Colligative
properties calculations are used for this type of problem. Calculations are as
follows:<span>
</span>
<span>ΔT(freezing point)
= (Kf)m
ΔT(freezing point)
= 1.86 °C kg / mol (0.705)
ΔT(freezing point) = 1.3113 °C
</span>
<span>
</span>
<span>Hope this answers the question. Have a nice day.</span>
HEY mate here your answer
CoBr2 . 6 h2o = Cobalt(II) Bromide Hexahydrate
The balanced reaction equation for the combustion of butane is as follows;
C₄H₁₀ + 13/2O₂ ---> 4CO₂ + 5H₂O
the limiting reactant in this reaction is C₄H₁₀ This means that all the butane moles are consumed and amount of product formed depends on the amount of C₄H₁₀ used up.
stoichiometry of C₄H₁₀ to H₂O is 1:5
mass of butane used - 6.97 g
number of moles - 6.97 g / 58 g/mol = 0.12 mol
then the number of water moles produced - 0.12 mol x 5 = 0.6 mol
Therefore mass of water produced - 0.6 mol x 18 g/mol = 10.8 g
Answer:
The correct answer is 0.3582 M.
Explanation:
Based on the given information, the volume of KOH given is 19.55 ml, the volume of sulfuric acid given is 20.20 ml, and the molarity of sulfuric acid is 0.3467 M. There is a need to find the molarity of KOH.
The formula to use in the given case is,
M1V1 = M2V2
Here M1 is the molarity of KOH, V1 is the volume of KOH, M2 is the molarity of sulfuric acid, and V2 is the volume of H2SO4.
Here V1 is 19.55 ml, M2 is 0.3467 M, and V2 is 2020 ml.
Now putting the values in the equation we get,
M1 * 19.55 = 0.3467 * 20.20
M1 * 19.55 = 7.00334
M1 = 7.00334/19.55
M1 = 0.3582 M
Hence, the molarity of the given KOH is 0.3582.
