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3 years ago

Who discovered atoms?

Chemistry

1 answer:

Kay [80]3 years ago

5 0

✧･ﾟ: *✧･ﾟ:* *:･ﾟ✧*:･ﾟ✧

Hello!

✧･ﾟ: *✧･ﾟ:* *:･ﾟ✧*:･ﾟ✧

❖ J.J Thomson was the first to discover atoms. He discovered atoms in 1897.

~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡

~ ᴄʟᴏᴜᴛᴀɴꜱᴡᴇʀꜱ

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<h3>What is a balanced chemical equation?</h3>

A balanced chemical equation is one in wgich the moles of atoms in the reactants side is equal to the moles of atoms on the product side.

The given equations of reaction is not clearly stated.

Therefore, the unbalanced equation is one in which the moles of atoms are not equal on both sides of the reaction.

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3 years ago

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3 years ago

A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor

AlexFokin [52]

Answer : The partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of $C_2HBrClF_3$ = 15.0 g

Mass of $O_2$ = 22.6 g

Molar mass of $C_2HBrClF_3$ = 197.4 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_2HBrClF_3$ and $O_2$ .

$\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole$

and,

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole$

Now we have to calculate the mole fraction of $C_2HBrClF_3$ and $O_2$ .

$\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971$

and,

$\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903$

Now we have to partial pressure of $C_2HBrClF_3$ and $O_2$ .

According to the Raoult's law,

$p^o=X\times p_T$

where,

$p^o$ = partial pressure of gas

$p_T$ = total pressure of gas

$X$ = mole fraction of gas

$p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T$

$p_{C_2HBrClF_3}=0.0971\times 862torr=84torr$

and,

$p_{O_2}=X_{O_2}\times p_T$

$p_{O_2}=0.903\times 862torr=778torr$

Therefore, the partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

6 0

3 years ago

Who discovered atoms?​

Who discovered atoms?