NH4I (aq) + KOH (aq) in chemical equation gives
NH4I (aq) + KOH (aq) = KI (aq) + H2O(l) + NH3 (l)
Ki is in aqueous state H2o is in liquid state while NH3 is in liquid state
from the equation above 1 mole of NH4I (aq) react with 1 mole of KOH(aq) to form 1mole of KI(aq) , 1mole of H2O(l) and 1 Mole of NH3(l)
Basis: 100 mL solution
From the given density, we calculate for the mass of the solution.
density = mass / volume
mass = density x volume
mass = (1.83 g/mL) x (100 mL) = 183 grams
Then, we calculate for the mass H2SO4 given the percentage.
mass of H2SO4 = (183 grams) x (0.981) = 179.523 grams
Calculate for the number of moles of H2SO4,
moles H2SO4 = (179.523 grams) / (98.079 g/mol)
moles H2SO4 = 1.83 moles
Molarity:
M = moles H2SO4 / volume solution (in L)
= 1.83 moles / (0.1L ) = 18.3 M
Molality:
m = moles of H2SO4 / kg of solvent
= 1.83 moles / (183 g)(1-0.983)(1 kg/ 1000 g) = 588.24 m
Find moles of MgSO4.7H2O
molar mass = 246
so moles = 32 / 246 = 0.13 moles.
When heated, all 7 H2O from 1 molecule will be gone.
total moles of H2O present = 7 x 0.13 = 0.91
mass of those H2O = 0.91 x 18 = 16.38g
so mass of anyhydrous MgSO4 remain = 32 - 16.38 = 15.62 g
Given reaction represents dissociation of bromine gas to form bromine atoms
Br2(g) ↔ 2Br(g)
The enthalpy of the above reaction is given as:
ΔH = ∑n(products)Δ
- ∑n(reactants)Δ
where n = number of moles
Δ
= enthalpy of formation
ΔH = [2*ΔH(Br(g)) - ΔH(Br2(g))] = 2*111.9 - 30.9 = 192.9 kJ/mol
Thus, enthalpy of dissociation is the bond energy of Br-Br = 192.9 kJ/mol
Answer:
b. 186 g
Explanation:
Step 1: Write the balanced equation.
4 NH₃(g) + 6 NO(g) → 5 N₂(g) + 6 H₂O(l)
Step 2: Calculate the moles corresponding to 145 g of N₂
The molar mass of nitrogen is 28.01 g/mol.
Step 3: Calculate the moles of NO required to produce 5.18 moles of N₂
The molar ratio of NO to N₂ is 6:5.
Step 4: Calculate the mass corresponding to 6.22 moles of NO
The molar mass of NO is 30.01 g/mol.
