Can you give me more explanation in the picture?

the element carbon has two common isotopes: C-12 (12 U) AND C-13 (13.003355 U). IF THE AVERAGE ATOMIC MASS OF CARBON IS 12.0107

oksian1 [2.3K]

Answer:

The percent isotopic abundance of C- 12 is 98.93 %

The percent isotopic abundance of C- 13 is 1.07 %

Explanation:

we know there are two naturally occurring isotopes of carbon, C-12 (12u) and C-13 (13.003355)

First of all we will set the fraction for both isotopes

X for the isotopes having mass 13.003355

1-x for isotopes having mass 12

The average atomic mass of carbon is 12.0107

we will use the following equation,

13.003355x + 12 (1-x) = 12.0107

13.003355x + 12 - 12x = 12.0107

13.003355x- 12x = 12.0107 -12

1.003355x = 0.0107

x= 0.0107/1.003355

x= 0.0107

0.0107 × 100 = 1.07 %

1.07 % is abundance of C-13 because we solve the fraction x.

now we will calculate the abundance of C-12.

(1-x)

1-0.0107 =0.9893

0.9893 × 100= 98.93 %

98.93 % for C-12.

7 0

3 years ago

If 20.0 g of NaOH is added to 0.750 L of 1.00 M Cd(NO₃)₂, how many grams of Cd(OH)₂ will be formed in the following precipitatio

bulgar [2K]

Answer:

$m_{Cd(OH)_2}=36.6 gCd(OH)_2$

Explanation:

Hello.

In this case, for the given chemical reaction, in order to compute the grams of cadmium hydroxide that would be yielded, we must first identify the limiting reactant by computing the yielded moles of that same product, by 20.0 grams of NaOH (molar mass = 40 g/mol) and by 0.750 L of the 1.00-M solution of cadmium nitrate as shown below considering the 1:2:1 mole ratios respectively:

$n_{Cd(OH)_2}^{by\ NaOH}=20.0gNaOH*\frac{1molNaOH}{40gNaOH} *\frac{1molCd(OH)_2}{2molNaOH} =0.25molCd(OH)_2\\\\n_{Cd(OH)_2}^{by\ Cd(NO_3)_2}=0.750L*1.00\frac{molCd(NO_3)_2}{L}*\frac{1molCd(OH)_2}{1molCd(NO_3)_2} =0.75molCd(OH)_2$

Thus, since 20.0 grams of NaOH yielded less of moles of cadmium hydroxide, NaOH is the limiting reactant, therefore the mass of cadmium hydroxide (molar mass = 146.4 g/mol) is:

$m_{Cd(OH)_2}=0.25molCd(OH)_2*\frac{146.4gCd(OH)_2}{1molCd(OH)_2} \\\\m_{Cd(OH)_2}=36.6 gCd(OH)_2$

Best regards.

4 0

3 years ago

Gallium has two naturally occurring isotopes: 69ga with a mass of 68.9256 amu and a natural abundance of 60.11% and 71ga. use th

Nataly [62]

There are two naturally occurring isotopes of gallium: mass of Ga-69 isotope is 68.9256 amu and its percentage abundance is 60.11%, let the mass of other isotope that is Ga-71 be X, the percentage abundance can be calculated as:

%Ga-71=100-60.11=39.89%

Atomic mass of an element is calculated by taking sum of atomic masses of its isotopes multiplied by their percentage abundance.

Thus, in this case:

Atomic mass= m(Ga-69)×%(Ga-69)+X×%(Ga-71)

From the periodic table, atomic mass of Ga is 69.723 amu.

Putting the values,

$69.723 amu=(68.9256 amu)(\frac{60.11}{100})+X(\frac{39.89}{100})$