What is the IUPAC name Of<br> CH3-CH2-CH2-CH=O<br> I<br> COOH

From the amount of NaOH added at the 1st equivalence point, calculate the original molarity of the acid. Carry out the same calc

gizmo_the_mogwai [7]

Answer:

Molarity of acid, Ca = Cb*Vb*A/Va*B

Explanation:

Using H2SO4 as acid, the reaction is as follow:

2NaOH + H2SO4 ⇒ Na2SO4 + 2H2O

Volume of acid = Va; Volume of base = Vb, Molar concentration of acid = Ca; Molar concentration of base = Cb; Molarity of acid = A and Molarity of base = B

Ca*Va/Cb*Vb =A/B

∴ Ca = Cb*Vb*A/Va*B

4 0

3 years ago

What letter on the model titration curve corresponds to the point where ph equals the numerical value of pka for hpr? what speci

irakobra [83]

Letter C on the model titration curve corresponds to the point where pH equals the numerical value of pKa for HPr

<h3>What is a titration curve?</h3>

A titration curve is a graph of the pH of a solution against increasing volumes of an acid or a base that is added to the solution.

The pH of a solution is the negative logarithm to base ten of the hydrogen ion concentration and is a measure of the acidity or alkalinity of the solution.

The pKa is the acid dissociation constant of an acid solution.

In a titration of a strong acid and strong base, the pH at equivalence point is equal to the pKa of the acid.

The equivalence point is the point when equal moles of acids and base has reacted.

In the given titration curve, pH = pKa at point C.

In conclusion, for a titration curve of strong acid and base, at equivalence point, pH is equal to pKa of acid.

Learn more about equivalence point at: brainly.com/question/23502649

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4 0

2 years ago

An element in Group 2 of the periodic table likely has which characteristics?

ser-zykov [4K]

Answer:

Shiny solid. Conducts heat and electricity. Easily bends

Explanation:

8 0

2 years ago

A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 64.0 mL of 0.0600 M EDTA . Titration of the excess unreacted EDTA req

tigry1 [53]

Answer:

the concentration of $Cd^{2+}$ in the original solution= 0.0088 M

the concentration of $Mn^{2+}$ in the original solution = 0.058 M

Explanation:

Given that:

The volume of the sample containing Cd2+ and Mn2+ = 50.0 mL; &

was treated with 64.0 mL of 0.0600 M EDTA

Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+

i.e the strength of the Ca2+ = 0.0310 M

Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+

To determine the concentrations of Cd2+ and Mn2+ in the original solution; we have the following :

Volume of newly freed EDTA = $\frac{Volume\ of \ Ca^{2+}* Sample \ of \ strength }{Strength \ of EDTA}$

= $\frac{14.2*0.0310}{0.0600}$

= 7.3367 mL

concentration of $Cd^{2+}$ = $\frac{volume \ of \ newly \ freed \ EDTA * strength \ of \ EDTA }{volume \ of \ sample}$

= $\frac{7.3367*0.0600}{50}$

= 0.0088 M

Thus the concentration of $Cd^{2+}$ in the original solution = 0.0088 M

Volume of excess unreacted EDTA = $\frac{volume \ of \ Ca^{2+} \ * strength \ of Ca^{2+} }{Strength \ of \ EDTA}$

= $\frac{16.1*0.0310}{0.0600}$

= 8.318 mL

Volume of EDTA required for sample containing $Cd^{2+}$ and $Mn^{2+}$ = (64.0 - 8.318) mL

= 55.682 mL

Volume of EDTA required for $Mn^{2+}$ = Volume of EDTA required for

sample containing $Cd^{2+}$ and

$Mn^{2+}$ -- Volume of newly freed EDTA

Volume of EDTA required for $Mn^{2+}$ = 55.682 - 7.3367

= 48.3453 mL

Concentration of $Mn^{2+}$ = $\frac{Volume \ of EDTA \ required \ for Mn^{2+} * strength \ of \ EDTA}{volume \ of \ sample}$

Concentration of $Mn^{2+}$ = $\frac{48.3453*0.0600}{50}$

Concentration of $Mn^{2+}$ in the original solution= 0.058 M

Thus the concentration of $Mn^{2+}$ = 0.058 M

6 0

3 years ago

You want to determine ΔH o for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) To do so, you first determine the heat capacity

Assoli18 [71]

Answer:

(A) The heat capacity of the calorimeter is therefore = −2.1428KJ÷13.5°C

= −0.1587KJ/°C

(B) ΔHo for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) = –15.42KJ

Explanation:

Solution

Calculate the heat actually evolved.

q = mcΔt

Finding the mass of the reactants in grams we have.

Use density. (50 mL + 50 mL ) = 100 mL of solution.

100 mL X 1.04g/mL = 104 grams of solution. (mass = Volume X Density)

Find the temperature change.

Δt =tfinal - tinitial = 30.4°C – 16.9°C = 13.5°C

q = mcΔt

= 104grams × 3.93J/g°C × 13.5°C = 5.51772×103J

= 5.51772 × 103 J

This is the heat lost in the reaction between HCl and NaOH, therefore q = -5.52 × 103 J.

this is an exothermic heat producing reaction.

To calculate the total heat of the reaction or heat per mole we have

50.0 mL of HCl X 2.00 mol HCl /(1000 mL HCl ) = 0.100 mol HCl

The same quantity of base, 0.100 mole NaOH, was used.

The energy per unit mole is given by

i.e. molar enthalpy = J/mol = -5.52 × 103J / 0.100 mol

= -5.52 × 104 J/mol

= -55177.2 J/mol

= -55.177 kJ/mol

Therefore, the enthalpy change for the neutralization of HCl and NaOH, that is the enthalpy, heat, of reaction is ΔH = -55.177 kJ/mol

Heat absorbed by the calorimeter = −57.32kJ − 55.177 kJ = −2.1428KJ

The heat capacity of the calorimeter is therefore = −2.1428KJ÷13.5°C

= −0.1587KJ/°C

(B) For the ZnCl we have

Calculate the heat actually evolved.

q = mcΔt

Finding the mass of the reactants in grams we have.

Use density. 100 mL of solution of HCl

100 mL X 1.015g/mL = 101.5 grams of solution. (mass = Volume X Density)

Find the temperature change.

Δt =tfinal - tinitial = 20.5°C – 16.8°C = 3.7 °C

q = mcΔt

= 101.5grams × 3.95J/g°C × 3.7°C = 1483.422×103J

= -1483.422×103J

This is the heat lost in the reaction between HCl and NaOH, therefore q = -1.483 × 103 J.

this is an exothermic heat producing reaction.

To calculate the total heat of the reaction or heat per mole we have

100.0 mL of HCl X 1.00 mol HCl /(1000 mL HCl ) = 0.100 mol HCl

The energy per unit mole is given by

i.e. molar enthalpy = J/mol = -1.483 × 103J / 0.100 mol

= -1.483 × 104 J/mol

= -14834.22 J/mol

= -14.834 kJ/mol

Therefore, the enthalpy change for the neutralization of HCl and NaOH, that is the enthalpy, heat, of reaction is ΔH = -14.834 kJ/mol

ΔHo for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

= -14.834 kJ –(0.1587KJ/°C×3.7°C) = -15.42KJ

ΔHo for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) = –15.42KJ

5 0

3 years ago

What is the IUPAC name Of CH3-CH2-CH2-CH=O I COOH