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3 years ago

The total number of sodium atoms in 46.0 grams of solution is?

Chemistry

2 answers:

USPshnik [31]3 years ago

7 0

Molar mass Na = 23g/mol
46g = 456/2 = 2mol
1mol = 6.022*10^23 atoms
2mol = 2*6.022*10623
= 1.204*10^24 atoms

madam [21]3 years ago

3 0

First step, is to find the Molar Mass.
If you go to the periodic table, the Molar Mass of the Sodium Na is 23g/mol

Second Step, you convert from grams to moles.
number of Mole = mass of sodium / molar mass = $\frac{46}{23} = 2 mol$

Third Step, you find the number of atoms.
As We know, Avogadro said that the number of atoms in 1 mole is about $6.022 * 10^{23}$ .
You use the Rule of three and you find the number of atoms!
1 mole = $6.022 * 10^{23}$ atoms
2 mole = x atomes
x = 2 * ( $6.022 * 10^{23}$ )
x= $12.044 * 10^{23}$ atomes

So, there is $12.044 * 10^{23}$ atoms of Sodium Na in 46g of Solution.

Hope this Helps! :D

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Ascorbic acid (H2C6H6O6; H2Asc for this problem), known as vitamin C, is a diprotic acid (Ka1= 1.0x10–5 and Ka2= 5x10–12) found

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Answer:

The concentrations are :

$[HAsc^-]=0.000702 M$

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Explanation:

$H_2Asc\rightleftharpoons HAs^-+H^+$ $K_{a1}=1.0\times 10^{-5}$

Initial

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Equilibrium

c-x x x

$K_{a1}=\frac{[HAs^-][H^+]}{[H_2Asc]}$

$1.0\times 10^{-5}=\frac{x\times x}{(c-x)}$

$1.0\times 10^{-5}=\frac{x^2}{(0.050-x)}$

Solving for x:

x = 0.000702 M

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Initially

x 0 0

At equilibrium ;

(x - y) y y

$K_{a2}=\frac{[As^{2-}][H^+]}{[HAsc^-]}$

$5\times 10^{-12}=\frac{y\times y}{(x-y)}$

$5\times 10^{-12}=\frac{y^2}{(x-y)}$

Putting value of x = 0.000702 M

$5\times 10^{-12}=\frac{y^2}{(0.000702 -y)}$

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The pH of the solution :

$pH=-\log[H^+]$

$pH=-\log[7.0206\times 10^{-4} M}=3.15$

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3 years ago

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Answer:

Explanation:

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We have the final answer as

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<h3>What is isotopes?</h3>

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