<h3><u>Answer;</u></h3>
True
<h3><u>Explanation</u>;</h3>
- The molecule NH3 contains all single bonds.
- NH3 has a three single covalent bond among its nitrogen and hydrogen atoms,because one valence electron of each of three atom of hydrogen is shared with three electron.
- There are three covalent bonds are in NH3 . Each hydrogen make a single bond with nitrogen and there is also a pair of electron which is unpaired from nitrogen.
Correct Question:
A chemist measures the enthalpy change ΔH during the following reaction: Fe(s) + 2HCl(g)-->FeCl2(s) + H2 ΔH=-157.0 kJ. Use this information to complete the table below. Round each of your answers to the nearest kJ/mol
Answer:
-314 kJ
+628 kJ
+157 kJ
Explanation:
The enthalpy change of a reaction measures the amount of heat that is lost or gained by it. If ΔH >0 the heat is gained, and the reaction is called endothermic, if ΔH<0, the heat is lost, and the reaction is called exothermic.
If the reaction is inverted, the value of ΔH is inverted too (the opposite endothermic reaction is exothermic), and if the reaction is multiplied by a constant, ΔH will be multiplied by it too.
1) 2Fe(s) + 4HCl --> 2FeCl2(s) + 2H2(g)
This reaction is the product of the given reaction by 2, so
ΔH = 2*(-157) = -314 kJ
2) 4FeCl2(s) + 4H2(g) --> 4Fe(s) + 8HCl(g)
This reaction is the inverted reaction given multiplied by 4, so
ΔH = 4*(157) = +628 kJ
3) FeCl2(s) + H2(g) --> Fe(s) + 2HCl
This reaction is the inverted reaction given, so
ΔH = +157 kJ
Answer:
a) heat it from 23.0 to 78.3
q = (50.0 g) (55.3 °C) (2.46 J/g·°C) =
b) boil it at 78.3
(39.3 kJ/mol) (50.0 g / 46.0684 g/mol) =
c) sum up the answers from the two calculations above. Be sure to change the J from the first calc into kJ
Explanation:
Both of you are overlooking a pretty big component of the question...the Group I cation isn't being dissociated into water. We're testing the solubility of the cation when mixed with HCl. And this IS a legitimate question, seeing as our lab manual is the one asking.
<span>By the way, the answer you're looking for is "Because Group I cations have insoluble chlorides". </span>
<span>"In order...to distinguish cation Group I, one adds HCl to a sample. If a Group I cation is present in the sample, a precipitate will form." </span>
