We are given that the balanced chemical reaction is:
cacl2⋅2h2o(aq) +
k2c2o4⋅h2o(aq) --->
cac2o4⋅h2o(s) +
2kcl(aq) + 2h2o(l)
We known that
the product was oven dried, therefore the mass of 0.333 g pertains only to that
of the substance cac2o4⋅h2o(s). So what we will do first is to convert this
into moles by dividing the mass with the molar mass. The molar mass of cac2o4⋅h2o(s) is
molar mass of cac2o4 plus the
molar mass of h2o.
molar mass cac2o4⋅h2o(s) = 128.10
+ 18 = 146.10 g /mole
moles cac2o4⋅h2o(s) =
0.333 / 146.10 = 2.28 x 10^-3 moles
Looking at
the balanced chemical reaction, the ratio of cac2o4⋅h2o(s) and k2c2o4⋅h2o(aq) is
1:1, therefore:
moles k2c2o4⋅h2o(aq) = 2.28
x 10^-3 moles
Converting
this to mass:
mass k2c2o4⋅h2o(aq) = 2.28
x 10^-3 moles (184.24 g /mol) = 0.419931006 g
Therefore:
The mass of k2c2o4⋅<span>h2o(aq) in
the salt mixture is about 0.420 g</span>
Answer:
Carbon=5, hydrogen=12, oxygen=16
Explanation:
Carbon=5, hydrogen=12, oxygen=16
In order to effectively count the number of atoms, we look at the equation closely and take note of the stoichiometric coefficients of each reactant as this influences the number of atoms of that element present.
For instance, oxygen is diatomic and has a stoichiometric coefficient of 8. This implies the there are sixteen atoms of oxygen altogether.
Note that the left hand side refers to the reactants side.
Answer:
Moles of KOH in 1000 mL solution = 0.255 moles
Moles of KOH in 1 mL solution = 0.255/1000 = 0.000255 moles
Moles in 95 mL solution = (95 * 255)/1000000 = 24225/1000000
Moles of KOH in 95 mL 0.255M solution = 0.024225 moles
The graph is needed to answer this question.
Solubility may increase or decrease with temperature depending on the properties of the solute and the solvent.
It is quite common that the solubility of the ionic compounds, like KBr, in water increases with temperature.
Use your solubility curve for the KBr and you wiil see a line that starts at a solubility a little greater than 50 grams of the salt in 100 grams of water for temperaute 0°C and increase linearly until almost 100 grams of the salt in 100 grams of water at 100°C.
So, in this case you can affirm that the solubility of KBr increases with the temperature.
Answer: the second option: the solubility increases.
