Answer:
23.92 g
Explanation:
Molar mass of H2SO4 = (2×1)+32+(16×4)= 2+32+48= 82g/mol
H2SO4 + 2NaOH ---> Na2SO4 + 2H2O
I mole of H2SO4 = 2 moles of NaOH
24.5/82 = 24.5/82 × 2
= 0.598 moles of NaOH will neutralize
Mass= mole× molar mass
Molar mass of NaOH= 23+16+1 = 40g/mol
Mass= 0.598 × 40 = 23.92g of NaOH
The percentage error is the error of the measured value to the true value. To find he percent error, the equation is as follows:
Percent error = |Measured Value - True Value|/True Value * 100
The || is needed to get the absolute value of the difference. Substituting the values,
Percent error = |(10.085 g/10 mL) - 0.9975 g/mL|/<span>0.9975 g/mL * 100
</span><em>Percent error = 1.1% </em>
<span>Average oxidation state = VO1.19
Oxygen is-2. Then 1.19 (-2) = -2.38
Average oxidation state of V is +2.38
Consider 100 formula units of VO1.19
There would be 119 Oxide ions = Each oxide is -2. Total charge = -2(119) = -238
The total charge of all the vanadium ions would be +238.
Let x = number of of V+2
Then 100 – x = number of V+3
X(+2) + 100-x(+3) = +238
2x + 300 – 3x = 238
-x = 238-300 = -62
x = 62
Thus 62/100 are V+2
62/100 * 100 = 62%
</span>62 % is the percentage of the vanadium atoms are in the lower oxidation state. Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.
Answer:
ΔTb = 0.66 C
Explanation:
Given
Mass of KBr = 185 g
Mass of water = 1.2 kg
Kb = 0.51 C/m
Explanation:
The change in boiling point (ΔTb) is given by the product of molality (m) of the solution and the boiling point constant (Kb)
[tex]\Delta T_{b}= 0.51 C/m * 1.296 m = 0.66 C[\tex]