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bazaltina [42]
3 years ago
15

What is an alkali metal with fewer than 10 protons in its nucleus?

Chemistry
1 answer:
Trava [24]3 years ago
6 0

Answer:

            Lithium

Explanation:

                  Alkali metals are group of metals which are present in first group of periodic table. As we know atomic number is equal to number of protons contained by a particular element. Therefore, the alkali metals along with there number of protons are listed below;

Alkali Metal                                         Number of Protons

Lithium                                                               3

Sodium                                                             11

Potassium                                                         19

Rubidium                                                         37

Cesium                                                             55

Francium                                                          87

Hence, it is cleared from above table that Lithium is having fewer protons than 10.

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2 years ago
Distinguish between zinc nitrate and lead nitrate ​
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3 years ago
What takes up space?<br> O A. Light<br> O B. Matter<br> O C. Sound<br> O D. Energy
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Answer:

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Explanation:

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6 0
3 years ago
Read 2 more answers
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-
Komok [63]

<u>Answer:</u> The cell potential of the cell is +0.118 V

<u>Explanation:</u>

The half reactions for the cell is:

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<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

5 0
3 years ago
How many atoms of hydrogen are present: 2(NaHCO3)
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Answer:

《HOPE IT WILL HELP YOU 》

Explanation:

2 Atom of Hydrogen are present in 2 (NaHCO3)

5 0
2 years ago
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