Answer:
Na₂CO₃ · 10H₂O
Explanation:
The formula for sodium carbonate hydrate is:
Na₂CO₃ · xH₂O
The unknown "x" is the number of water molecules contained in the hydrate.
To find "x" we have to use the hydrogen percentage in the sample, 7.05 % H.
First we calculate the molecular weight of Na₂CO₃ · xH₂O:
molecular weight of Na₂CO₃ · xH₂O = 23 × 2 + 12 + 16 × 3 + 18x
molecular weight of Na₂CO₃ · xH₂O = 106 + 18x g/mole
Now we devise the fallowing reasoning tanking in account 1 mole of Na₂CO₃ · xH₂O:
if in 106 + 18x grams of Na₂CO₃ · xH₂O we have 2x grams of hydrogen
then in 100 grams of Na₂CO₃ · xH₂O we have 7.05 grams of hydrogen
106 + 18x = (100 × 2x) / 7.05
106 + 18x = 28.4x
106 = 28.4x - 18x
106 = 10.4x
x = 106 / 10.4
x = 10.2 ≈ 10
The formula for the washing soda is Na₂CO₃ · 10H₂O.
The k is the proportionality constant of the reaction. Graphically, this is the slope of the graph. Since the graph is linear, then there is only 1 value of k. To calculate this, choose two random points in the line. Suppose we use (0.15,10) and (0.30,20), calculate for the slope.
Slope = k = (10 - 20)/(0.15 - 0.30) = 66.67 mL CO₂/g CaCO₃
Answer:
59.2 grams
Explanation:
We are given that 70.4% of the weight of the total 200 g of the concentration is made up of nitric acid, the remaining information is not required to solve the problem. Since water and nitric acid are the only components of the solution, the total weight of water is given by:

There are 59.2 grams of water in this solution.
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
Answer:
A --- (E)-oct-2-en-1-o1
B ----(E)-oct-2-enal
Explanation:
See the attached file for the structure.