The number of formula units in 2.50 mol of the compound is 15.1 * 10^23.
The question is unclear whether NaNO2 or NaNO3 is implied. However,in either case, the solution applies equally.
6.02 * 10^23 formula units of the compound are contained in 1 mole
x formula units are contained in 2.5 moles of the compound
x = 6.02 * 10^23 formula units * 2.5 moles/ 1 mole
x = 15.1 * 10^23 formula units of the compound.
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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 : The partial pressure of
at equilibrium is, 1.0 × 10⁻⁶
Explanation :
The partial pressure of
= 
The partial pressure of
= 
The partial pressure of
= 

The balanced equilibrium reaction is,

Initial pressure 1.0×10⁻² 2.0×10⁻⁴ 2.0×10⁻⁴
At eqm. (1.0×10⁻²-2p) (2.0×10⁻⁴+p) (2.0×10⁻⁴+p)
The expression of equilibrium constant
for the reaction will be:

Now put all the values in this expression, we get :


The partial pressure of
at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶
Therefore, the partial pressure of
at equilibrium is, 1.0 × 10⁻⁶
The answer is D; Mass and Velocity