88 Grams because the principal of conservation of mass states total mass in equals total out so 200 in 122+88=200 out
Answer:
20 mL OF 6 M HYDROCHLORIC ACID WILL BE NEEDED
Explanation:
M1 V1 = M2 V2
M1 = Molarity of sodium hydroxide = 3 M
V1 = volume of sodium hydroxide = 40 mL
M2 = Molarity of hydrochloric acid = 6 M
V2 = Volume of hydrochloric acid = unknown
Rearranging the equation, we have:
V2 = M1 V1 / M2
V2 = 3 * 40 mL / 6
V2 = 120 / 6
V2 = 20 mL
To precipitate the benzoic acid by 6 M of hydrochloric acid, 20 mL volume will be needed.
This is a tricky question. a mole of any compound contains the same number of molecules of that certain compound. so, one mole of chlorine gas has the same number of molecules as one mole of glucose, which is 6.02 x 10^23.
this is avogadro's number and it applies for any mole of molecules.
the question is tricky because it is like asking. " what weighs more, a pound of feathers or a pound of rocks?" the both weigh the same, a pound. when ewe talking about moles, same as pounds, it is a quantity unit. one mole will aways be equal to 6.02 x 10^23 molecules.
Answer:
6ΔG°(f) H₂O = -229 Kj/mol
Explanation:
4NH₃(g) + 5O₂(g) => 4NO(g) + 6H₂O(g)
ΔG°(f) 4mol(-16.66Kj/mol) | 5mol(0Kj/mol) || 4mol(+86.71Kj/mol) | 6ΔG°(f) H₂O
Hess's Law
ΔG°(Rxn) = ∑ΔG°(f) Products - ∑ΔG°(f) Reactants
-957.9 Kj = [(4mol(+86.71Kj/mol)) + 6ΔG°(f) H₂O(g)] - [4mol(-16.66Kj/mol) + 5mol(0Kj/mol)]
-957.9 Kj = [4(86.7)Kj + 6ΔG°(f) H₂O] - [4(-16.66)Kj] = 346.84Kj + 6ΔG°(f) H₂O + 66.64Kj
ΔG°(f) H₂O = ((-957.9 - 346.84 -66.64)/6)Kj = -228.56 Kj ≅ -228.6 Kj*
*Verified with Standard Heat of Formation Table
Answer:
10.81, Boron
Explanation:
I love the explanation Khan Academy gives.
