Answer:
884.56 torr
Explanation:
Formula: 
P = Pressure
V = Volume
T = Temperature in kelvin (Celsius + 273.15)


P = 884.56169
1) Write the balanced equation to state the molar ratios:
<span>3H2(g) + N2(g) → 2NH3(g)
=> molar ratios = 3 mol H2 : 1 mol N2 : 2 mol NH3
What volume of nitrogen is needed to produce 250.0 L of ammonia gas at STP?
First, convert the 250.0 L of NH3 to number of moles at STP .
Use the fact that 1 mole of gas at STP occupies 22.4 L
=> 250.0 L * 1mol/22.4 L = 11.16 L
Second, use the molar ratio to find the number of moles of N2 that produces 11.16 L of NH3
=> 11.16 L NH3 * [1 mol N2 / 2 mol NH3] = 5.58 mol N2
Third, convert 5.58 mol N2 into liters at STP
=> 5.58 mol N2 * [22.4 L/mol] = 124.99 liters
Answer: 124,99 liters
What volume of hydrogen is needed to produce 2.50 mol NH3 at STP?
First, find the number of moles of H2 that produce 2.50 mol by using the molar ratios:
2.50 mol NH3 * [3mol H2 / 2 mol NH3] = 3.75 mol H2
Second, convert the number of moles to liters of gas at STP:
3.75 mol * 22.4 L/mol = 84 liters of H2
Answer: 84 liters
</span>
Answer:
Hydrogen: -141 kJ/g
Methane: -55kJ/g
The energy released per gram of hydrogen in its combustion is higher than the energy released per gram of methane in its combustion.
Explanation:
According to the law of conservation of the energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.
Qc + Qb = 0
Qc = -Qb [1]
We can calculate the heat absorbed by the bomb calorimeter using the following expression.
Q = C . ΔT
where,
C is the heat capacity
ΔT is the change in the temperature
<h3>Hydrogen</h3>
Qc = -Qb = -C . ΔT = -(11.3 kJ/°C) . (14.3°C) = -162 kJ
The heat released per gram of hydrogen is:

<h3>Methane</h3>
Qc = -Qb = -C . ΔT = -(11.3 kJ/°C) . (7.3°C) = -82 kJ
The heat released per gram of methane is:

Answer
× 10²³ molecules are in 41.8 g of sulfuric acid
Explanation
The first step is to convert 41.8 g of sulfuric acid to moles by dividing the mass of sulfuric acid by its molar mass.
Molar mass of sulfuric acid, H₂SO₄ = 98.079 g/mol

Finally, convert the moles of sulfuric acid to molecules using Avogadro's number.
Conversion factor: 1 mole of any substance = 6.022 × 10²³ molecules.
Therefore, 0.426187053 moles of sulfuric acid is equal

Thus, 2.57 × 10²³ molecules are in 41.8 g of sulfuric acid.
The answer is D, reactant.