The amount of heat that could be removed by 20.0 g of ethyl chloride is 8.184 kJ.
<h3>How do we calculate required heat?</h3>
Required amount of heat which can be removed for the vaporization will be calculated as:
Q = (n)(ΔHv), where
- n = moles of ethyl chloride
- ΔHv = heat of vaporization = 26.4 kj/mol
Moles will be calculated as:
n = W/M, where
- W = given mass of ethyl chloride = 20g
- M = molar mass of ethyl chloride = 64.51 g/mol
n = 20 / 64.51 = 0.31 mol
On putting all these values in the above equation, we get
Q = (0.31)(26.4) = 8.184 kJ
Hence involved amount of heat is 8.184 kJ.
To know more about heat of vaporization, visit the below link:
brainly.com/question/13106213
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Hydrogen burns with oxygen as in combustion reaction to form water.
H₂ + O₂ → H₂O
For the answer to the questions above,
a) Ag2CO3(s) => Ag2O(s)+CO2(g)
<span>b) Cl2(g)+2(KI)(aq) => I2(s)+2(KCl)(aq) (coefficients are for balanced equation) </span>
<span>net ionic is Cl2(g)+2I- => I2(s)+2Cl-(aq) </span>
<span>c) I2(s)+3(Cl2)(g)=>2(ICl3)
</span>I hope I helped you with your problem
I think it's hydrogen. Hope this helps :)

<span>2 NaOH(aq)+ 2 Al(s)+ 2 H</span>₂<span>O → 2 NaAlO</span>₂<span>(aq)+ 3 H</span>₂<span>(g)
</span> 2mol : 2mol : 3mol
2,14mol : 1,89mol : 2,835mol
remains completely consumed
2,14-1,89=0,25mol
A) Al
B)

C)