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3 years ago

How would understanding the properties of matter be helpful in other fields of study besides chemistry?

Chemistry

2 answers:

Vika [28.1K]3 years ago

8 0

Answer:

Matter is the basic unit of all things living or non living.Living thing or non- living things are made from matter.

Explanation:

Matter : It is defined as the substance which occupy space and has mass.

In other subject:

Matter is the basic unit of all things living or non living.Living thing or non- living things made from matter.

Matter can be found in different form such as liquid , solid and gases.

Matter can be any type of material.

In physics

Matter consist of various types of particles ,each with mass and size.

Matter is that substance which has inertia and occupies physical space.

erma4kov [3.2K]3 years ago

5 0

Matter is the basic unit of all things in the world, whether living or nonliving. It is not only useful in chemistry, but also to other fields of science like physics. Physics deals with motion, so it has something to do with matte especially the movement of molecules for solids and fluids.

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What mass of sucrose c12h22o11 is needed to make 300 ml of a 0.50m solution?

zepelin [54]

Use the concentration to obtain the moles. I am assuming you mean to write capital M. because little m means molality.

So, first convert the ml into Liters and then into moles, then moles to grams using the molar mass (just adding the values of each atom from the periodic table. )

Molar mass= 12 (12.0) + 22 (1.01)+ 11 (16.0)= 342 grams/mole

300 ml (1 liter/ 1000 mL) x (0.50 moles/ 1 Liter) x (342 grams/ 1 mole)= 51.3 grams

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3 years ago

In a reaction between 6.0 g of oxygen gas, 4.0 g of hydrogen gas, and 5.0 g of solid sulfur at standard temperature and pressure

nikitadnepr [17]

Answer:

The limiting reagent is the O₂

Explanation:

We can think, this reaction

2O₂(g) + H₂(g) + S(s) → H₂SO₄

Mole of each = Mass / molar mass

6 g / 32 g/m = 0.187 mole O₂

4g / 2 g/m = 2 mole H₂

5g / 32.06 g/m = 0.156 mole S

Ratio between reactants is 2:1:1, 1:2:1, 1:1:2

For 2 mole of O₂, I need to react 1 mol of H₂ and 1 mol of S

0.187 mole of O₂, I need (the half)

0.093 mole of H₂ and 0.093 mole of S

For 1 mole of H₂, I need to react 2 mole of O₂ and 1 mol of S

2 mole of H₂, I need (the double of O₂ and the same for S)

4 mole of O₂ ; 2 mole of S

For 1 mol of S, I need to react 1 mol of H₂ and 2 mole of O₂

0.156 mole I need the same amount for H₂ and the double for O₂

0.156 mole of H₂ and 0.312 mole of O₂

In both cases, I can't make react, all the mass of oxygen, so this is the limiting reagent.

6 0

4 years ago

Ammonia NH3 may react with oxygen to form nitrogen gas and water.4NH3 (aq) + 3O2 (g) \rightarrow 2 N2 (g) + 6H2O (l)If 2.15g of

bagirrra123 [75]

Answer:

NH3 is the limiting reactant

The % yield is 36.1 %

Explanation:

Step 1: Data given

Mass of NH3 = 2.15 grams

Mass of O2 = 3.23 grams

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

volume of N2 produced = 0.550 L

Temperature = 295 K

Pressure = 1.00 atm

Step 2: The balanced equation:

4NH3 (aq) + 3O2 (g) → 2 N2 (g) + 6H2O (l)

Step 3: Calculate moles of NH3

Moles NH3 = Mass NH3 / Molar Mass NH3

Moles NH3 = 2.15 grams / 17.03 g/mol

Moles NH3 = 0.126 moles

Step 4: Calculate moles of O2

Moles O2 = 3.23 grams / 32 g/mol

Moles O2 = 0.101 moles

Step 5: Calculate the limiting reactant

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

NH3 is the limiting reactant. It will completely be consumed ( 0.126 moles).

O2 is in excess, there will be 3/4 * 0.126 = 0.0945 moles consumed

There will remain 0.101 - 0.945 = 0.0065 moles of O2

Step 6: Calculate moles of N2

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

For 4 moles NH3 , we'll have 2 moles of N2 produced

For 0.126 moles NH3 consumed, we'll have 0.063 moles of N2 produced.

Step 7: Calculate volume of N2 produced

p*V = n*R*T

⇒ with p = the pressure of the gas = 1.00 atm

⇒ with V = the volume = TO BE DETERMINED

⇒ with n = the number of moles N2 = 0.063 moles

⇒ with R = the gasconstant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 295

V = (nRT)/p

V = (0.063*0.08206*295)/1

V = 1.525 L = theoretical yield

Step 8: Calculate the % yield

% yield = actual yield / theoretical yield

% yield = (0.550 L / 1.525 L)*100%

% yield = 36.1 %

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