Answer:
4.7 kJ/kmol-K
Explanation:
Using the Debye model the specific heat capacity in kJ/kmol-K
c = 12π⁴Nk(T/θ)³/5
where N = avogadro's number = 6.02 × 10²³ mol⁻¹, k = 1.38 × 10⁻²³ JK⁻¹, T = room temperature = 298 K and θ = Debye temperature = 2219 K
Substituting these values into c we have
c = 12π⁴Nk(T/θ)³/5
= 12π⁴(6.02 × 10²³ mol⁻¹)(1.38 × 10⁻²³ JK⁻¹)(298 K/2219 K)³/5
= 9710.83(298 K/2219 K)³/5
= 1942.17(0.1343)³
= 4.704 J/mol-K
= 4.704 × 10⁻³ kJ/10⁻³ kmol-K
= 4.704 kJ/kmol-K
≅ 4.7 kJ/kmol-K
So, the specific heat of diamond in kJ/kmol-K is 4.7 kJ/kmol-K
Answer: 4.7432 L
Explanation:
Use stoichiometry: .4235 mol CuCl2 (1 mol I2 / 2 mol CuCl2)(22.4 L / 1 mol I2) = 4.7432 L :)
Answer:
You didn't show which element it is. The proton is the atomic number, the electron is the same number of protons, and the neutron is the atomic mass rounded to the nearest whole number minus the proton.
Explanation:
100ml volume of 0.0150m hcl solution is requires to titrate 150ml of a 0.0100m caoh2 solution.
Dilution is a solution of decreasing the concentration of a solute in the solution by adding more solvent to the solution. We can use the expression for dilute formula,
C1 V1 =C2 V2
where C1 is the initial concentration,C2 is the final concentration,V1 is the initial volume and V2 is the final volume. Here given, volume of 0.0150M(C1) HCL solution is required to titrate 150ml(V2) of a 0.0100M(C1) Caoh2 solution.
While diluting a solution from a high concentration substance to a low concentration substance we always use the formula of dilution.so, putting all value give in the expression we get the volume of the final concentration.
V1= C2 V2/ C1
= 0.0100m . 150ml /0.0150M
= 100ml
The volume of the hcl solution is 100ml.
To learn more about dilution formula please visit:
brainly.com/question/7208939
#SPJ4
The balanced equation for the above reaction is;
2K + Cl₂ ---> 2KCl
Stoichiomtery of K to KCl is 2:2
Potassium is the limiting reactant which is fully consumed in the reaction. The amount of product formed depends on amount of limits reactant present.
Number of moles of K reacted - 6.75 g/ 39 g/mol = 0.17 mol
Therefore number of KCl moles formed - 0.17 mol
Mass of KCl formed - 0.17 mol x 74.5 g/mol = 12.67 g
