Answer:
e−(Ea/RT): the fraction of the molecules present in a gas which have energies equal to or in excess of activation energy at a particular temperature
Answer:
Single replacement and Zinc Sulfate
the 2nd one is double replacement and potassium nitrate
Answer:
3
Explanation:
The mass has 4 sig digits.
The volume has 3 sig digits
The density is = 11.50/9.03 = 1.2735
To 3 sig digits (determined by 9.03) the answer would be 1.27 gram/mL
Answer:
ΔH = -470.4kJ
Explanation:
It is possible to sum 2 or more reactions to obtain the ΔH of the reaction you want to study (Hess's law). Using the reactions:
1. CaC2(s) + 2H2O(l) → C2H2(g) + Ca(OH)2(s)ΔH = −414kJ
2. 6C2H2(g) + 3CO2(g) + 4H2O(g) → 5CH2CHCO2H(g)ΔH = 132kJ
6 times the reaction 1.
6CaC2(s) + 12H2O(l) → 6C2H2(g) + 6Ca(OH)2(s)ΔH = −414kJ*6 = -2484kJ
This reaction + 2:
6CaC2(s) + 3CO2(g) + 16H2O(l) → + 6Ca(OH)2(s) + 5CH2CHCO2H(g) ΔH = -2484kJ + 132kJ = -2352kJ
As we want to calculate the net change enthalpy in the formation of just 1 mole of acrylic acid we need to divide this last reaction in 5:
6/5CaC2(s) + 3/5CO2(g) + 16/5H2O(l) → + 6/5Ca(OH)2(s) + CH2CHCO2H(g) ΔH = -2352kJ / 5
<h3>ΔH = -470.4kJ</h3>
Answer:
Explanation:
Assume you are using 1 L of water.
Then you are washing 4 L of salty oil.
1. Calculate the mass of the salty oil
Assume the oil has a density of 0.86 g/mL.
2. Calculate the mass of salt in the salty oil
3. Calculate the mass of salt in the spent water
4. Mass of salt remaining in washed oil
Mass = 172 g - 150 g = 22 g
5. Concentration of salt in washed oil
