Answer:
Average atomic mass of carbon = 12.01 amu.
Explanation:
Given data:
Abundance of C¹² = 98.89%
Abundance of C¹³ = 1.11%
Atomic mass of C¹² = 12.000 amu
Atomic mass of C¹³ = 13.003 amu
Average atomic mass = ?
Solution:
Average atomic mass of carbon = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass of carbon = (12.000×98.89)+(13.003×1.11) /100
Average atomic mass of carbon= 1186.68 + 14.43333 / 100
Average atomic mass of carbon = 1201.11333 / 100
Average atomic mass of carbon = 12.01 amu.
count each atom of the reactants and the products
Answer : The value of reaction quotient, Q is 0.0625.
Solution : Given,
Concentration of 
= 2.00 M
Concentration of 
= 2.00 M
Concentration of 
= 1.00 M
Reaction quotient : It is defined as a concentration of a chemical species involved in the chemical reaction.
The balanced equilibrium reaction is,
The expression of reaction quotient for this reaction is,
![Q=\frac{[Product]^p}{[Reactant]^r}\\Q=\frac{[NH_3]^2}{[N_2]^1[H_2]^3}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BProduct%5D%5Ep%7D%7B%5BReactant%5D%5Er%7D%5C%5CQ%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5E1%5BH_2%5D%5E3%7D)
Now put all the given values in this expression, we get
Therefore, the value of reaction quotient, Q is 0.0625.
Answer:
D
Explanation:
I explained how to do it on your other problem so look there :)
Answer:
-177.9 kJ.
Explanation:
Use Hess's law. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2Ca(s) + O2(g) → 2CaO(s) ΔH = -1269.8 kJ We need to get rid of the Ca and O2 in the equations, so we need to change the equations so that they're on both sides so they "cancel" out, similar to a system of equations. I changed the second equation. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ The sign changes in the second equation above since the reaction changed direction. Next, we need to multiply the first equation by two in order to get the coefficients of the Ca and O2 to match those in the second equation. We also multiply the enthalpy of the first equation by 2. 2Ca(s) + 2CO2(g) + O2(g) → 2CaCO3(s) ΔH = -1625.6 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ Now we add the two equations. The O2 and 2Ca "cancel" since they're on opposite sides of the arrow. Think of it more mathematically. We add the two enthalpies and get 2CaO(s) + 2CO2(g) → 2CaCO3(s) and ΔH = -355.8 kJ. Finally divide by two to get the given equation: CaO(s) + CO2(g) → CaCO3(s) and ΔH = -177.9 kJ.
