Answer:
The 2292 moles of CO are needed to react completely with 122 Kg of Fe₂O₃.
Explanation:
Given data:
Mass of Fe₂O₃ = 122 Kg ( 122×1000 = 122000 g)
Moles of CO = ?
Solution:
Chemical equation:
Fe₂O₃ + 3CO → 2Fe + 3CO₂
Number of moles of Fe₂O₃:
Number of moles = mass/ molar mass
Number of moles = 122000 g /159.69 g/mol
Number of moles = 764 mol
Now we will compare the moles of Fe₂O₃ with CO.
Fe₂O₃ : CO
1 : 3
764 : 3×764 =2292 mol
The 2292 moles of CO are needed to react completely with 122 Kg of Fe₂O₃.
Answer:
Qm = -55.8Kj/mole
Explanation:
NaOH(aq) + HNO₃(aq) => NaNO₃(aq) + H₂O(l)
Qm = (mc∆T)water /moles acid
Given => 100ml(0.300M) NaOH(aq) + 100ml(0.300M)HNO₃(aq)
=> 0.03mole NaOH(aq) + 0.03mole HNO₃(aq)
=> 0.03mole NaNO₃(aq) + 0.03mole H₂O(l)
ΔH⁰rxn = [(200ml)(1.00cal/g∙°C)(37 – 35)°C]water / 0.03mole HNO₃
= 13,333 cal/mole x 4.184J/cal = 55,787J/mol = 55.8Kj/mole (exothermic)*
Heat of reactions comes from formation of H-Oxy bonds on formation of water of reaction and heats the 200ml of solvent water from 35⁰C to 37⁰C.
Answer:
0.50 M
Explanation:
Given data
- Mass of sodium sulfate (solute): 7.1 g
- Volume of solution: 100 mL
Step 1: Calculate the moles of the solute
The molar mass of sodium sulfate is 142.04 g/mol. The moles corresponding to 7.1 grams of sodium sulfate are:
Step 2: Convert the volume of solution to liters
We will use the relation 1 L = 1000 mL.
Step 3: Calculate the molarity of the solution
Aromatic compound has continuous cyclic structure with( 4n+2)π electrons (Huckels rule), where n = 0,1,2…
Here number of pi electron are 6, where 4 from two double bond and 2 from nitrogen non-bonding electrons, hence it has total 6 pi electrons, therefore
6= ( 4n+2)π
4 = 4n
n =1
Hence it is an aromatic compound
Answer:
c) 387g
Explanation:
Water;
Mass = 250g
Specific heat = 4.184
Initial Temp, T1 = 25 + 273 = 298K
Final Temp, T2 = 35 + 273 = 308K
Heat = ?
H = mc(T2 - T1)
H = 250 * 4.184 (308 - 298)
H = 10460 J
Iron;
Initial Temp, T2 = 95 + 273 = 368K (Upon converting to kelvin temperature)
Mass = ?
Final Temp, T1 = 35 + 273 = 308
Heat = 10460 (Heat lost by iron is qual to heat gained by water)
Specific heat = 0.45
H = mc(T2-T1)
M = 10460 / [0.45 (308 - 368)]
M = 10460 / 27
M = 387g
