Answer:
42.5 g
Explanation:
Calculate the mass of the soft drink given the density and volume:
355 mL × 1.04 g/mL = 369.2 g
Now calculate the mass of sucrose given the percentage:
0.115 × 369.2 g = 42.46 g
Rounded to 3 significant figures, the mass is 42.5 g.
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>
The question is incomplete, the complete question is;
The table above summarizes data given to a student to evaluate the type of change that took place when substance X was mixed with water. The student claimed that the data did not provide enough evidence to determine whether a chemical or physical change took place and that additional tests were needed. Which of the following identifies the best way to gather evidence to support the type of change that occurred when water and Xwere mixed?
A. Measuring the melting point of the mixture of water and X
B. Adding another substance to the mixture of water and X to see whether a solid forms
C Measuring and comparing the masses of the water, X, and the mixture of water and X
D Measuring the electrical conductivities of X and the mixture of water and X
Answer:
D Measuring the electrical conductivities of X and the mixture of water and X
Explanation:
Unfortunately, I am unable to reproduce the table here. However, from the table, the temperature of the of the mixture of the solid X and water was 101.6°C. This is above the boiling point of water and way below the temperature of the solid X.
This goes a long way to suggest that there was some kind of interaction between the water and X which accounted for the observed temperature of the system of X in water.
The only way we can be able to confirm if X actually dissolved in water is to measure the conductivity of the water. dissolved solids increase the conductivity of water.
This is an incomplete question, the table is attached below.
Answer : The correct ranking of the solution from most exothermic to most endothermic will be: A, B and C.
Explanation :
As we know that the intermolecular force of attraction play an important role in the interaction of solute-solute, solute-solvent and solvent solvent solution.
In the solution A, the solute-solute and solvent-solvent interactions are weak. So, their solute-solvent interaction will be strong. That means, the solution will be more exothermic.
In the solution C, the solute-solute and solvent-solvent interactions are strong. So, their solute-solvent interaction will be weak. That means, the solution will be more endothermic.
Thus, the correct ranking of the solution from most exothermic to most endothermic will be: A, B and C.
Answer:
The new molar concentration of CO at equilibrium will be :[CO]=1.16 M.
Explanation:
Equilibrium concentration of all reactant and product:
![[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M](https://tex.z-dn.net/?f=%5BCO_2%5D%20%3D%200.24%20M%2C%20%5BH_2%5D%20%3D%200.24%20M%2C%20%5BH_2O%5D%20%3D%200.48%20M%2C%20%5BCO%5D%20%3D%200.48%20M)
Equilibrium constant of the reaction :
![K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BH_2O%5D%5BCO%5D%7D%7B%5BCO_2%5D%5BH_2%5D%7D%3D%5Cfrac%7B0.48%20M%5Ctimes%200.48%20M%7D%7B0.24%20M%5Ctimes%200.24%20M%7D)
K = 4

Concentration at eq'm:
0.24 M 0.24 M 0.48 M 0.48 M
After addition of 0.34 moles per liter of
and
are added.
(0.24+0.34) M (0.24+0.34) M (0.48+x)M (0.48+x)M
Equilibrium constant of the reaction after addition of more carbon dioxide and water:


Solving for x: x = 0.68
The new molar concentration of CO at equilibrium will be:
[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M