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3 years ago

What elements fill the d orbitals on the periodic table?

Chemistry

1 answer:

lions [1.4K]3 years ago

8 0

The elements that fill the d orbitals on the periodic table are called transition metals.

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What is the process of using one or more of your five senses to measure or collect data?

sveta [45]

Answer:

B no observation pls mark me branilest

7 0

3 years ago

Read 2 more answers

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

Osmium is the most dense element we know of. A 22 g sample of Osmium has a volume of 100 cL. Calculate the density of Osmium in

kakasveta [241]

Answer:

a. V = 1000 mL

b. Denisty = 0.022 g/mL

Explanation:

First we need to convert the volume of the Osmium into mL. For that purpose we are given the conversion unit as:

1 mL = 0.1 cL

Hence, the given volume of Osmium will be:

V = Volume of Osmium = 100 cL = (100 cL)(1 mL/0.1 cL) = 1000 mL

V = 1000 mL

The density of Osmium is given by the following formula:

Density = mass/Volume

Denisty = 22 g/1000 mL

Denisty = 0.022 g/mL

5 0

3 years ago

I'm in a rush please help

Alexeev081 [22]

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4 0

3 years ago

If 5.100 g of c6h6 is burned and the heat produced from the burning is added to 5691 g of water at 21 °c, what is the final temp

emmasim [6.3K]

C6H12 = 6x12 + 6x1 = 78.
The equation indicates that 2x78 = 156g benzene will produce 6542kJ.
Using proportions you can then calculate that
x/6542kJ = 7.9g / 156g
x = 331.3kJ = 331300J.

heat = mass x ΔT x 4.18J/g°
ΔT = 331300J / (5691g x 4.18J/g°) = 13.9°

final temp = 21 + 14° = 35°C

4 0

3 years ago