Answer:
H₂: 0.48, N₂: 0.43; Ar: 0.09
Explanation:
First of all, sum all the pressures to know the total pressure in the mixture.
434 Torr + 389.9 Torr + 77.9 Torr = 901.8 Torr
Mole fraction = Pressure gas / Total Pressure
Mole Fraction H₂: 434 Torr /901.8 Torr = 0.48
Mole Fraction N₂: 389.9 /901.8 Torr =0.43
Mole Fraction Ar: 77.9 /901.8 Torr = 0.09
Remember: <u>SUM OF MOLE FRACTION = 1</u>
Answer:
3.94 L
Explanation:
From the question given above, the following data were obtained:
Mass of O₂ = 5.62 g
Volume of O₂ =?
Next, we shall determine the number of mole present in 5.62 g of O₂. This can be obtained as follow:
Mass of O₂ = 5.62 g
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mole of O₂ =?
Mole = mass / molar mass
Mole of O₂ = 5.62 / 32
Mole of O₂ = 0.176 mole
Finally, we shall determine the volume of 5.62 g (i.e 0.176 mole) of O₂ at STP. This can be obtained as follow:
1 mole of O₂ occupied 22.4 L at STP.
Therefore, 0.176 mole of O₂ will occupy = 0.176 × 22.4 = 3.94 L at STP.
Thus 5.62 g (i.e 0.176 mole) of O₂ occupied 3.94 L at STP
Answer:
52.2 g
Explanation:
Step 1: Write the balanced equation
3 KOH + H₃PO₄ ⟶ K₃PO₄ + 3 H₂O
Step 2: Calculate the moles corresponding to 89.7 g of KOH
The molar mass of KOH is 56.11 g/mol.
89.7 g × 1 mol/56.11 g = 1.60 mol
Step 3: Calculate the moles of H₃PO₄ needed to react with 1.60 moles of KOH
The molar ratio of KOH to H₃PO₄ is 3:1. The moles of H₃PO₄ needed are 1/3 × 1.60 mol = 0.533 mol.
Step 4: Calculate the mass corresponding to 0.533 moles of H₃PO₄
The molar mass of H₃PO₄ is 97.99 g/mol.
0.533 mol × 97.99 g/mol = 52.2 g
<span>Involuntary Muscles
</span>⇒ Involuntary Muscles <span>are muscles that contract without any conscious control.
The most important </span><span>Involuntary Muscle in the body is the heart. </span>
