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Ray Of Light [21]
3 years ago
9

How many moles are present in 45.0 g of lithium oxide (Li2O)?

Chemistry
1 answer:
BigorU [14]3 years ago
4 0
Molar mass Li2O = 7.0 x 2 + 16 => 30 g/mol

1 mole Li2O ---------- 30 g
( moles Li2O ) -------- 45.0 g

moles Li2O = 45.0 x 1 / 30

moles Li2O = 45.0 / 30

= 1.50 moles of Li2O 

hope this helps!


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PLEASE ANSWER THIS QUESTION ASAP 50 POINTS
quester [9]

Answer:

A very large amount of energy is produced from a series of chemical reactions.

Explanation:

Nuclear fission is the process of splitting apart nuclei (usually large nuclei). When large nuclei, such as uranium-235, fissions, energy is released. So much energy is released that there is a measurable decrease in mass, from the mass-energy equivalence. This means that some of the mass is converted to energy.

4 0
3 years ago
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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
In the cathode ray tube experiment, J. J. Thomson passed an electric current through different gases inside a cathode ray tube i
Ber [7]

It showed that atoms can be divided into smaller parts.

It showed that all atoms contain electrons.

Explanation:

The experiment carried out by J.J Thomson on the gas discharge tube by passing electric current through a tube filled with many different gases provided a good insight into the structure of an atom.

This experiment led to the development of the plum pudding model of the atom.

  • Cathode rays and it properties were discovered in this set up.
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  • The cathode rays which were later termed electrons became a fundamental particles known for every atom.

learn more:

Rutherford's model of the atom brainly.com/question/1859083

#learnwithBrainly

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Answer:

4057.85 g/mol

Explanation:

Hello, the numerical procedure is shown in the attached file.

- In this case, since we don't have the density of the protein, we must assume that the volume of the solution is solely given by the benzene's volume, in order to obtain the moles of the solute (protein).

-Van't Hoff factor is assumed to be one.

Best regards.

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3 years ago
Why cant fungi make there own food using photosynthesis
dangina [55]
Fungi cannot make their food from sunlight, water and carbon dioxide as plants do, in the process known as photosynthesis
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