What do I possibly answer here?
ask a full question pls!
Answer:
0.1066 hours
Explanation:
A common pesticide degrades in a first-order process with a rate constant (k) of 6.5 1/hours. We can calculate its half-life (t1/2), that is, the times that it takes for its concentration to be halved, using the following expression.
t1/2 = ln2/k
t1/2 = ln2/6.5 h⁻¹
t1/2 = 0.1066 h
The half-life of the pesticide is 0.1066 hours.
Answer:
357.475
Explanation:
First you need periodic table and you have to look for mass
Fe = 3 x 55.845 = 167.535
P = 2 x 30.97 = 61.94
o = 4 x 2 so 8 oxygen = 8 x 16 = 128
add all and you get 357.475
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
