Answer:
37.3
263.5
Explanation:
The scale measures hundreds of units, tens of units, units, and parts of units (1 decimal place.
Scale 1
Hundreds 0 * 100 = 0
Tens: 3 * 10 = 30
Units: 7 * 1 = 7
1/10 unit = 3* 0.1 = 0.3
Total 30 + 7 + 0.3 = 37.3
Scale 2
Hundreds 2 * 100 = 200
Tens: 6 * 10 = 60
Units: 3 * 1 = 3
1/10 unit = 5* 0.1 = 0.5
Total = 200 + 60 + 3 + 0.5 = 263.5
Answer:

Explanation:
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In this case, according to the rules for the oxidation states in chemical reactions, it is possible to realize that lone elements have 0 and since magnesium is in group 2A, it forms the cation Mg⁺² as it loses electrons and oxygen is in group 6A so it forms the anion O⁻²; therefore resulting oxidation numbers are:

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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:
the mass number of the atom or ion is 6
Answer:
463.0 g.
Explanation:
- We can use the following relation:
<em>n = mass/molar mass.</em>
where, n is the mass of copper(ii) fluoride (m = 4.56 mol),
mass of copper(ii) fluoride = ??? g.
molar mass of copper(ii) fluoride = 101.543 g/mol.
∴ mass of copper(ii) fluoride = (n)(molar mass) = (4.56 mol)(101.543 g/mol) = 463.0 g.