All categories

4 years ago

How many grams of nitrogen are in a sample of Al(NO3)3 that contains 364.0 g of aluminum? Answer: ________ grams of nitrogen.

Chemistry

1 answer:

saw5 [17]4 years ago

7 0

<u>Answer:</u> The mass of nitrogen that is present for given amount of aluminium is 566.22 g.

<u>Explanation: </u>

We are given:

Mass of aluminium = 364 grams

The chemical formula of aluminium nitrate is $Al(NO_3)_3$

Molar mass of nitrogen = 14 g/mol

Molar mass of aluminium = 27 g/mol

In 1 mole of aluminium nitrate, 27 grams of aluminium is combining with 42 grams of nitrogen.

So, 364 grams of aluminium will combine with = $\frac{42}{27}\times 364=566.22g$ of nitrogen.

Hence, the mass of nitrogen that is present for given amount of aluminium is 566.22 g.

You might be interested in

A gem has a mass of 5. 50 g. When the gem is placed in a graduated cylinder containing 9. 500 ml of water, the water level rises

Katena32 [7]

The density in grams per milliliter 7.975.

<h3>What is density?</h3>

Density is the mass of per unit volume. Density is equals to mass into volume.

D = M x V

The mass of a gem is given, M = 5.50

The volume is 10.45 - 9.00 = 1.45

Thus, the density is

D = 5.50 × 1.45 = 7.975

Thus, the density in grams per milliliter is 7.975.

Learn more about density

brainly.com/question/19320982

#SPJ4

6 0

2 years ago

The coinage metals -- copper, silver, and gold -- crystallize in a cubic closest packed structure. Use the density of silver (10

lara31 [8.8K]

Answer:

The answer to the question is

The approximate atomic radius for silver is 1.445×10⁻⁸ cm

Explanation:

To solve the question we list out the known variables as follows

The density of silver = 10.5 g/cm³

Molar mass of silver = 107.9 g/mol

Number of moles in one cm³ of silver = 10.5/107.9 moles or 0.0973 moles

Avogadro's law states that equal volumes of all substances contain equal number of particles that is one mole of any substance contain 6.022 × 10²³ elementary particles

Therefore one mole of silver contains 6.022 × 10²³ silver atoms

number of moles of silver in a unit cell = 4/6.022 × 10²³ or 6.642 × 10⁻²⁴ mol which has a mass = 6.642 × 10⁻²⁴ mol × molar mass of silver or 7.167× 10⁻²² g

Therefore the volume of a unit cell is given by Volume = mass/density =

7.167× 10⁻²² g/10.5 g/cm³ = 6.83× 10⁻²³ cm³

The diagonal of the face of a unit cell contains four atomic silver radius therefore

That is 4 × silver radius = diagonal of cubic unit cell face

= √2 × ∛(6.83 × 10⁻²³ cm³)

The approximate atomic radius for silver = (1/4) × √2 × ∛(6.83× 10⁻²³ cm³)

= 1.445×10⁻⁸ cm

8 0

4 years ago

What are the two components of a solution

7nadin3 [17]

The components of a solution are solute and solvent

8 0

3 years ago

Help please?? I’m confused

Nady [450]

It’s c. O stands for oxygen and h stands for hydrogen. They both do play in the roll

3 0

3 years ago

What metalic element is sometimes used in flarres and flashbulbs

hjlf

Magnesium bc it burns a bright light when used

7 0

4 years ago