Oxidation reaction
In ---> In³⁺ + 3e ---1)
reduction reaction
Cd²⁺ + 2e ---> Cd ---2)
when balancing the reactions, electrons have to be balanced. to balance the electrons multiple 1st reaction by 2 and 2nd reaction by 3
1) x 2
2) x 3
2In ---> 2In³⁺ + 6e
3Cd²⁺ + 6e ---> 3Cd
add the 2 equations to obtain the overall reaction

2In + 3Cd²⁺ ---> 2In³⁺ + 3Cd

5 0

4 years ago

A water molecule consists of an oxygen atom and two hydrogen atoms covalently bonded in a bent shape. Oxygen however, attracts t

77julia77 [94]

Answer:

Explanation:

Water is polar, for one thing. Polar mixes with polar, nonpolar mixes wih nonpolar. This leaves D.

6 0

3 years ago

A chemist titrates 150.0 mL of a 0.2653 M carbonic acid (H2CO3) solution with 0.2196 M NaOH solution at 25 °C. Calculate the pH

xxTIMURxx [149]

Answer:

9.3

Explanation:

This is long and complicated so get ready

We are going to use the conjugate base of carbonic acid with water to make carbonic acid and OH- (Na is simply a spectator ion and is irrelavent here)

Let the conjugate base be A- and Carbonic acid be HA

A- + H20 ⇄ HA + OH-

To find the concentration of A- we must find the concentration of the reactants given. We know this will be equal because it is a strong base and all of it disassociates.

to get moles of acid we take the concentration and multiply by liters to cancel

.2653 x .150 = .039795 mol HA

Because it is at equivalence point we know the moles will be equal. We are given the concentration so we only have to solve for liters

We plug it into the equation and found: .181 L

Now use moles and combined volums to fins concentrarion which is .120 M

Now plug that use the Ka converted to Kb to find the cincentrations of HA and OH-

Ka is (10^-3.60) = 2.4E-4

Kb x Ka is 10^-14

Kb = 3.98E-11

Now we know Kb = [HA] [OH] / [A-]

Solve for this through algebra by using x for the values you dont know

youll find x^2 = 3.3E-10

X = 1.8 E -5

this is the OH- concentration

-log [oh] = pOH

pOH = 4.73

We know 14-pOH = ph so pH= 9.3

6 0

3 years ago