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3 years ago

What is the purpose of the periotic table?

Chemistry

1 answer:

Alexus [3.1K]3 years ago

6 0

Answer:

The periodic table was first designed by Mendeleev. It is designed to systematically categorise elements according to atomic number, electron configuration and recurring chemical properties. This allows identification of elements through their characteristics simply by analysing its position on the table.

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How would the freezing point of seawater compare with that of pure water? a)higher b) lower c) the same d) ten times higher

maksim [4K]

The answer would be B) Lower, because pure water freezes at 32 degrees Fahrenheit whereas salt water freezes at 28.4 degrees Fahrenheit. Hopefully this helps!

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3 years ago

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Ludmilka [50]

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2 years ago

What is the overall reaction order for this rate law: rate = k[A]2[B][C]?

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3 years ago

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A soluti

ella [17]

Answer :

The mass of excess mass of $Na_2CO_3$ , $AgNO_3,Ag_2CO_3\text{ and }NaNO_3$ are, 1.908 g, 0 g, 12.144 g and 3.74 g respectively.

Explanation : Given,

Mass of $Na_2CO_3$ = 4.25 g

Mass of $AgNO_3$ = 7.50 g

Molar mass of $Na_2CO_3$ = 106 g/mole

Molar mass of $AgNO_3$ = 170 g/mole

Molar mass of $Ag_2CO_3$ = 276 g/mole

Molar mass of $NaNO_3$ = 85 g/mole

First we have to calculate the moles of $Na_2CO_3$ and $AgNO_3$ .

$\text{Moles of }Na_2CO_3=\frac{\text{Mass of }Na_2CO_3}{\text{Molar mass of }Na_2CO_3}=\frac{4.25g}{106g/mole}=0.040moles$

$\text{Moles of }AgNO_3=\frac{\text{Mass of }AgNO_3}{\text{Molar mass of }AgNO_3}=\frac{7.50g}{170g/mole}=0.044moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Na_2CO_3+2AgNO_3\rightarrow Ag_2CO_3+2NaNO_3$

From the balanced reaction we conclude that

As, 2 moles of $AgNO_3$ react with 1 mole of $Na_2CO_3$

So, 0.044 moles of $AgNO_3$ react with $\frac{0.044}{2}=0.022$ moles of $Na_2CO_3$

From this we conclude that, $Na_2CO_3$ is an excess reagent because the given moles are greater than the required moles and $AgNO_3$ is a limiting reagent and it limits the formation of product.

The excess mole of $Na_2CO_3$ = 0.040 - 0.022 = 0.018 mole

Now we have to calculate the mass of excess mole of $Na_2CO_3$ .

$\text{Mass of }Na_2CO_3=\text{Moles of }Na_2CO_3\times \text{Molar mass of }Na_2CO_3=(0.018mole)\times (106g/mole)=1.908g$