Ask question

Ask question

All categories

geniusboy [140]

3 years ago

7

Ag+(aq) + e- → Ag(s) E° = +0.800 V AgBr(s) + e- → Ag(s) + Br-(aq) E° = +0.071 V Br2(l) + 2 e- → 2 Br-(aq) E° = +1.066 V Use some

of the data above to calculate Ksp at 25°C for AgBr. Enter your answer in exponential format (sample 1.23E-4) with two decimal places and no units.

1 answer:

Gekata [30.6K]3 years ago

5 0

Answer:

$k_{sp}=4.7 \times 10^{-13}.$

Explanation:

Now equation of tqo halves are:

Oxidation : $Ag(s)-->Ag^+(aq)+e^-$

Reduction : $AgBr(s)+e^--->Ag(s)+Br^-(aq)$

We know,

$E^o_{cell}=0.071-(0.8)=-0.729\ V.$

$\Delta G^o=-n\times F \times E^o= -R\times T\times ln(k_s_p)\\-1\times 96485\times (-0.729)=-8.314\times 298\times ln(k_s_p)\\k_s_p=4.7\times 10^{-13}.$

$k_{sp}=4.7 \times 10^{-13}.$

Hence, this is the required solution.

You might be interested in

What happens to hydrogen during hydrogen bonding?

loris [4]

Answer:

The electron spends a lot of time away from hydrogen and hydrogen's nucleus is exposed for

interaction

Explanation:

I got the question right

7 0

3 years ago

What is the total number of valence electrons in a germanium atom in the ground state?(1) 22 (3) 32(2) 2 (4) 4

larisa86 [58]

Answer is (4) - 4.

The atomic number of Germanium (Ge) is 32 and it is placed in 14th group of periodic table. In the neutral atom, 32 electrons are present. The electron configuration of Ge is [Ar] 3d¹⁰ 4s² 4p².

Valence electrons are the electrons which are in outer shell of the atom. There are 4 electrons in outer shell of Ge atom at ground state.

5 0

4 years ago

Read 2 more answers

A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen (by mass). calculate the empirical formula.

Eva8 [605]

C:H:O = 40/12 : 6,7/1 : 53,3/16 = 3,33 : 6,7 : 3,33 ≈ 1 : 2 : 1

CH₂O

3 0

4 years ago

POH is a measure of the ________________ in solution. *

Digiron [165]

Answer:

<u><em>Both acids </em></u>and<u><em> bases</em></u> can measured using the pH or pOH scale. Both scales provide a measure of either the H+ concentration or the OH- concentration. Notice that each scale shows were acids and bases both are located. When acids are measured, the pH is less than 7, but the pOH is greater than 7.

Explanation:

3 0

3 years ago

18. How many milliliters of water at 23 °C with a density of 1.00 g/mL must be mixed with 180 mL (about 6 oz) of coffee at 95 °C

sveticcg [70]

Answer:

The number of water that must be mixed in the solution is 170.27 mL

Explanation:

given information :

Temperature of water ( $T_{1}$ ) = $23^{o}C$

density of water (ρ) = 1.00 g/mL

Temperature of coffee ( $T_{1}$ ) = $95^{o}C$

volume of coffee ( $V_{2}$ ) = 180 mL

Mixed Temperature ( $T_{mix}$ ) = $60^{o}C$

to calculate the heat, w use the formula :

Q = m x c x ΔT

where

m = mass of the substance (g)

c = specific heat (J/ $g^{o}C$ )

ΔT = the temperature change ( $^{o}C$ )

in the mixture solution, the heat of the water ( $Q_{1}$ ) should be the same as the heat of coffee ( $Q_{2}$ ). Thus,

$Q_{1}$ = $Q_{2}$

$m_{1}$ x $c_{1}$ x Δ $T_{1}$ = $m_{2}$ x $c_{2}$ x Δ $T_{2}$

where

$m_{1}$ is the mass of water

$m_{2}$ is the mass of coffee

$c_{1}$ is the specific heat of water

$c_{2}$ is the specific heat of coffee

Assume coffee and water have the same specific heat. So,

$c_{1}$ = $c_{2}$ , we can remove it from the equation.

Hence.

$m_{1}$ x Δ $T_{1}$ = $m_{2}$ x Δ $T_{2}$

we know that

ρ = $\frac{m}{V}$

m = ρ x V, subtitute it to the equation:

ρ $_{1}$ x V $_{1}$ x Δ $T_{1}$ = ρ $_{2}$ x V $_{2}$ x Δ $T_{2}$

$V_{1}$ is the volume of water

coffee and water have the same density, so we can remove the formula

V $_{1}$ x Δ $T_{1}$ = V $_{2}$ x Δ $T_{2}$

V $_{1}$ = (V $_{2}$ x Δ $T_{2}$ ) / Δ $T_{1}$

V $_{1}$ = V $_{2}$ x ( $\frac{(T_{2} - T_{mix}) }{(T_{mix} - T_{1})}$

V $_{1}$ = 180 mL x $\frac{(95-60)^{o}C}{(60-23)^{o}C}$

V $_{1}$ = 180 mL x $\frac{(35)^{o}C}{(37)^{o}C}$

V $_{1}$ = 170.27 mL

3 0

3 years ago