Ask question

Ask question

All categories

3 years ago

15

ilicon has three naturally occurring isotopes (Si-28, Si-29, and Si-30). Masses and natural abundances for two isotopes are list

ed here. Isotope Mass (amu) Abundance (%) Si-28 27.9769 92.2 Si-29 28.9765 4.67 Si-30 ? ? Part A Find the natural abundance of Si-30 . Express your answer using two significant figures.

1 answer:

sergejj [24]3 years ago

5 0

Answer:

Part A

The natural abundance of Si-30 = 3.1

Part B

The mass of Si-30 = 29.9551 amu

Explanation:

Part A

The sum of the natural abundances = 100

Si-28 has a mass of 27.9769 amu and a natural abundance of 92.2%.

Si-29 has a mass of 28.9765 amu and a natural abundance of 4.67%

Si-30's mass and natural abundance are unknown.

Natural abundance for Si-30 = 100% - 92.2% - 4.67% = 3.13% = 3.1% to 2 s.f.

Part B

The total atomic mass of an element is an addition combination of the mass and natural abundances of all the isotopes of that element.

Molar mass of Silicon normally = 28.0855 amu

Let the mass of Si-30 be m

28.0855 = (27.9769×0.922) + (28.9765×0.0467) + (m×0.0313)

28.0855 = 27.14790435 + 0.0313m

0.0313m = 28.0855 - 27.14790435 = 0.93759565

m = (0.93759565/0.0313) = 29.9551325879 amu = 29.9551 amu

Hope this Helps!!

You might be interested in

Consider these two entries from a fictional table of standard reduction potentials.

Vika [28.1K]

Answer:

1.89 V

Explanation:

To calculate the standard cell potential, subtract the reduction potential of the anode from the reduction potential of the cathode.

So for your calculation,

-0.22 V - (-2.11 V)= 1.89 V

8 0

3 years ago

What is the molar mass of a gas with a density of 3.50 g/L at a

Kruka [31]

Answer:

<u>104 g/mol</u>

Explanation:

Use the ideal gas equation:

$pV=nRT$

Where:

p is pressure: 0.950atm
V is volume: unknown
n is number of moles: unknown
R is the universal constat of gases: 0.08206 atm.liter/ (K.mol)
T is the absolute temperature: 345K

Use the <em>molar mass</em> of the gas to include the density in the formula:

molar mass = mass in grams / number of moles

⇒ mass in grams = number of moles × molar mass

density = mass in grams / volume

⇒ density = number of moles × molar mass / volume

density = (n/V) × molar mass

⇒ n/V = density / molar mass

Clear n/V from the gas ideal equation and subsittute with density/molar mass:

n/V = p/(RT)

density / molar mass = n/V

density/molar mass = p/(RT)

molar mass = density × RT / p

Now you can subsitute the data:

molar mass = (3.50g/liter) × 0.08206 atm.liter/(K.mol) × 345K / 0.950 atm

molar mass = 104.3 g/mol

Round to the nearest whole number: 104g/mol ← answer

3 0

3 years ago

Explain how camaflouge is an adaption

suter [353]

Camouflage is an adaptation because you change yourself for better survival.

For example, an Octopus can change the color of its body to blend in with its surroundings which makes it hard for predators and prey to see them.

3 0

3 years ago

1.00 M CaCl2 Density = 1.07 g/mL

Lesechka [4]

Explanation:

Molarity of solution = 1.00 M = 1.00 mol/L

In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

$m = 1 mol\times 111 g/mol = 111 g$

Mass of solution = M

Volume of solution = V = 1L = 1000 mL

Density of solution , d= 1.07 g/mL

$M=d\times V=1.07 g/mL\times 1000 mL=1,070 g$

1) The value of %(m/M):

$\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%$

2) The value of %(m/V):

$\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%$

$Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}$

$Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}$

n = Equivalent mass

n = $\frac{\text{molar mass of ion}}{\text{charge on an ion}}$

3) Normality of calcium ions:

Moles of calcium ion = 1 mol (1 $CaCl_2$ mole has 1 mole of calcium ion)

$n=\frac{40 g/mol}{2}=20$

$=\frac{1 mol}{20 g/mol\times 1L}=0.050 N$

4) Normality of chlorine ions:

Moles of chlorine ion = 2 mol (1 $CaCl_2$ mole has 2 mole of chlorine ion)

$n=\frac{35.5 g/mol}{1}=35.5$

$=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N$

Moles of calcium chloride = 1.00 mol

Mass of solvent = Mass of solution - mass of solute

= 1,070 g - 111 g = 959 g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

$\frac{1 mol}{0.959 kg}=1.043 mol/kg$

Moles of calcium chloride = $n_1=1mol$

Mass of solvent = 959 g

Moles of water = $n_2=\frac{959 g}{18 g/mol}=53.28 mol$

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842$

7) Mole fraction of water =

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816$

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

$m'=d\times v=1.07 g/ml\times 100 g= 107 g$

Mass of 100 mL of this solution 107 grams of solution.

9) Volume of solution = V = 100 mL

Mass of solution = M'' = 107 g

Mass of solute = m

The value of %(m/V) of solution = 11.1%

$11.1\%=\frac{m}{100 mL}\times 100$

m = 11.1 g

Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g

95.9 grams of water was present in 100 mL of given solution.

3 0

3 years ago

What makes the periodic table such a useful tool? *

Sidana [21]

Answer:

I think it is D. It allows you to compare the properties of the elements. I might be wrong.

5 0

4 years ago