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3 years ago

What is the mass in grams of 1.24 mol of water

1 answer:

4 0

Mass of H2O=2.01568g+15.9994g=18.01508gmol H2O.
Step 2: Multiply 1.24 mol H2O with the molecular mass. 1.24 mol H2O x 18.01508gmol H2O = 22.3 g H2O.

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Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has

Ymorist [56]

Answer:

$\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

Explanation:

Given that:

The Half-life of $^{235}U$ = $7.13 \times 10^8 \ years$ is less than that of $^{238} U = 4.51 \times 10^9 \ years$

Although we are not given any value about the present weight of $^{235}U$ .

So, consider the present weight in the percentage of $^{235}U$ to be y%

Then, the time elapsed to get the present weight of $^{235}U$ = $t_1$

Therefore;

$N_1 = N_o e^{-\lambda \ t_1}$

here;

$N_1$ = Number of radioactive atoms relating to the weight of y of $^{235}U$

Thus:

$In( \dfrac{N_1}{N_o}) = - \lambda t_1$

$In( \dfrac{N_o}{N_1}) = \lambda t_1$ --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of $^{235}U$ to be = $t_2$

Then:

$In( \dfrac{N_o}{N_2}) = \lambda t_2$ ---- (2)

here;

$N_2$ = Number of radioactive atoms of $^{235}U$ relating to 3.0 a/o weight

Now, equating equation (1) and (2) together, we have:

$In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) = \lambda( t_1-t_2)$

replacing the half-life of $^{235}U$ = $7.13 \times 10^8 \ years$

$In( \dfrac{N_2}{N_1}) = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)$ ( since $\lambda = \dfrac{In 2}{t_{1/2}}$ )

∴

$\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}$

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is $\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

8 0

3 years ago

What sublevels are contained in the hydrogen atoms first four energy levels what orbitals are related to each?

erik [133]

1st level = s
2nd level = s,p 
3rd level = s,p,d 
4th level = s,p,d,f

6 0

3 years ago

Question 6

I am Lyosha [343]

Answer:

A. is the correct point.

Explanation:

This is true because no matter how many mL of water is added, the solution only gets more height; the concentration in everything else stays the same, and water doesn't have any concentration. Very confusing, I know. Good luck!

4 0

2 years ago

If you wanted to change the polarity of hydrogen bromide (HBr) by substituting

Tcecarenko [31]

Chlorine. Electronegativity generally increases up and across the periodic table

7 0

3 years ago

Can somebody help me, please!

zavuch27 [327]

Answer:

Rubidium-85=61.2

Rubidium-87=24.36

Atomic Mass=85.56 amu

Explanation:

To find the atomic mass, we must multiply the masses of the isotope by the percent abundance, then add.

Rubidium-85

This isotope has an abundance of 72%.

Convert 72% to a decimal. Divide by 100 or move the decimal two places to the left.

72/100= 0.72 or 72.0 --> 7.2 ---> 0.72

Multiply the mass of the isotope, which is 85, by the abundance as a decimal.

mass * decimal abundance= 85* 0.72= 61.2

Rubidium-85=61.2

Rubidium-87

This isotope has an abundance of 28%.

Convert 28% to a decimal. Divide by 100 or move the decimal two places to the left.

28/100= 0.28 or 28.0 --> 2.8 ---> 0.28

Multiply the mass of the isotope, which is 87, by the abundance as a decimal.

mass * decimal abundance= 87* 0.28= 24.36

Rubidium-87=24.36

Atomic Mass of Rubidium:

Add the two numbers together.

Rb-85 (61.2) and Rb-87 (24.36)

61.2+24.36=85.56 amu

4 0

3 years ago