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3 years ago

A mixture contains 20% iron ions by mass. What mass of iron chloride (FeCl) would

Chemistry

1 answer:

omeli [17]3 years ago

8 0

Mass of iron chloride (FeCl₂) : 22.68 g

<h3>Further explanation</h3>

Proust stated the Comparative Law that compounds are formed from elements with the same Mass Comparison so that the compound has a fixed composition of elements

In the same compound, although from different sources and formed by different processes, it will still have the same composition/comparison

With this law, we can calculate how many grams an element is needed to make a compound with a certain mass, as desired

<h3 />

$\tt mass~Fe=\dfrac{Ar~Fe}{MW~FeCl_2}\times mass~FeCl_2$

The iron ions in 50 g mixture :

$\tt 20\%\times 50~g=10~g$

$\tt 10=\dfrac{56}{127}\times mass~FeCl_2\\\\mass~FeCl_2=22.68~g$

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nignag [31]

The answer is D higher potential energy and is unstable

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2 years ago

Which of the following characteristics describes aerobic organisms.

horsena [70]

Answer: Option (3) is the correct answer.

Explanation:

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4 years ago

How many mole of ZnCl2 will be produced from 55.0 g of Zn, assuming HCL is available in excess

Ilia_Sergeevich [38]

Answer:

0.84 mol

Explanation:

Given data:

Moles of ZnCl₂ produced = ?

Mass of Zn = 55.0 g

Solution:

Chemical equation:

2HCl + Zn → ZnCl₂ + H₂

Number of moles of Zn:

Number of moles = mass / molar mass

Number of moles = 55.0 g/ 65.38 g/mol

Number of moles = 0.84 mol

Now we will compare the moles of Zn with ZnCl₂ from balance chemical equation.

Zn : ZnCl₂

1 : 1

0.84 : 0.84

So from 55 g of Zn 0.84 moles of zinc chloride will be produced.

8 0

4 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 $O_2$ .

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