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Furkat [3]
3 years ago
13

Freshwater biomes contain less salt than ocean biomes. true or false

Chemistry
1 answer:
Nadya [2.5K]3 years ago
4 0

true...as it gives home to many of the organisms

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During which time interval does the substance exist as both a liquid and a solid
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The Chemistry Regents is one of the four science Regents exams. The other three are Earth Science, Living Environment, and Physics. You'll need to pass at least one of these four exams to graduate high school.
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3 years ago
Mendeleev created this table as he noticed that a
Nimfa-mama [501]

Answer:

Mendeleev had left the noble gases out of his periodic table.

Explanation:

Mendeleev's periodic table is pictured in the image attached to the question.

Mendeleev's table obviously lacked the noble gases. The reason for this grave omission is simple; the noble gases were not known as at the time when he formulated his periodic table. There weren't any known elements  whose properties were similar to the  properties of the noble gases. This would have lead him to suspect their existence.

4 0
3 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
3 years ago
Which part of the atom cannot have its location accurately determined and is modeled by a cloud around the center of the atom? A
Tju [1.3M]

Answer:

Electron

Explanation:

The answer would be the electron because it is constantly moving so its location cannot be accurately determined

3 0
3 years ago
Read 2 more answers
Given the standard enthalpy changes for the following two reactions:
lawyer [7]

Answer:

ΔH° = -186.2 kJ

Explanation:

Hello,

This case in which the Hess method is applied to compute the required chemical reaction. Thus, we should arrange the given first two reactions as:

(1) it is changed as:

SnCl2(s) --> Sn(s) + Cl2(g)...... ΔH° = 325.1 kJ

That is why the enthalpy of reaction sign is inverted.

(2) remains the same:

Sn(s) + 2Cl2(g) --> SnCl4(l)......ΔH° = -511.3 kJ

Therefore, by adding them, we obtain the requested chemical reaction:

(3) SnCl2(s) + Cl2(g) --> SnCl4(l)

For which the enthalpy change is:

ΔH° = 325.1 kJ - 511.3 kJ

ΔH° = -186.2 kJ

Best regards.

7 0
3 years ago
Read 2 more answers
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