Answer:
Percent by mass of water is 56%
Explanation:
First of all calculate the mass of hydrated compound as,
Mass of Sodium = Na × 2 = 22.99 × 1 = 45.98 g
Mass of Sulfur = S × 1 = 32.06 × 1 = 32.06 g
Mass of Oxygen = O × 14 = 16 × 14 = 224 g
Mass of Hydrogen = H × 20 = 1.01 × 20 = 20.2 g
Mass of Na₂S0₄.10H₂O = 322.24 g
Secondly, calculate mass of water present in hydrated compound. For this one should look for the coefficient present before H₂O in molecular formula of hydrated compound. In this case the coefficient is 10, so the mass of water is...
Mass of water = 10 × 18.02
Mass of water = 180.2 g
Now, we will apply following formula to find percent of water in hydrated compound,
%H₂O = Mass of H₂O / Mass of Hydrated Compound × 100
Putting values,
%H₂O = 180.2 g / 322.24 g × 100
%H₂O = 55.92 % ≈ 56%
Answer is: 39,083kJ.
m(coal) = 2,00g.
m(water) = 500g.
ΔT = 43,7°C - 25°C = 18,7°C, <span>difference at temperatures.</span>
c(water) = 4,18 J/g·°C, <span>specific heat of water
</span>Q = m(water)·ΔT·c(water), heat of reaction.
Q = 500g·18,7°C·4,18J/g·°C.
Q = 39083J = 39,083kJ.
Answer:
Explanation:
The first one is CrO. The Chromium has the same charge as the oxygen so mol numbers are dropped.
The Second one is CrO2 The two oxygens have a charge of 2(-2) = -4. To balance this, the Chromium must have a charge of +4 Cr(Iv)O2
The third one is can be set up like this
Cr + 3(-2) = 0
Cr - 6 = 0
Cr = 6
Therefore the formula is Cr(vi)O3
The last one is a bit tricky. Follow this carefully. There are 2 Crs and 3Os.
The formula looks like this
2Cr + 3(-2) = 0
2Cr - 6 = 0
2Cr = 6
Cr = 3
The formula is Cr(iii)2 O3
The swimming pools pH is below 7, meaning it is slightly acidic. If you want to make the pH higher, you must add a base which by definition has a pH higher than 7.
D. Add base
A: Trial 1, because the average rate of the reaction is lower.
The rate of reaction is the speed with which reactants are converted into products. It is also the rate at which reactants disappear and products appear. The higher the rate of reaction, the greater the amount of product formed in a reaction.
If we look at the graph, we will realize that trial 1 produces a lesser amount of product than trial 2. This implies that the average rate of the reaction in trial 1 is lower than in trial 2.
Lower average rate of reaction implies lower concentration of the reactants since the rate of reaction depends on the concentration of reactants.
Hence trial 1 has a lower concentration of reactants because the average rate of the reaction is lower.
