Here is the correct question
You mix 125 mL of 0.170 M CsOH with 50.0 mL of 0.425 M HF in a coffee-cup calorimeter, and the temperature of both solutions rises from 20.20 °C before mixing to 22.17 °C after the reaction. What is the enthalpy of reaction per mole of ? Assume the densities of the solutions are all 1.00 g/mL, and the specific heat capacities of the solutions are 4.2 J/g · K. Enthalpy of reaction = kJ/mol
Answer:
75.059 kJ/mol
Explanation:
The formula for calculating density is:
Making mass the subject of the formula; we have :
mass = density × volume
which can be rewritten as:
mass of the solution = density × volume of the solution
= 1.00 g/mL × (125+ 50 ) mL
= 175 g
Specific heat capacity = 4.2 J/g.K
∴ the energy absorbed is = mcΔT
= 175 × 4.2 × (22.17 - 20.00) ° C
= 1594.95 J
= 1.595 J
number of moles of CsOH = 
= 0.2125 mole
Therefore; the enthalpy of the reaction = 
= 
= 75.059 kJ/mol
Answer:
There is 61.538% oxygen in Al2(SO4)3.
Explanation:
Wt Of oxygen in the compound = 12*16 = 192 amu.
Total Wt. Of the compound = 2*12+3*32+12*16 = 312 amu.
Thus, percent of oxygen = Wt of oxygen/total Wt. Of compound *100
= 192/312 * 100=61.538 %
As temperature increases, the molecules and atoms move faster. :^)
Answer:
ΔH = -110.5kJ
Explanation:
It is possible to obtain enthalpy of combustion of a particular reaction by the algebraic sum of similar reactions (Hess's law). Using:
1. C(s) + O₂(g) → CO₂(g) ΔH₁ = -393.5kJ
2. CO(g) + 1/2O₂(g) → CO₂(g) ΔH₂ = -283.0kJ
The sum of 1 -2 gives:
C(s) + <u>O₂(g)</u> + <u>CO₂(g)</u> → <u>CO₂(g)</u> + CO(g) + <u>1/2O₂(g)</u>
C(s) + 1/2O₂(g) → CO(g) ΔH = -393.5kJ - (-283.0kJ) =
<h3>ΔH = -110.5kJ</h3>
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Answer:
C
) 2, 1, 2
Explanation:
The given reaction is synthesis reaction in which lithium and bromine react to form lithium bromide.
Chemical equation:
Li + Br₂ → LiBr
Balanced chemical equation:
2Li + Br₂ → 2LiBr
Step 1:
Li + Br₂ → LiBr
left hand side Right hand side
Li = 1 Li = 1
Br = 2 Br = 1
Step 2:
Li + Br₂ → 2LiBr
left hand side Right hand side
Li = 1 Li = 2
Br = 2 Br = 2
Step 3:
2Li + Br₂ → 2LiBr
left hand side Right hand side
Li = 2 Li = 2
Br = 2 Br = 2
