a) when adding to 24 mL of 0.19 M NaOH:first, we need to get moles of HCl = molarity * volume
= 0.19 M * 0.019 L
= 0.00361 moles
then, moles of NaOH = molarity * volume
= 0.19 M * 0.024 L
= 0.00456 moles
NaOH remaining = 0.00456 - 0.00361
= 0.00095 moles
when the total volume = 24mL+19 mL = 43 mL
∴ molarity of NaOH = moles / total volume
= 0.00095 / 0.043L
= 0.0221 M
when POH = -㏒[OH-]
= -㏒0.0221 M
= 1.66
∴PH = 14 - POH
= 14- 1.66
= 12.34
b) when adding to 29 mL of 0.24 M NaOH:when moles of HCl = 0.00361 moles
then, moles of NaOH = molarity * volume
= 0.24 M * 0.029 L
= 0.00696 moles
∴ NaOH remaining = 0.00696 - 0.00361
= 0.00335 moles
the total volume = 29 mL + 19 mL = 48 mL
molarity of NaOH = moles / total volume
= 0.00335 / 0.048L
= 0.0698 M
∴POH = -㏒[OH-]
= -㏒ 0.0698
= 1.16
∴ PH = 14- POH
= 14- 1.16
= 12.84
Answer:
We should fill the acetylene tank to a pressure of 100 atm
Explanation:
Step 1: Data given
Volume tank 1 = 7.00 L
Tank 1 is filled with oxygen and has a pressure of 125 atm
Volume tank 2 = 3.50 L
Step 2: The balanced equation
2 C2H2 + 5 O2 → 4 CO2 + 2 H2O
Step 3: Calculate the pressure of acetylene
p(oxygen) * V(Oxygen) = n(Oxygen) * R * T
p(Acetylene) * V(Acetylene) = n(Acetylene) * R * T
We can assume both tanks are at the same temperature, so we can write this as followed:
p(Oxygen)* V(Oxygen) / p(Acetylene)*V(Acetylene) = n(Oxygen) / n(Acetylene)
⇒with n(Oxygen) / n(Acetylene) = 5/2
⇒with p(Oxygen) = 125atm
⇒V(Oxygen) =7L
⇒V(Acetylene) = 3.5L
⇒ this gives us: (125 * 7) / (P(Acetylene) * 3.5) = 5/2
p(Acetylene) = 100 atm
We should fill the acetylene tank to a pressure of 100 atm
<span>the approximate angle between two hydrogen-oxygen bonds in water is </span>105˚.