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mars1129 [50]
3 years ago
6

The open spaces in water's crystal structure make it possible for

Chemistry
1 answer:
SashulF [63]3 years ago
6 0
The open spaces in water's crystal structure make it possible for <span>aquatic life to exist at the north pole.

It is good to know that, at the north pole, when the temperature drops below zero, only the top layer of the water freezes. Underneath this freezing layer, water remains in its liquid form, thus, allowing marine life to exist. As for oxygen, it is also trapped underneath the ice layer. </span>
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If aluminum is heated from 650°C to 670°C, then it will change from a A.liquid to a gas. B.liquid to a solid. C.solid to a liqui
kirill115 [55]
Answer: C

Explanation:
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3 years ago
Is it possible for an object not to be in motion and still have force acting upon it?
RSB [31]
Yes,this is an example:
An apple is sitting on a desk he is not in motion but one of the forces acting upon it is gravitational pull because is not floating around, the earth gravitational pull is pulling it down to the earth's surface.

3 0
3 years ago
Pleasee help me I really need an answer help
Juliette [100K]

Answer:

This is hard for me also but I am trying to find out answer

Explanation:

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6 0
3 years ago
A chemist prepares a solution of silver perchlorate by measuring out of silver perchlorate into a volumetric flask and filling t
djverab [1.8K]

Complete Question:

A chemist prepares a solution of silver (I) perchlorate (AgCIO4) by measuring out 134.g of silver (I) perchlorate into a 50.ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the silver (I) perchlorate solution. Round your answer to 2 significant digits.

Answer:

13 mol/L

Explanation:

The concentration in mol/L is the molarity of the solution and indicates how much moles have in 1 L of it. So, the molarity (M) is the number of moles (n) divided by the volume (V) in L:

M = n/V

The number of moles is the mass (m) divided by the molar mass (MM). The molar mass of silver(I) perchlorate is 207.319 g/mol, so:

n = 134/207.319

n = 0.646 mol

So, for a volume of 50 mL (0.05 L), the concentration is:

M = 0.646/0.05

M = 12.92 mol/L

Rounded to 2 significant digits, M = 13 mol/L

7 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
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