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3 years ago

A tool might be made of an alloy because the alloy _____.

Chemistry

2 answers:

zheka24 [161]3 years ago

7 0

<h2>~Hello Wonderful Person~</h2>

Your answer is:

B. Resists rust.

A tool might be made of an alloy because the alloy resists rust.

Why?

Iron is a great building material but steel (an alloy made by adding small amounts of non-metallic carbon to iron) is stronger, harder, and <u>rustproof</u><u>.</u> Aluminum is a very light metal but it's also very soft in its pure form.

Alloy steels have a wide range of special properties, such as <u>hardness</u>, toughness,<u> corrosion resistance,</u> magnetizability, and ductility. Nonferrous alloys, mainly copper–nickel, bronze, and aluminum alloys, are much used in the coinage.

____________________________________________________________

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Basile [38]3 years ago

3 0

I assume the answer is: A

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Explanation:

From the given riddle. Potassium is the right answer. This because on the periodic table, potassium has the smallest atomic number of any element in its period with the electronic configuration 1s2 2s2 2p6 3s2 3p6 4s1, it has the largest atomic radius within its group because the radius has no d-electron shell filled. Potassium is a soft metal but contains high reactivity.

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

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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 $Na_2CO_3$ .

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

Now we have to calculate the moles of $Ag_2CO_3$ .

As, 1 moles of $AgNO_3$ react to give 1 moles of $Ag_2CO_3$

So, 0.044 moles of $AgNO_3$ react to give 0.044 moles of $Ag_2CO_3$

Now we have to calculate the mass of $AgCO_3$ .

$\text{Mass of }Ag_2CO_3=\text{Moles of }Ag_2CO_3\times \text{Molar mass of }Ag_2CO_3=(0.044mole)\times (276g/mole)=12.144g$

Now we have to calculate the moles of $NaNO_3$ .

As, 2 moles of $AgNO_3$ react to give 2 moles of $NaNO_3$

So, 0.044 moles of $AgNO_3$ react to give 0.044 moles of $NaNO_3$

Now we have to calculate the mass of $NaNO_3$ .

$\text{Mass of }NaNO_3=\text{Moles of }NaNO_3\times \text{Molar mass of }NaNO_3=(0.044mole)\times (85g/mole)=3.74g$

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Answer:

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To balance the equation it should read NaO+H2O-->2NaOH

Explanation:

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3 years ago

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Answer:

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Explanation:

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3 years ago