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2 years ago

Why is it important to convert into units of kelvin before using the ideal gas law?.

1 answer:

iogann1982 [59]2 years ago

8 0

It is important to convert into units of kelvin before using the ideal gas law because
Temperature is the denominator in the combined gas laws and can be derived in the idea gas law to the denominator.

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Why are gold rings not made from 100% gold?

aleksandr82 [10.1K]

Gold is actually very soft. If rings were made of pure gold they would bend and become disfigured. Other elements are added to give the ring its stability.

3 0

3 years ago

Read 2 more answers

Identify the right

Basile [38]

.answer:

so the first one is analogous

the second is homologous

and the third is analogous

explanation:

analogous. this means they share a similar function (flight) but do not have the same embryonic origin.

6 0

3 years ago

Group 18 noble gases are relatively inert because *

choli [55]

Answer:I believe its A. There noble gasses so there happy with there electron count!

Explanation:

3 0

3 years ago

When hydrated copper(II) sulfate crystals are heated, anhydrous copper(II) sulfate forms. A mass of 12.5 g of hydrated copper(II

pychu [463]

The formula of hydrated copper(II) sulfate is CuSO4.10H2O

<h3>What is the formula of the hydrated copper (ii) sulfate salt?</h3>

The formula of the hydrated copper (ii) sulfate is determined as follows:

Mass of hydrated salt = 12.5 g

Mass of anhydrous salt = 8.0 g

Mass of water = 12.5 - 8 = 4.5 g

mole ratio of water and anhydrous salt is;

4.5/18 : 8.0/159.5

0.562 : 0.05

10 : 1

Water of crystallization (n) = 10.

Therefore, the formula of hydrated copper(II) sulfate is CuSO4.10H2O

Learn more about water of crystallization at: brainly.com/question/26146814

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7 0

2 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