<span>Important information to solve the exercise :
Substance ΔHf (kJ/mol):
HCl(g)= −92.0 </span><span>kJ/mol
Al(OH)3(s)= −1277.0 </span><span><span>kJ/mol
</span> H2O(l)= −285.8 </span><span>kJ/mol
AlCl3(s) =−705.6 </span><span>kJ/mol
</span><span>Al(OH)3(s)+3HCl(g)→AlCl3(s)+3H2O(l)
reactants products
products- reactants:</span><span>
(−705.6) + (3 x −285.8) - ( −1277.0 ) - (3 x −92.0 ) = - 10.0 </span>kJ per mole at 25°C
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</span>
Answer:
If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia
Explanation:
Step 1: Data given
Mass of nitrogen gas (N2) = 13.4 grams
Molar mass of N2 = 28 g/mol
Molar mass of NH3 = 17.03 g/mol
Step 2: The balanced equation
N2 + 3H2 → 2NH3
Step 3: Calculate moles of N2
Moles N2 = Mass N2 / molar mass N2
Moles N2 = 13.4 grams / 28.00 g/mol
Moles N2 = 0.479 moles
Step 4: Calculate moles of NH3
For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3
For 0.479 moles N2 we'll produce 2*0.479 = 0.958 moles
Step 5: Calculate mass of NH3
Mass of NH3 = moles NH3 * molar mass NH3
Mass NH3 = 0.958 moles * 17.03 g/mol
Mass NH3 = 16.3 grams
If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia
Answer: the boiling point is = 137.325°C
Explanation:
From the formula: ∆Tb= Kb*m
From the question, Kb= 0.95, m= 27.5, T1= 111.2°C
Substitute into ∆Tb= Kb*m
∆Tb= 0.95*27.5= 26.125
∆Tb= T2-T1
Hence
T2- 111.2=26.125
T2= 26.125+ 111.2= 137.325°C
Answer:
It’s false
Explanation:
it could be true if the question mentioned alkaline solution
Answer:
The amount of space an object occupies.
Explanation:
