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Ilya [14]
2 years ago
14

SOMEONE PLZZ help me on this chemistry test I need to pass it to be able to get my grade up :(((

Chemistry
1 answer:
iVinArrow [24]2 years ago
6 0

Answer:

s-2 p-6 d-10 f-14

Explanation:

s orbital max can fit 2 electrons

p orbital max can fit 6 electrons

d orbital max can fit 10 electrons

f orbital max can fit 14 electrons

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How does latitude affect climate
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Cobalt-63 has a half-life of 5.3 years. If a pellet that has been in storage for 15.9 years contains 40.0g of Cobalt-63, how muc
Cerrena [4.2K]

Answer:

320 g  

Step-by-step explanation:

The half-life of Co-63 (5.3 yr) is the time it takes for half of it to decay.  

After one half-life, half (50 %) of the original amount will remain.  

After a second half-life, half of that amount (25 %) will remain, and so on.  

We can construct a table as follows:  

  No. of               Fraction         Mass

half-lives   t/yr   Remaining   Remaining/g

      0        0              1

      1         5.3           ½

     2        10.6           ¼

     3        15.9           ⅛                 40.0

     4        21.2           ¹/₁₆

We see that 40.0 g remain after three half-lives.

This is one-eighth of the original mass.

The mass of the original sample was 8 × 40 g = 320 g

5 0
3 years ago
A concentration cell is constructed using two Ni electrodes with Ni2+ concentrations of 1.0 M and 1.00 � 10�4 M in the two half-
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<u>Answer:</u> The cell potential of the cell is +0.118 V

<u>Explanation:</u>

The half reactions for the cell is:

<u>Oxidation half reaction (anode):</u>  Ni(s)\rightarrow Ni^{2+}+2e^-

<u>Reduction half reaction (cathode):</u>  Ni^{2+}+2e^-\rightarrow Ni(s)

In this case, the cathode and anode both are same. So, E^o_{cell} will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Ni^{2+}_{diluted}]}{[Ni^{2+}_{concentrated}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

[Ni^{2+}_{diluted}] = 1.00\times 10^{-4}M

[Ni^{2+}_{concentrated}] = 1.0 M

Putting values in above equation, we get:

E_{cell}=0-\frac{0.0592}{2}\log \frac{1.00\times 10^{-4}M}{1.0M}

E_{cell}=0.118V

Hence, the cell potential of the cell is +0.118 V

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