Answer: 2) 2HCl(sq) + CaCO3(s) CaCl2(sq) + CO2(g) + H2O (l) No of moles of CaCO3 = amount of the CaCO3 (g)/mw of CaCO3 (g/mole)= 0.8085 g/100 g/mole = 0.008085
Explanation:
At the end of the reaction, the catalyst is UNCHANGED.
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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:
T2 = 260 K
Explanation:
<em>Given data:</em>
P1 = 150.0 k Pa
T1 = (-23+ 273.15) K = 250.15 K
V1 = 1.75 L
P2 = 210.0 kPa
V2 = 1.30 L
<em>To find:</em>
T2 = ?
<em>Formula:</em>


<em>Calculation:</em>
T2 = (210.0 kPa) x (1.30 L) x (250.15 K) / (150.0 kPa) x (1.75 L)
T2 = 260 K