The name for the ionic compound Cs2S is Cesium sulfide.
The water is formed from oxygen gas and...hydrogen gas, I'm assuming? It would have been nice for the question to have been a bit more explicit (not blaming you, of course).
Assuming that's the case, our chemical reaction would be:
2H₂(g) + O₂(g) → 2H₂O(l).
We are told that 1 mol of a gas has a volume of 24.0 dm³ at RTP. We can use this relation to determine the number of moles of O₂ gas that reacts given its initial volume, 33.5 dm³.
33.5 dm³ O₂(g)/24.0 dm³/mol = 1.396 mol O₂(g).
Since we are not given any information about H₂(g), or any other reactant for that matter, I am assuming that the O₂(g) is the limiting reactant. According to the equation, the stoichiometric ratio between O₂ and H₂O is 1:2. That is, for every one mole of O₂ that is consumed, two moles of H₂O are formed (i.e., the number of moles of H₂O formed is double the number moles of O₂).
Since 1.396 moles of O₂ reacts, 2(1.396) = 2.792 moles of H₂O are produced. To convert moles of water to grams, we multiply the number of moles of H₂O by the molar mass of H₂O:
(2.792 moles H₂O)(18.015 g/mol) = 50.3 g H₂O.
So, approximately 50.3 grams of water are formed from 33.5 dm³ of oxygen gas at RTP.
<u>Answer:</u> The heat required for the process is 4.24 kJ
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
Given mass of benzene = 24.8 g
Molar mass of benzene = 78.11 g/mol
Putting values in above equation, we get:
To calculate the enthalpy change of the reaction, we use the equation:
where,
= amount of heat absorbed = ?
n = number of moles = 0.318 moles
= enthalpy change of the reaction = 30.7 kJ/mol
Putting values in above equation, we get:
Hence, the heat required for the process is 4.24 kJ
The answer is D or 4,3,2!