Ammonia undergoes combustion with oxygen to produce nitric oxide and water. The volume of the oxygen required to react with 720 ml of ammonia is 900 ml.
<h3>What is volume?</h3>
Volume is the area occupied by the substance and is the ratio of the mass to the density.
At STP, 1 mole of gas occupies 22.4 L of volume
Given,
Volume of ammonia reacted = 0.720 L
The combustion reaction is shown as,
From the stoichiometry of the reaction, it can be said that,
L of ammonia reacts with 
L of oxygen gas.
So, 0.720 L of ammonia will react with:
Therefore, the volume of oxygen required is 900 mL.
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Answer:
semipermeability
Explanation:
partially but not freely or wholly permeable specifically : permeable to some usually small molecules but not to other usually larger particles a semipermeable membrane.
Answer:
Explanation:
Hello,
In this case, since the molarity is computed by:
Whereas the solute is the hydrochloric acid, we compute the corresponding moles with its molar mass (36.45 g/mol):
Next, since the solution contains both HCl and water, we compute the volume in liters by using its density:
Therefore, the molarity turns out:
Regards.
Since hydrogen bonding is a stronger intermolecular force than van der Waals forces, more energy is required to separate the molecules of ethanol than the molecules of ethane. Thus ethanol has a higher melting point than ethane.
Answer:
1.88 × 10²³ particles
Explanation:
Given data:
Volume of H₂ = 0.7 L
Number of particles at STP = ?
Solution:
First of all we will calculate the number of moles of H₂.
PV = nRT
n = PV / RT
n = 1 atm . 7 L / 0.0821 L. atm. mol⁻¹. k⁻¹ × 273.15 K
n = 7 atm. L / 22.43 L. atm. mol⁻¹
n = 0.312 mol
it is known that,
2 g H₂ = 1 mol = 6.022 × 10²³ particles
For 0.312 mol
0.312 × 6.022 × 10²³ particles
1.88 × 10²³ particles
