Answer:
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Explanation:
Answer:
3.84 Ω
Explanation:
From the question given above, the following data were obtained:
Electrical power (P) = 150 W
Voltage (V) = 24 V
Resistance (R) =?
P = IV
Recall:
V = IR
Divide both side by R
I = V/R
P = V/R × V
P = V² / R
Where:
P => Electrical power
V => Voltage
I => Current
R => Resistance
With the above formula (i.e P = V²/R), we can calculate resistance as illustrated below:
Electrical power (P) = 150 W
Voltage (V) = 24 V
Resistance (R) =?
P = V²/R
150 = 24² / R
150 = 576 / R
Cross multiply
150 × R = 576
Divide both side by 150
R = 576 / 150
R = 3.84 Ω
Thus, the resistance is 3.84 Ω
15.63 mol. You need 15.63 mol HgO to produce 250.0 g O_2.
<em>Step 1</em>. Convert <em>grams of O_2 to moles of O_2</em>
Moles of O_2 = 250.0 g O_2 × (1 mol O_2/32.00 g O_2) = 7.8125 mol O_2
<em>Step 2</em>. Use the molar ratio of HgO:O_2 to convert <em>moles of O_2 to moles of HgO
</em>
Moles of HgO = 0.8885 mol O_2 × (2 mol HgO/1 mol O_2) = <em>15.63 mol HgO</em>
<h2>
Answer: 6 moles</h2>
<h3>
Explanation:</h3>
3 H₂ + N₂ → 2 NH₃
↓ ↓
4 mol 3 mol
Since the moles of N₂ is the smaller of the two reactants, then N₂ is the limiting factor (the reactant that will decide how much ammonia is produced since it has the smaller amount of moles). ∴ we have to use it in calculating the number of moles of ammonia
The mole ratio of N₂ to NH₃ based on the balanced equation is 1 to 2.
∴ the moles of NH₃ = moles of N₂ × 2
= 3 moles × 2
= 6 moles