She will most likely observe that the temperature
does not change during melting because the heat absorbed is used to overcome
intermolecular forces rather than to increase the kinetic energy of the
particles if she measures the temperature of the water in the beaker.
Answer : The resulting concentrations of CV and NaOH are 0.0027 M and 0.025 M respectively.
Explanation :
Step 1 : Find moles of crystal violet and NaOH.
The molarity formula is
Molarity of crystal violet =
The volume of crystal violet solution is 18 mL which is 0.018 L.
Moles of crystal violet =
Moles of crystal violet = 5.4 x 10⁻⁵
Moles of NaOH =
Moles of NaOH = 5.00 x 10⁻⁴
Step 2 : Find total volume of the solution
The total volume of the solution after mixing NaOH and crystal violet is
0.018 L + 0.00200 = 0.020 L
Step 3 : Use molarity formula to find final concentrations
Molarity of crystal violet =
Final concentration of CV = 0.0027 M
Molarity of NaOH=
NaOH is a strong base and dissociates completely as follows.
The mole ratio of NaOH and OH⁻ is 1:1 . Therefore the concentration of OH⁻ is same as that of NaOH.
Concentration of OH⁻ = 0.025 M
The energy of the carbide released is 7262.5MJ.
<h3>What is the energy?</h3>
We know that the reaction between calcium oxide and carbon occurs in accordance with the reaction; . The reaction is seen to produce 464.8kJ of energy per mole of carbide produced.
Number of moles of produced = 1000 * 10^3 g/64 g/mol
= 15625 moles of calcium carbide
If 1 mole of transfers 464.8 * 10^3 J
15625 moles of calcium carbide transfers 15625 moles * 464.8 * 10^3 J/ 1 mol
= 7262.5MJ
Learn more about reaction enthalpy:brainly.com/question/1657608
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Answer:
a. MgO(s) + H2CO3(aq) → MgCO3(s) + H2O(l) DOUBLE DISPLACEMENT REACTION.
b. 2KNO3(s)→2KNO2 (s) + O2(8) DESCOMPOSITION REACTION.
c. H2(g) + CuO(s) → Cu(s) + H2O(1) SINGLE DISPLACEMENT REACTION.
d. NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(1) COMBUSTION REACTION.
e. H2(8) + Cl2(g) → 2HCl(8) SYNTHESIS REACTION.
f. SO3(g) + H2O(1)→ H2SO4(aq) SYNTHESIS REACTION.
2MnO4-(aq) + 16H+ + 10Cl-(aq) ⇔ 2Mn2+(aq) + 8H2O + 5Cl2 (g)
<h3>Further explanation</h3>
Given
Reaction(unbalanced)
Mno4- (aq) + Cl- (aq) → Mn2+ + Cl2 (g)
Required
Half reaction
Solution
1. Add coefficient(equalizing atoms in reaction)
2. Adding H₂O on the O-deficient side.
3. Adding H⁺ on the H-deficient side.
4. Adding e⁻
5. Equalizing the number of electrons and sum the all the half-reaction
1.MnO₄⁻(aq) = Mn²⁺(aq) reduction
2.MnO4(aq) = Mn2+(aq) + 4H2O
3. MnO4-(aq) + 8H+ = Mn2+(aq) + 4H2O
4. MnO4-(aq) + 8H+ + 5e- = Mn2+(aq) + 4H2O
Cl⁻(aq) = Cl₂(g) oxidation
1. 2Cl-(aq) = Cl2 (g)
2-3 none
4. 2Cl-(aq) = Cl2 (g) + 2e-
5.
MnO4-(aq) + 8H+ + 5e- = Mn2+(aq) + 4H2O x2
2Cl-(aq) = Cl2 (g) + 2e- x5
2MnO4-(aq) + 16H+ + 10e- = 2Mn2+(aq) + 8H2O
10Cl-(aq) = 5Cl2 (g) + 10e-
<em>2MnO4-(aq) + 16H+ + 10Cl-(aq) ⇔ 2Mn2+(aq) + 8H2O + 5Cl2 (g)</em>