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3 years ago

15

A material that cannot be broken down farther by chemical means

1 answer:

antoniya [11.8K]3 years ago

8 0

Answer:

<em>An element is the material that cannot be broken down further by chemical means.</em>

Explanation:

An element can be described as a material which is made up of only one kind of atom. For example, argon, hydrogen etc. An element cannot be broken down further by physical or chemical means.

Compounds, on the other hand, are made of different kinds of atoms. They can be broken down to simpler molecules by chemical means. For example, water is a compound made up of hydrogen and oxygen.

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2 years ago

s) Suppose we now collect hydrogen gas, H2(g), over water at 21◦C in a vessel with total pressure of 743 Torr. If the hydrogen g

Elenna [48]

This is an incomplete question, here is a complete question.

Suppose we now collect hydrogen gas, H₂(g), over water at 21°C in a vessel with total pressure of 743 Torr. If the hydrogen gas is produced by the reaction of aluminum with hydrochloric acid:

$2Al(s)+6HCl(aq)\rightarrow 2AlCl_3(aq)+3H_2(g)$

what volume of hydrogen gas will be collected if 1.35 g Al(s) reacts with excess HCl(aq)? Express your answer in liters.

Answer : The volume of hydrogen gas that will be collected is 1.85 L

Explanation :

First we have to calculate the number of moles of aluminium.

Given mass of aluminium = 1.35 g

Molar mass of aluminium = 27 g/mol

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Moles of aluminium}=\frac{1.35g}{27g/mol}=0.05mol$

The given chemical reaction is:

$2Al(s)+6HCl(aq)\rightarrow 2AlCl_3(aq)+3H_2(g)$

As, hydrochloric acid is present in excess. So, it is considered as an excess reagent.

Thus, aluminium is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of hydrogen gas

So, 0.005 moles of aluminium will produce = $\frac{3}{2}\times 0.05=0.0750mol$ of hydrogen gas

Now we have to calculate the mass of helium gas by using ideal gas equation.

PV = nRT

where,

P = Pressure of hydrogen gas = 743 Torr

V = Volume of the helium gas = ?

n = number of moles of hydrogen gas = 0.075 mol

R = Gas constant = $62.364\text{ L Torr }mol^{-1}K^{-1}$

T = Temperature of hydrogen gas = $21^oC=[21+273]K=294K$

Now put all the given values in above equation, we get:

$743Torr\times V=0.075mol\times 62.364\text{ L Torr }mol^{-1}K^{-1}\times 294K\\\\V=1.85L$

Hence, the volume of hydrogen gas that will be collected is 1.85 L

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