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2 years ago

12

Pls helppp ASAP I’ll brainlest

2 answers:

docker41 [41]2 years ago

5 0

The Answer is b (noble)

My name is Ann [436]2 years ago

4 0

Answer:

The answer is B

Explanation:

Hope that helps

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What is a solution in chemistry

Hatshy [7]

Solutions are a type of mixture.

A solution is a homogeneous mixture of particles so small that they cannot be seen and cannot be filtered out.

Definition: Homogeneous mixture looks the same.

I grabbed these from my science notes, hope this helps.

4 0

2 years ago

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

2 years ago

What is the pH of a substance with a hydrogen concentration of 1.0 x 10-13?

Nikolay [14]

Answer:

Favorite Answer

1.0 x10^-14 = (1.0 x 10^-13) (x)

x = 1.0 x 10^-1 = 0.1 M (this is the [OH-])

pOH = -log 0.1 = 1.0

Explanation:

I hope this helps :) sorry if not :(

4 0

2 years ago

Please write thoughtful, concise summaries of your findings in paragraph form. Include key calculation values and point out what

dedylja [7]

I am not sure how to do this

7 0

3 years ago

HELP SCIENCE QUIZ I WILL MARK YOU BRAINLIEST

Goryan [66]

Answer:

i want to say nuclues but again i also think that it is electron shells

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2 years ago

Read 2 more answers