Answer:
The answer to your question is n = 0.838 moles
Explanation:
Data
Volume = 2.15 l
Pressure = 8.91 atm
Temperature = 5.81°C
moles = ?
Gas constant = 0.082 atm l / mol°K
Process
1.- Convert temperature to °K
Temperature = 5.81 + 273 = 278.81°K
2.- Write the Ideal gas formula
PV = nRT
-Solve for moles (n)
n = PV / RT
3.- Substitution
n = (8.91 x 2.15) / (0.082 x 278.81)
4.- Simplification
n = 19.16 / 22.86
5.- Result
n = 0.838 moles
Answer:
Have no valence electrons because they are stable.
Thank you.
BY GERALD GREAT
An atom whose valence electron shell is nearly full is highly reactive, if the atom requires only 1 and or 2 electrons to complete its octet and or have 8 valence electrons in its electron shell.
<u>Answer:</u> The activation energy of the reverse reaction is 47 kJ/mol
<u>Explanation:</u>
The chemical equation for the decomposition of dinitrogen pentaoxide follows:
We are given:
Activation energy of the above reaction (forward reaction) = 102 kJ/mol
Enthalpy of the reaction = +55 kJ/mol
As, the enthalpy of the reaction is positive, the reaction is said to be endothermic in nature.
To calculate the activation energy for the reverse reaction, we use the equation:
where,
= Activation energy of the forward reaction = 102 kJ/mol
= Activation energy of the backward reaction = ?
= Enthalpy of the reaction = +55 kJ/mol
Putting values in above equation, we get:
Hence, the activation energy of the reverse reaction is 47 kJ/mol
Answer:
The answer to your question is 58.5 g of NaCl
Explanation:
Data
mass of NaCl = ?
mass of HCl = 40 g
mass of NaOH = 40 g
Balanced chemical reaction
HCl + NaOH ⇒ H₂O + NaCl
-Calculate the molar mass of the reactants
HCl = 1 + 35.5 = 36.5 g
NaOH = 23 + 16 + 1 = 40 g
-Calculate the limiting reactant
theoretical proportion = HCl/NaOH = 36.5 / 40 = 0.9125
experimental proportion = HCl / NaOH = 40 / 40 = 1
As the experimental proportion was higher than the theoretical proportion, the limiting reactant is NaOH.
-Calculate the molar mass of NaCl
NaCl = 23 + 35.5 = 58.5
-Calculate the experimental mass of NaCl
40 g of NaOH -------------------- 58.5 g of NaCl
40 g of NaOH ------------------- x
x = (40 x 58.5) / 40
x = 58.5 g of NaCl