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3 years ago

6

How many grams of chromium metal are plated out when a constant current of 8.00 A is passed through an aqueous solution containi

ng Cr 3+ ions for 40.0 minutes?

1 answer:

Vsevolod [243]3 years ago

6 0

Answer:

fijhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

Explanation:

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Are many common foods we eat more likely to be acidic or basic

myrzilka [38]

Consumption of the acidic and the basic is equal during human's life. For example, acidic ones add sourness to things we eat and what about basic - the taste is quite bitter that conidered to be less enjoyable. Every our dish consists of both listed elements but our bodies are already accustomed to this fact.
Hope that helps.

7 0

3 years ago

If I initially have 4.0 L of a gas at a pressure of 1.1 atm and a temperature of 298 K, what will the volume be if I increase th

DochEvi [55]

Answer:

1.34L

Explanation:

1 torr = 0.00132atm

P2= 2584torr = 3.41atm

T2= 37°C = 273+37 = 310K

Using combined gas law

P1V1/T1 = P2V2/T2

(1.1×4.0)/298 = (3.41 × V2)/310

V2= (1.1×4×310)/(298×3.41)

V2= 1.34L

4 0

3 years ago

Calculate the cell potential E at 25°C for the reaction 2 Al(s) + 3 Fe2+(aq) → 2 Al3+(aq) + 3 Fe(s) given that [Fe 2+] = 0.020 M

Elodia [21]

Answer:

1.18 V

Explanation:

The given cell is:

$Al(s)/Al^{3+}(0.10M)||Fe^{2+}(0.020M)/Fe(s)$

Half reactions for the given cell follows:

Oxidation half reaction: $Al(s)\rightarrow Al^{3+}(0.10M)+2e^-;E^o_{Al^{3+}/Al}=-1.66V$

Reduction half reaction: $Fe^{2+}(0.020M)+2e^-\rightarrow Fe(s);E^o_{Fe^{2+}/Fe}=-0.45V$

Multiply Oxidation half reaction by 2 and Reduction half reaction by 3

Net reaction: $2Al(s)+3Fe^{2+}(0.020M)\rightarrow 2Al^{3+}(0.10M)+3Fe(s)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=-0.45-(-1.66)=1.21V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}$

where,

$E_{cell}$ = electrode potential of the cell = ?V

$E^o_{cell}$ = standard electrode potential of the cell = +1.21 V

n = number of electrons exchanged = 6

Putting values in above equation, we get:

$E_{cell}=1.21-\frac{0.059}{6}\times \log(\frac{0.10^2}{0.020^3})\\\\E_{cell}=1.18V$

5 0

4 years ago

The Percentage of salt-water vs. fresh-water:???%

I am Lyosha [343]

Answer:

<em><u>Over 96 percent is saline. Of total freshwater, over 68 percent is locked up in ice and glaciers. Another 30 percent of freshwater is in the ground.</u></em>

6 0

2 years ago

the pressure exerted by water at the bottom of the mariana trench, the deepest part of the world's ocean is 1070 atm. how many T

deff fn [24]

Answer:

i wanna say its 16000 pounds per sq inch

Explanation:

6 0

3 years ago