The Answer would be e: 3.2 g/L
<span>At STP : p = 1 atm and T = 273.15 K
molar mass Cl2 = 70.906 g/mol
d = molar mass x p/RT = 70.906 x 1 atm / 0.08206 x 273.15=3.16 g/L</span>
Answer:
Explanation:
We can use the Ideal Gas Law to calculate the density of the gas.
pV = nRT
n = m/M Substitute for n
pV = (m/M)RT Multiply both sides by M
pVM = mRT Divide both sides by V
pM = (m/V) RT
ρ = m/V Substitute for m/V
pM = ρRT Divide each side by RT
Data:
p = 1.00 bar
M = 49 g/mol
R = 0.083 14 bar·L·K⁻¹mol⁻¹
T = 0 °C = 273.15 K
Calculation:
ρ = (1.00 × 49)/(0.083 14 × 273.15) = 2.2 g/L
The density of the gas is .
The substitution product that is supposed to be favoured. The product with retention of configuration is expected to predominate slightly because the leaving group blocks the nucleophile.
<h3>What is a substitution product?</h3>
A substitution product is a product that is gotten from a reaction in which a group of reactants is replaced by another group in the reaction.
Therefore, the substitution product is supposed to be favoured. The product with retention of configuration is expected to predominate slightly because the leaving group blocks the nucleophile.
Learn more about substitution reaction below.
brainly.com/question/10143438
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<u>Given:</u>
Mass of Ag = 1.67 g
Mass of Cl = 2.21 g
Heat evolved = 1.96 kJ
<u>To determine:</u>
The enthalpy of formation of AgCl(s)
<u>Explanation:</u>
The reaction is:
2Ag(s) + Cl2(g) → 2AgCl(s)
Calculate the moles of Ag and Cl from the given masses
Atomic mass of Ag = 108 g/mol
# moles of Ag = 1.67/108 = 0.0155 moles
Atomic mass of Cl = 35 g/mol
# moles of Cl = 2.21/35 = 0.0631 moles
Since moles of Ag << moles of Cl, silver is the limiting reagent.
Based on reaction stoichiometry: # moles of AgCl formed = 0.0155 moles
Enthalpy of formation of AgCl = 1.96 kJ/0.0155 moles = 126.5 kJ/mol
Ans: Formation enthalpy = 126.5 kJ/mol
Hey there!
Given the reaction:
4 AI + 3O2 ------> 2 AI2O3
4 moles Al --------- 3 moles O2
9.30 moles Al --------- ?? moles O2
9.30 * 3 / 4 => 6.975 moles of O2
Molar mass O2 => 32.0 g
Therefore:
6.975 * 32.0 => 223.3 g of Oxygen
Hope that helps!
