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2 years ago

How many moles of aluminum oxide al2o3 are in a sample with a mass of 204.0

Chemistry

2 answers:

Leviafan [203]2 years ago

6 0

Answer:

2 moles

Explanation

Molecular mass of Al : 26.981539 u

Molecular mass of O: 15.999 u

The molecular mass of Al₂O₃ is:

= 101.960078 u

This can be approximated to 102 u.

1mole weighs 102 u

So, 2moles will weigh 2*102 = 204 u

erik [133]2 years ago

3 0

Answer:

2 moles

Explanation:

Let us first start by calculating the molecular mass of Al₂O₃.

The mass of a mole of any compound is called it's molar mass. 1 molar mass 6.02 X 10²³, or Avogadro's number, of compound entities.

Say, 1 mole of Al₂O₃ has 6.02 X 10²³ of Al₂O₃ molecules/atoms. It also has 2*6.02 X 10²³ number of Al atoms and 3*6.02 X 10²³ number of O atoms.

Molecular mass of Al : 26.981539 u

Molecular mass of O: 15.999 u

Therefore, molecular mass of Al₂O₃ is:

= $(2*26.981539) + (3*15.999)$ u

= 101.960078 u

This can be approximated to 102 u.

1mole weighs 102 u

So, 2moles will weigh 2*102 = 204 u

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What is energy conversion

strojnjashka [21]

Answer:

Energy transformation,also known as energy conversion,is the process of changing energy from one form to another.

Explanation:

For example,to heat a home,the fornace burns fuel,whose chemical potential energy is converted into thermal energy,which is then transferred to the homes air to raise its temperature.

6 0

3 years ago

A 30ml sample of a liquid has a mass of 50 grams. What is the density of the liquid?

mash [69]

Answer:

<h2>Density = 1.67 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

$Density = \frac{mass}{volume}$

From the question

mass = 50 g

volume = 30 mL

Substitute the values into the above formula and solve for the density

That's

$Density = \frac{50}{30} \\ = \frac{5}{3} \\ = 1.66666...$

Wr have the final answer as

<h3>Density = 1.67 g/mL</h3>

Hope this helps you

5 0

3 years ago

Describe the appearance of the material. Include its color and state (solid or liquid). Pour ¼ cup of water into a small, clear

Alex777 [14]

<u>Answer:</u>

<em>Here the given material is taken and mixed with water.</em>

<u>Explanation:</u>

The amount of material and water taken are same. Hence if it is not soluble in water it should make a dense and flowy paste like material and if it is soluble in water it should this and thicker density of water should remain.

If the amount of water that we are taking is more than the material will float in water if it is not soluble and lighter than water or would sink if it is heavier than water.

5 0

3 years ago

Read 2 more answers

What is the balanced NET ionic equation for the reaction when aqueous Cs₃PO₄ and aqueous AgNO₃ are mixed in solution to form sol

rusak2 [61]

Answer:

$PO_4^{3-}(aq)+3Ag^+(aq)\rightarrow Ag_3PO_4(s)$

Explanation:

Hello!

In this case, since the net ionic equation of a chemical reaction shows up the ionic species that result from the simplification of the spectator ions, which are those at both reactants and products sides, we take into account that aqueous species ionize into ions whereas liquid, solid and gas species remain unionized. In such a way, for the reaction of cesium phosphate and silver nitrate we can write the complete molecular equation:

$Cs_3PO_4(aq)+3AgNO_3(aq)\rightarrow Ag_3PO_4(s)+3CsNO_3(aq)$

Whereas the three aqueous salts are ionized in order to write the following complete ionic equation:

$3Cs^+(aq)+PO_4^{3-}(aq)+3Ag^+(aq)+3NO_3^-(aq)\rightarrow Ag_3PO_4(s)+3Cs^+(aq)+3NO_3^-(aq)$

In such a way, since the cesium and nitrate ions are the spectator ions because of the aforementioned, the net ionic equation turns out:

$PO_4^{3-}(aq)+3Ag^+(aq)\rightarrow Ag_3PO_4(s)$

Best regards!

7 0

2 years ago

A mouse is placed in a sealed chamber with air at 769.0 torr. This chamber is equipped with enough solid KOH to absorb any CO2 a

svlad2 [7]

<u>Answer:</u> The amount of oxygen gas consumed by mouse is 0.202 grams.

<u>Explanation:</u>

We are given:

Initial pressure of air = 769.0 torr

Final pressure of air = 717.1 torr

Pressure of oxygen = Pressure decreased = Initial pressure - Final pressure = (769.0 - 717.1) torr = 51.9 torr

To calculate the amount of oxygen gas consumed, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 51.9 torr

V = Volume of the gas = 2.20 L

T = Temperature of the gas = 292 K

R = Gas constant = $62.364\text{ L. Torr }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$51.9torr\times 2.20L=n\times 62.364\text{ L. Torr }mol^{-1}K^{-1}\times 292K\\\\n=\frac{51.9\times 2.20}{62.364\times 292}=0.0063mol$

To calculate the mass from given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of oxygen gas = 0.0063 moles

Molar mass of oxygen gas = 32 g/mol

Putting values in above equation, we get:

$0.0063mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(0.0063mol\times 32g/mol)=0.202g$

Hence, the amount of oxygen gas consumed by mouse is 0.202 grams.

5 0

2 years ago