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andriy [413]
3 years ago
7

------------------------------------------------

Chemistry
2 answers:
Black_prince [1.1K]3 years ago
8 0

:-( ;-) :-\ O:-) :-! B-) :O l..!..!............................Answer:

Explanation:

.

zavuch27 [327]3 years ago
5 0

Answer:

what would u like me to do on this?

Explanation:

do you have a tablel of elements? if not then look at on online and search them all.

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Enter the electron configuration for the ion most likely formed by phosphorus. Express your answer in the order of orbital filli
Bingel [31]

Answer:

[Ne] 2s2 2p3

Explanation:

Phosphorus will most likely have an ion that will be 3- because it wants to have a full outer shell. Thus, the elctron configuration is: 1s2 2s2 2p6 3s2 3p3.

4 0
3 years ago
What would be the pH of a solution of a solution with a hydrogen ion concentration of 0.001 M?
son4ous [18]

0001 M HCl is the same as saying that 1 *10-4 moles of H+ ions have been added to solution. The -log[. 0001] =4, so the pH of the solution =4.

7 0
3 years ago
For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia
Fittoniya [83]

Answer : The value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction:  Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction:  Cu^{2+}+2e^-\rightarrow Cu

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu

To calculate the E^o_{(Cu^{2+}/Cu)} of the reaction, we use the equation:

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

E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}

Putting values in above equation, we get:

1.10V=E^o_{(Cu^{2+}/Cu)}-(-0.76V)

E^o_{(Cu^{2+}/Cu)}=0.34V

Hence, the value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

8 0
3 years ago
What is the total number of liters of NH3 formed when 20 liters of N2 reacts completely
goblinko [34]

Answer:

N2 + H2 ----------》NH3

On balancing it

N2. + 3.H2------->2.NH3

( 1 mol) (3 mol) (2 mol)

1 mol of nitrogen reacts with 3 mol of hydrogen to give 2 mol of ammonia.

Likewise,

20 litres of nitrogen reacts with 60 litres of hydrogen to give 40 litres of Ammonia.

Hence, the answer is 40 Litres.

5 0
3 years ago
After a product is released from its active site, a(n) ______ can be used again in another reaction because it is not consumed o
3241004551 [841]

Answer:

Enzyme.

Explanation:

Enzymes act as a catalyst in all reaction but remain unchanged by the process.

7 0
3 years ago
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