0.001245 = 0.0498g
⇒0.0498g of NaOH was used during the titration
<u><em>∴Mass concentration (g/L) = 0.0498g ÷ 0.02L</em></u>
<u><em>= 2.49g/L.</em></u>
Explanation:
The more reactive element replaces less reactive element during chemical reaction.
Since, potassium is more reactive than beryllium. When potassium reacts with beryllium choride, it replaces beryllium and forms potassium chloride and produces beryllium.
