Answer: Mass of 
required to form 930 kg of iron is 1328 kg
Explanation:
To calculate the number of moles, we use the equation:
.....(1)
For iron:
Given mass of iron = 930 kg = 930000 g (1kg=1000g)
Molar mass of iron = 56 g/mol
Putting values in equation 1, we get:
The chemical equation for the production of iron follows:
By Stoichiometry of the reaction:
2 moles of iron are produced by = 1 mole of
So, 16607 moles of iron will be produced by = 
of
Now, calculating the mass of from equation 1, we get:
Mass of = 
Thus mass of required to form 930 kg of iron is 1328 kg
Answer:
0.83 mL
Explanation:
Given data
- Initial concentration (C₁): 12 M
- Final concentration (C₂): 1.0 M
- Final volume (V₂): 10.0 mL
We can calculate the initial volume of HCl using the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 1.0 M × 10.0 mL / 12 M
V₁ = 0.83 mL
The required volume of the initial solution is 0.83 mL.
Answer:
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Answer:
Q = 90,000 J
Explanation:
Given data:
Mass skillet = 2000 g
Specific heat capacity = 0.450 J/g.°C
Energy required to raise temperature = ?
Initial temperature = 25°C
Final temperature = 125°C
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 125°C - 25°C
ΔT = 100°C
Q = 2000 g × 0.450 J/g.°C × 100°C
Q = 90,000 J
Answer:
6.78 × 10⁻³ L
Explanation:
Step 1: Write the balanced equation
Mg₃N₂(s) + 3 H₂O(g) ⇒ 3 MgO(s) + 2 NH₃(g)
Step 2: Calculate the moles corresponding to 10.2 mL (0.0102 L) of H₂O(g)
At STP, 1 mole of H₂O(g) has a volume of 22.4 L.
0.0102 L × 1 mol/22.4 L = 4.55 × 10⁻⁴ mol
Step 3: Calculate the moles of NH₃(g) formed from 4.55 × 10⁻⁴ moles of H₂O(g)
The molar ratio of H₂O to NH₃ is 3:2. The moles of NH₃ produced are 2/3 × 4.55 × 10⁻⁴ mol = 3.03 × 10⁻⁴ mol.
Step 4: Calculate the volume corresponding to 3.03 × 10⁻⁴ moles of NH₃
At STP, 1 mole of NH₃(g) has a volume of 22.4 L.
3.03 × 10⁻⁴ mol × 22.4 L/mol = 6.78 × 10⁻³ L
