Answer:
1913. Henry Moseley determined the atomic number of each of the known elements. He realized that, if the elements were arranged in order of increasing atomic number rather than atomic weight, they gave a better fit within the 'periodic table'.
Explanation:
GOO
Answer:
CHO
Explanation:
Carbon = 41%, Hydrogen = 4.58%, oxygen = 54.6%
Step 1:
Divide through by their respective relative atomic masses
41/ 12, 4.58/1, 54.6/16
3.41 4.58 3.41
Step 2:
Divide by the lowest ratio:
3.41/3.41, 4.58/3.41, 3.41/3.41
1, 1, 1
Hence the empirical formula is CHO
Answer:
⇒ We have Na2O + H2O --> NaOH. We have 2 sodiums and 2 oxygens and 2 hydrogens on the left side, but only one of each on the right side.
Sodium Oxide + Water → Sodium Hydroxide
⇒ Na2O + H2O → 2NaOH .
Sodium oxide is used in ceramics and glasses. Sodium oxide reacts exothermically with cold water to produce sodium hydroxide solution.
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
The answer to this question would be: 3.125%
Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%
