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stira [4]
3 years ago
8

Write the complete electron configuration for sodium ion

Chemistry
1 answer:
Katena32 [7]3 years ago
7 0
Sodium ion has 10 electrons. So: 1s^2 2s^2 2p^6
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A 4.36 g sample of an unknown alkali metal hydroxide is dissolved in 100.0 ml of water. an acid-base indicator is added and the
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Answer:

(a) 102.6g/mol

(b) Rubidium

Explanation:

Hello,

This titration is carried out by assuming that the volume of base doesn't have a significant change when the mass is added, thus, we state the following data a apply the down below formula to compute the molarity of the base solution:

V_{base}=0.1L; M_{acid}=2.5M, V_{acid}=0.017L\\V_{base}M_{base}=V_{acid}M_{acid}

Solving for the molarity of base we've got:

M_{base}=\frac{M_{acid}*V_{acid}}{V_{base}}=\frac{2.50M*0.017L}{0.1L} =0.425M=0.425mol/L

Now, we can compute the moles of the base as:

n_{base}=0.425mol/L*0.1L=0.0425mol

(a) Now, one divides the provided mass over the previously computed moles to get the molecular mass of the unknown base:

\frac{4.36g}{0.0425mol} =102.6g/mol

(b) Subtracting the atomic mass of oxygen and hydrogen, the metal's atomic mass turns out into:

102.6g/mol-16g/mol-1g/mol=85.6g/mol

So, that atomic mass dovetails to the Rubidium's atomic mass.

Best regards.

8 0
3 years ago
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