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3 years ago

How adding tin to copper makes the bronze alloy harder?

Chemistry

2 answers:

andrey2020 [161]3 years ago

8 0

It makes the bronze stronger and harder than either of the other two medals

nika2105 [10]3 years ago

4 0

The molecules are different sizes so they can’t move past each other in the structure.

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Which pair of elements are in the same period?Hg & PbLi & NaB & ClO & As

Alexeev081 [22]

Answer:

Hg Pb are in one, Li and B and O are in one, and NaCl are in one, As is alone

Explanation:

Periods are horizontal

Hope this helps!

4 0

2 years ago

Which describes an covalent bond?

Ilya [14]

Answer:

is there choices you have to pick from

Explanation:

or do you have to describe a covalent bond ?

8 0

3 years ago

Read 2 more answers

In a solution of a carbonated beverage the water is what? Solute, saturated, solvent or precipitate

Scilla [17]

Https://us-static.z-dn.net/files/d49/33d4ec86853ef95e6f6c14242c663be4.png

5 0

3 years ago

.1 Write 'a' 'an' or 'the' where is necessary.

kolezko [41]

The answers

A) a
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D) the

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3 years ago

For the following reaction, 4.07 grams of aluminum oxide are mixed with excess sulfuric acid. The reaction yields 10.4 grams of

torisob [31]

Answer:

Theoretical yield = 13.7 g

% yield =76 %

Explanation:

For $Al_2O_3$

Mass of $Al_2O_3$ = 4.07 g

Molar mass of $Al_2O_3$ = 101.96 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{4.07\ g}{101.96\ g/mol}$

$Moles\ of\ Al_2O_3= 0.0399\ mol$

According to the reaction:

$Al_2O_3+3H_2SO_4\rightarrow Al_2(SO_4)_3+3H_2O$

1 mole of $Al_2O_3$ on reaction produces 1 mole of $Al_2(SO_4)_3$

So,

0.0399 mole of $Al_2O_3$ on reaction produces 0.0399 mole of $Al_2(SO_4)_3$

Moles of $Al_2(SO_4)_3$ obtained = 0.0399 mole

Molar mass of $Al_2(SO_4)_3$ = 342.2 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.0399= \frac{Mass}{342.2\ g/mol}$

$Mass= 13.7\ g$

<u>Theoretical yield = 13.7 g</u>

The expression for the calculation of the percentage yield for a chemical reaction is shown below as:-

$\%\ yield =\frac {Experimental\ yield}{Theoretical\ yield}\times 100$

Given , Values from the question:-

Theoretical yield = 13.7 g

Experimental yield = 10.4 g

Applying the values in the above expression as:-

$\%\ yield =\frac{10.4}{13.7}\times 100$

<u>% yield =76 %</u>

6 0

3 years ago