Answer:
13.4 milimoles.
Explanation:
The following data were obtained from the question:
Volume = 1.88 L
Molarity = 0.00713 M
Millimoles of NaCN =?
Next, we shall determine the number of mole NaCN in the solution. This can be obtained as follow:
Molarity = mole /Volume
Volume = 1.88 L
Molarity = 0.00713 M
Mole of NaCN =?
Molarity = mole /Volume
0.00713 = moles of NaCN /1.88
Cross multiply
Mole of NaCN = 0.00713 × 1.88
Mole of NaCN = 0.0134 mole
Finally, we shall convert 0.0134 mole to Millimoles. This can be obtained as follow:
1 mole = 1000 millimoles
Therefore,
0.0134 mole = 0.0134 × 1000
0.0134 mole = 13.4 milimoles
Therefore, the millimoles of the solute, NaCN in the solution is 13.4 milimoles
Answer:
a) the final kilocalories per gram for food will be less because the mass was reduced
b)the final kilocalories per gram for food will be less since
c) the final kilocalories per gram for food will be less since the reaction will eventually go to completion
d) the final kilocalories per gram for food will be more.
Explanation:
a) the final kilocalories per gram for food will be less because the mass was reduced from 110.3 to 101.3g
b)the final kilocalories per gram for food will be less since some marshmallow fell off before the reaction
c) the final kilocalories per gram for food will be less since the reaction will eventually go to completion
d) the final kilocalories per gram for food will be more since the thermometer that got stuck will add to the value of final kilocalories per gram
According to the reaction equation:
and by using ICE table:
CN- + H2O ↔ HCN + OH-
initial 0.08 0 0
change -X +X +X
Equ (0.08-X) X X
so from the equilibrium equation, we can get Ka expression
when Ka = [HCN] [OH-]/[CN-]
when Ka = Kw/Kb
= (1 x 10^-14) / (4.9 x 10^-10)
= 2 x 10^-5
So, by substitution:
2 x 10^-5 = X^2 / (0.08 - X)
X= 0.0013
∴ [OH] = X = 0.0013
∴ POH = -㏒[OH]
= -㏒0.0013
= 2.886
∴ PH = 14 - POH
= 14 - 2.886 = 11.11
Answer:
Explanation:
1 . Write the chemical equation.
2 . Calculate the moles of product from the first reactant.
3 . Calculate the moles of product from the second reactant.
4 . Identify the limiting reactant and the excess reactant.
5 . Calculate the mass of excess reactant used up.
6 . Calculate the mass of unused excess reactant.
7 . Calculate the mass of limiting reactant needed to react with the 8 . unused excess reactant.
