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3 years ago

We (get) up at 4:30 yesterday morning

Chemistry

2 answers:

ioda3 years ago

4 0

Answer:

oop oof

Explanation:

saw5 [17]3 years ago

4 0

That’s tuff. Could never be me .

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A chemist designs a galvanic cell that uses these two half-reactions:

Nonamiya [84]

Answer :

(a) Reaction at anode (oxidation) : $4Fe^{2+}\rightarrow 4Fe^{3+}+4e^-$

(b) Reaction at cathode (reduction) : $O_2+4H^++4e^-\rightarrow 2H_2O$

(c) $O_2+4H^++4Fe^{2+}\rightarrow 2H_2O+4Fe^{3+}$

(d) Yes, we have have enough information to calculate the cell voltage under standard conditions.

Explanation :

The half reaction will be:

Reaction at anode (oxidation) : $Fe^{2+}\rightarrow Fe^{3+}+e^-$ $E^0_{anode}=+0.771V$

Reaction at cathode (reduction) : $O_2+4H^++4e^-\rightarrow 2H_2O$ $E^0_{cathode}=+1.23V$

To balance the electrons we are multiplying oxidation reaction by 4 and then adding both the reaction, we get:

Part (a):

Reaction at anode (oxidation) : $4Fe^{2+}\rightarrow 4Fe^{3+}+4e^-$ $E^0_{anode}=+0.771V$

Part (b):

Reaction at cathode (reduction) : $O_2+4H^++4e^-\rightarrow 2H_2O$ $E^0_{cathode}=+1.23V$

Part (c):

The balanced cell reaction will be,

$O_2+4H^++4Fe^{2+}\rightarrow 2H_2O+4Fe^{3+}$

Part (d):

Now we have to calculate the standard electrode potential of the cell.

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

$E^o=(1.23V)-(0.771V)=+0.459V$

For a reaction to be spontaneous, the standard electrode potential must be positive.

So, we have have enough information to calculate the cell voltage under standard conditions.

4 0

3 years ago

Please help!! my grades dying <br> 15 points

AysviL [449]

Answers-in-bold:

There are two common temperature scales. On the Fahrenheit scale, water freezes at 32 degrees. The Celsius scale divides the interval between the freezing and boiling points of water into 100 degrees.

4 0

3 years ago

What molarity is made when 36.5 g HCl is mixed into 250 mL of solution?

Dominik [7]

Answer:

The molarity of the HCl solution is 4M.

Explanation:

$Molarity=\frac{number\:of\:moles}{Volume\:of\:solution\:in\:L}\\ \\The\:volume\:of\:solution=250mL=0.25L\\\\Number\:of\:moles=\frac{weight\:of\:HCl\:in\:solution}{molecular\:weight\:of\:HCl}\\ \\Molecular\:weight\:of\:HCl=36.5g\\\\The\:weight\:of\:HCl\:in\:solution=36.5g\\\\Number\:of\:moles\:of\:HCl=\frac{36.5}{36.5}=1\:mole\\\\The\:molarity\:of\:HCl\:solution=\frac{1}{0.25}=4M$

Hence, the molarity of the HCl solution = 4 M

5 0

3 years ago

This is a science question. <br> What is its acceleration?

trapecia [35]

Your answer is C
Hope this helped!

6 0

3 years ago

Uranium (VIII) Sulfide formula

tresset_1 [31]

Answer:

US₂

Explanation:

Uranium sulfide (US₂)

Uranium atomic symbol = U

Sulfur atomic symbol = S

Uranium valency = +4

Sulfur valency = -2

So;

Uranium sulfide (US₂)

8 0

4 years ago

We (get) up at 4:30 yesterday morning​

We (get) up at 4:30 yesterday morning