All categories

2 years ago

WHO CAN HELP ME ON THIS NEED OF HELP! Due soon !

Chemistry

1 answer:

Otrada [13]2 years ago

7 0

Answer:

Runoff from a cattle ranch

You might be interested in

It is desired that an acetic acid sodium acetate buffered solution have a pH of 5.27. You have a solution that has an acetic aci

SVETLANKA909090 [29]

Answer:

0.034 M is the molarity of sodium acetate needed.

Explanation:

The pH of the buffer solution is calculated by the Henderson-Hasselbalch equation:

$pH = pK_a + \log \frac{[A^-]}{[HA]}$

Where:

pK_a= Negative logarithm of the dissociation constant of a weak acid

$[Ac^-]$ = Concentration of the conjugate base

[HA] = Concentration of the weak acid

According to the question:

$HAc(aq)\rightleftharpoons Ac^-(aq)+H^+(aq)$

The desired pH of the buffer solution = pH = 5.27

The pKa of acetic acid = 4.74

The molarity of acetic acid solution = [HAc] = 0.01 M

The molarity of acetate ion = $[Ac^-] = ?$

Using Henderson-Hasselbalch equation:

$5.27= 4.74 + \log \frac{[Ac^-]}{[0.01 M]}$

$[Ac^-]=0.0339 M\approx 0.034M$

Sodium acetate dissociates into sodium ions and acetate ions when dissolved in water.

$NaAc(aq)\rightarrow Na^+(aq)+Ac^-(aq)\\$

$[Ac^-]=[Na^+]=[NaAc]= 0.034M$

0.034 M is the molarity of sodium acetate needed.

3 0

3 years ago

Calculate the equilibrium constant for the reaction: 2 Cr + 3 Pb2+ ----> 3 Pb + 2 Cr3+ at 25oC. Eocell = 0.61 V

sattari [20]

Answer:

The value is $K = 8*10^{61}$

Explanation:

From the question we are told that

The equation is $2 Cr + 3Pb^{2+} \to 3Pb + 2Cr^{3+}$

The temperature is $T = 25^oC = 298 K [room \ temperature ]$

The emf at standard condition is $E^o_{cell} = 0.61 \ V$

Generally at the cathode

$3Pb^{2+}(aq) + 6 e- --> 3Pb(s)$

At the anode

$2Cr^{3+} + 6e^- \to 2Cr$

Generally for an electrochemical reaction, at room temperature the Gibbs free energy is mathematically represented as

$G = n* F * E^o_{cell}$

Here n is the no of electron with value n = 6

F is the Faraday's constant with value 96487 J/V

=> $G = 6 * 96487 * 0.61$

=> $G = 3.5 *10^{5} \ J$

This Gibbs free energy can also be represented mathematically as

$G = RTlogK$

Here R is the cell constant with value 8.314J/K

K is the equilibrium constant

From above

=> $K = antilog^{\frac{G}{ RT} }$

Generally antilog = 2.718

=> $K = 2.718^{\frac{3.5 *10^5}{ 8.314* 298} }$

=> $K = 8*10^{61}$

6 0

3 years ago

Please help Me I really need it

butalik [34]

Can you show the question that goes with those answer pls

7 0

2 years ago

Find the relative molecular mass of hydrated iron (II) sulphate FeSO4.7H2O

Pepsi [2]

Answer:

the relative molecular mass of hydrated iron (II) sulfate FeSO4.7H2O is 278.02

Explanation:

5 0

2 years ago

What is the role of blood in the transportation of materials throughout the body

scZoUnD [109]

Blood carries hormones to specific organs.

7 0

3 years ago