Q= mcΔT
1623 = 33.69g x c x (110.8 - 29.4)
1623 = 2742.366 g•°C x c
c = 0.59j/g•°C
Answer:
V₂ = 15.3
Explanation:
Given data:
Initial volume = 12.0 L
Initial temperature = 20°C
Final temperature =100°C
Final volume = ?
Solution:
First of all we will convert the temperature into kelvin.
20°C + 273 = 293 K
100°C + 273 = 373 K
Formula:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 12.0 L × 373 K / 293 k
V₂ = 4476 L.K /293 k
V₂ = 15.3
V₂ = 1566 L.K / 298 K
V₂ = 5.3 L
Answer:
C. Lithium is most easily oxidized of the metals listed on the activity series and therefore it will most easily give electrons to metal cations
Explanation:
"Lithium" is a type of alkali metal that has a "single valence electron." Since it is a reactive element, it easily gives up an electron when it is combined with other elements. Such giving up of electron is meant to create compounds or bonds.
Among the common metals listed, "lithium" is the most easily oxidized. This means that it donates its electrons immediately. Such combination makes it exist as a<em> "cation"</em> or <em>"positively-charged."</em>
So, this explains the answer.
<span>100 g of KClO3 @ 122.55 g/mol = 0.816 moles of KClO3
by the reaction
2 KClO3 --> 2 KCl & 3 O2
0.816 moles of KClO3 @ 3 moles O2 / 2 moles KClO3 = 1.224 moles of O2 can be made
using molar mass
1.224 moles of O2 @ 32.0 g/mol =
39.2 grams of O2 can be made</span>