Answer:
-21 kJ·mol⁻¹
Explanation:
Data:
H₃O⁺ + OH⁻ ⟶ 2H₂O
V/mL: 50 50
c/mol·dm⁻³: 1.0 1.0
ΔT = 4.5 °C
C = 4.184 J·°C⁻¹g⁻¹
C_cal = 50 J·°C⁻¹
Calculations:
(a) Moles of acid
So, we have 0.050 mol of reaction
(b) Volume of solution
V = 50 dm³ + 50 dm³ = 100 dm³
(c) Mass of solution
(d) Calorimetry
There are three energy flows in this reaction.
q₁ = heat from reaction
q₂ = heat to warm the water
q₃ = heat to warm the calorimeter
q₁ + q₂ + q₃ = 0
nΔH + mCΔT + C_calΔT = 0
0.050ΔH + 100×4.184×4.5 + 50×4.5 = 0
0.050ΔH + 1883 + 225 = 0
0.050ΔH + 2108 = 0
0.050ΔH = -2108
ΔH = -2108/0.0500
= -42 000 J/mol
= -42 kJ/mol
This is the heat of reaction for the formation of 2 mol of water
The heat of reaction for the formation of mol of water is -21 kJ·mol⁻¹.
I believe it is NaCl(aq) + H2O
when we convert 32.5 lb/in² to atmosphere, the result obtained is 2.21 atm
<h3>Conversion scale</h3>
14.6959 lb/in² = 1 atm
<h3>Data obtained from the question</h3>
- Pressure (in lb/in²) = 32.5 lb/in²
- Pressure (in ATM) =?
<h3>How to convert 32.5 lb/in² to atm</h3>
14.6959 lb/in² = 1 atm
Therefore
32.5 lb/in² = 32.5 / 14.6959
32.5 lb/in² = 2.21 atm
Thus, 32.5 lb/in² is equivalent to 2.21 atm
Learn more about conversion:
brainly.com/question/2139943
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Explanation:
Mg(s) + Cr(C2H3O2)3 (aq)
Overall, balanced molecular equation
Mg(s) + Cr(C2H3O2)3(aq) --> Mg(C2H3O2)3(aq) + Cr(s)
To identify if an element has been reduced or oxidized, the oxidation number is observed in both the reactant and product phase.
An increase in oxidation number denotes that the element has been oxidized.
A decrease in oxidation number denotes that the element has been reduced.
Oxidation number of Mg:
Reactant - 0
Product - +3
Oxidation number of Cr:
Reactant - +3
Product - 0
Note: C2H3O2 is actually acetate ion; CH3COO- The oxidatioon number of C, H and O do not change.
Oxidized : Mg
Reduced : Cr
<span>Express the answer in scientific notation and with the correct number of significant figures:
(6.32 x 10-4) ÷ 12.64
5.00 x 10^-5</span>
