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V125BC [204]
3 years ago
7

Silicon has three naturally occurring isotopes. Silicon-28 has a mass of 27.98 and a relative abundance of 92.23%. Silicon-29 ha

s a mass of 28.98 and a relative abundance of 4.68%. Silicon-30 has a mass of 29.97 and a relative abundance of 3.09%. What is the weighted average atomic mass of silicon?
Chemistry
2 answers:
Advocard [28]3 years ago
5 0
The weighted average atomic mass of Silicon is 28.1

(27.98 x .9223) + (28.98 x .0468) + (29.97 x .0309) 
= 25.81 + 1.36 + .93
= 28.1
skad [1K]3 years ago
4 0

Answer : The atomic mass of the silicon is, 28.09 amu

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i

As we are given that,

Mass of isotope, Silicon-28 = 27.98 amu

Percentage abundance of isotope, Silicon-28 = 92.23 %

Fractional abundance of isotope, Silicon-28 = 0.9223

Mass of isotope, Silicon-29 = 28.98 amu

Percentage abundance of isotope, Silicon-29 = 4.68 %

Fractional abundance of isotope, Silicon-29 = 0.0468

Mass of isotope, Silicon-30 = 29.97 amu

Percentage abundance of isotope, Silicon-30 = 3.09 %

Fractional abundance of isotope, Silicon-30 = 0.0309

Now put all the given values in above formula, we get:

\text{Average atomic mass of element}=\sum[(27.98\times 0.9223)+(28.98\times 0.0468)+(29.97\times 0.0309)]

\text{Average atomic mass of element}=28.09amu

Therefore, the atomic mass of the silicon is, 28.09 amu

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Answer:

a) = 0.704%

b) = 1.30%

c) = 2.60%

Explanation:

Given that:

K_a= 1.34*10^{-5

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Percentage Ionization(∝)  \alpha = \sqrt{\frac{K_a}{C} }

\alpha = \sqrt{\frac{1.34*10^{-5}}{0.270} }

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\alpha = 7.044*10^{-3

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For Part B; where Concentration of B = 7.84*10^{-2 M

\alpha = \sqrt{\frac{1.34*10^{-5}}{7.84*10^{-2}} }

\alpha = \sqrt{1.709*10^{-4} }

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For Part C; where Concentration of C= 1.92*10^{-2} M

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3 years ago
A solution contains 15.27 grams of NaCl in 0.670 kg water at 25 °C. What is the vapor pressure of the solution?
Anvisha [2.4K]

Answer:

The vapor pressure of the solution is 23.636 torr

Explanation:

P_{solution} = X_{solvent}*P_{solvent}

Where;

P_{solution is the vapor pressure of the solution

X_{solvent is the mole fraction of the solvent

P_{solvent is the vapor pressure of the pure solvent

Thus,

15.27 g of NaCl = [(15.27)/(58.5)]moles = 0.261 moles of NaCl

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Mole fraction of solvent (water) = (number of moles of water)/(total number of moles present in solution)

Mole fraction of solvent (water) = (37.222)/(37.222+0.261)

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Therefore, the vapor pressure of the solution = 0.993 * 0.0313 atm

the vapor pressure of the solution = 0.0311 atm = 23.636 torr

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