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Zigmanuir [339]
3 years ago
14

Please help me with number 43

Chemistry
1 answer:
Lana71 [14]3 years ago
8 0

Answer:

The reason why atomic mass is usually not a whole number is because it is a weighted average of the mass numbers of isotopes

Explanation:

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How many dL in 1,000 mL
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Calculate the molarity of sodium ion in a solution made
Arada [10]

Answer:

0.1035 M

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Moles =Molarity \times {Volume\ of\ the\ solution}

Sodium chloride will furnish Sodium ions as:

NaCl\rightarrow Na^{+}+Cl^-

Given :

For Sodium chloride :

Molarity = 0.288 M

Volume = 3.58 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 3.58×10⁻³ L

Thus, moles of Sodium furnished by Sodium chloride is same the moles of Sodium chloride as shown below:

Moles =0.288 \times {3.58\times 10^{-3}}\ moles

Moles of sodium ions by sodium chloride = 0.00103104 moles

Sodium sulfate will furnish Sodium ions as:

Na_2SO_4\rightarrow 2Na^{+}+SO_4^{2-}

Given :

For Sodium sulfate :

Molarity = 0.001 M

Volume = 6.51 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 6.51 ×10⁻³ L

Thus, moles of Sodium furnished by Sodium sulfate is twice the moles of Sodium sulfate as shown below:

Moles =2\times 0.001 \times {6.51\times 10^{-3}}\ moles

Moles of sodium ions by Sodium sulfate = 0.00001302 moles

Total moles = 0.00103104 moles + 0.00001302 moles = 0.00104406 moles

Total volume = 3.58 ×10⁻³ L + 6.51 ×10⁻³ L = 10.09 ×10⁻³ L

Concentration of sodium ions is:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity_{Na^+}=\frac{0.00104406}{10.09\times 10^{-3}}

<u> The final concentration of sodium anion = 0.1035 M</u>

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3 years ago
The electron configuration below applies to which element? 1s2 2s2 2p6 3s2 3p1
zvonat [6]
Aluminum! hope this helps
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3 years ago
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