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3 years ago

Which of the following sentences describes oxygen at room temperature? (check all that apply)

Chemistry

2 answers:

stealth61 [152]3 years ago

7 0

Hey there!!

Correct answer is

It has no definite volume and takes the shape of its container.

Hope it helps you.

Degger [83]3 years ago

5 0

At room temperature, O2 is in gaseous state.

a gas has no definite volume or definite shape. It occupies volume of container and attains shape of container only.

Thus

It has no definite volume and takes the shape of its container.

Its particles move fast enough to overcome the attraction between them.: the gas molecules have minimum intermolecular interactions and have high kinetic energy.

It has more energy than it would at a cooler temperature: the kinetic energy of gas molecules increases with increase in temperature. Thus the energy increases with temperature and decreases with decrease in temperature.

You might be interested in

Give the major force between ethanol and rubbing alcohol.

insens350 [35]

The major force between ethanol and rubbing alcohol is hydrogen bond. Hydrogen bond are intermolecular force that are weaker than covalent bond but holds atoms together in a molecules. For an hydrogen bond to be formed, a molecule must contain an hydrogen atom that will be bonded to one of the most electronegative element.

7 0

3 years ago

1: Scrieţi şi egalaţi ecuaţiile reacţiilor chimice de schimb prezentate mai jos: a) carbonat de sodiu + clorură de calciu = b) c

prisoha [69]

Answer:

AMBANTOT MO MALIGO KANA

5 0

3 years ago

A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The initial concentrations of Pb2+ and Cu2+ are

VARVARA [1.3K]

Answer:

a) Ecell = 0.5123 V

b) Ecell = 0.4695 V

c) [Pb2 +] = 4.75 M

Explanation:

The reaction at the cathode is represented as follows:

Cu2 + + 2e- -> Cu (s) Eocathode = 0.34 V

The reaction at the anode is equal to:

Pb (s) -> Pb2 + + 2e- Eoanode = -0.13 V

The number of moles of the electrons that are involved is equal to n = 2

Standard cell potential equals Eo = Eocathode - Eoanode = 0.34 V- (-0.13 V) = 0.47 V

The initial cell potential can be calculated with the following formula:

Ecell = Eocell - - 0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V

The reaction in the cell is equal to:

Cu2 + + Pb (s) -> Cu (s) + Pb2 +

The concentration of Cu2 that gives the exercise is equal 0.2 M

Therefore, the change in concentration for Cu2 + is equal to:

Cu2 + = 1.4 M - 0.2 M = 1.2 M

We use the formula from part a)

Ecell = Eocell - (0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V

To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:

Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))

0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)

Clearing Pb2 +:

[Pb2 +] = 4.75 M

8 0

3 years ago

A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving 0.250 g of the substance in 40.0 g of ccl4. the b

Finger [1]

Answer : The molar mass of the solute will be 87.90 g/mol.

Explanation : We know the formula for elevation in boiling point, which is

Δt = i $K_{b}$ m

given that, Δt = 0.357, $K_{b}$ = 5.02 and mass of $CCl _{4}$ = 40,

on substituting the value we get,
0.357 = (1) X (5.02) X (x/ 0.044), on solving we get x = 2.844 X $10^{-3}$ .
Now, 0.250/ 2.844 X $10^{-3}$ = 87.90 g/mol. which is the weight of unknown component.

3 0

4 years ago

One of the commercial uses of sulfuric acid is the production of calcium sulfate and phosphoric acid. If 26.8 g of Ca₃(PO₄)₂ rea

d1i1m1o1n [39]

Answer:

The percent yield reaction is 64.3%

Explanation:

This is the ballanced reaction

Ca₃(PO₄)₂ (s) + 3H₂SO₄ (aq) → 2H₃PO₄ (aq) + 3CaSO₄ (aq)

Let's determine the moles of our reactants:

Mass / Molar mass = Mol

26.8 g / 310.18 g/m = 0.0864 moles of phosphate.

54.3 g / 98.06 g/m = 0.554 moles of sulfuric

1 mol of phosphate reacts with 3 mol of sulfuric so

0.0864 mol of PO₄⁻³ will react with (0.0864 .3)/1 = 0.259 moles

I have 0.554 of sulfuric, so this is the reactant in excess.

The limiting reagent is the Phosphate.

1 mol of PO₄⁻³ produces 2 mol of phosphoric

0.0864 of PO₄⁻³ will produce the double amount (0.0864 .2) = 0.173 moles

Mol . molar mass = Mass

0.173 m . 97.98g/m = 16.95 g (This is the theoretical yield)

Percent yield = (Produced / Theoretical) .100

(10.9 g / 16.95 g) . 100 = 64.3 %

5 0

3 years ago

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