Why does split petrol form a stick on top of water?

Answers

Answer 1

Answer: Because oil or petrol is less dense than water therefore, instead of being absorbed by the water, it separates leaving the petrol on top of the water instead of in it or underneath it

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Please help me. Chemistry

Indicate whether each of the following substances would be more soluble in H2O or C6H14:(a) KI

(b) C8H18

(c) grease

(d) CH3(CH2)12OH

(e) NaC2H3O2

Answers

Answer:

H2O-a,e, C6H14 - b,c,d

Explanation:

H2O - (a) KI, e(NaC2H3O2) because they are ionic compounds

C6H14 -(b) C8H18, (c) grease, (d) CH3(CH2)12OH because they are mostly un-polar.

Final answer:

KI and NaC2H3O2 are ionic and therefore more soluble in polar solvents like H2O, while C8H18, grease, and CH3(CH2)12OH are nonpolar and more soluble in nonpolar solvents like C6H14.

Explanation:

The solubility of a substance generally follows the principle of 'like dissolves like'. That means polar substances are more soluble in polar solvents (like H2O) and nonpolar substances are more soluble in nonpolar solvents (like C6H14).

KI is ionic, therefore it is polar and more soluble in H2O.
C8H18 is nonpolar, hence more soluble in C6H14.
Grease is nonpolar, so it is more soluble in C6H14.
CH3(CH2)12OH is nonpolar due to the long hydrocarbon chain, making it more soluble in C6H14, despite the polar OH group at the end.
NaC2H3O2 is ionic, hence it is polar and more soluble in H2O.

Learn more about Solubility here:

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Potassium has an atomic number of 19.Which of a potassium atom's electrons are its valence electrons?
Select one:
a. Those in the third shell from the nucleus.
b. Those in the second shell from the nucleus.
c. Those in the outermost shell.
d. Those in the innermost shell.

Answers

The correct response would be C. Those in the outermost shell, and of highest electron energy are the valence electrons for potassium.

Please help me!!!30 ptsShow all work and box in your answers.
1. A 15.0 gram sample of a compound is found to contain 8.83 grams of sodium and 6.17 grams of
sulfur. Calculate the empirical formula of this compound.

2. Analysis of a 10.150 gram sample of a compound known to contain only phosphorus and oxygen
indicates a phosphorus content of 4.433 grams. What is the empirical formula of this compound?

3. A compound is found to contain 36.48%Na, 25.41% S, and 38.11% O. Find its empirical formula.

4. A compound is found to contain 63.52% iron and 36.48% sulfur. Find its empirical formula.

5. Qualitative analysis shows that a compound contains 32.38%sodium, 22.65% sulfur and 44.99 %
oxygen. Find the empirical formula of this compound.

6. Analysis of a 20.0 gram sample of a compound containing only calcium and bromine indicates that
4.00 grams of calcium are present. What is the empirical formula of the compound formed?

7. A 60.00 gram sample of tetraethyl lead, a gasoline additive, is found to contain 38.43grams of
lead, 17.83 grams of carbon, and 3.74 grams of hydrogen. Find its empirical formula.

8. Determine the molecular formula for the compound with empirical formula P2O5 and molar mass of
284 g/mol.

9. Determine the molecular formula for the compound with empirical formula OCNCl and molar mass
of 232.41 g/mol.

10. A compound is found to be 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen. Its molar mass is
60.0 g/mol. What is its molecular formula?

11. A compound is 64.9%carbon, 13.5% hydrogen and 21.6% oxygen. Its molar mass is 74.0 g/mol.
What is its molecular formula?

12. A compound is 54.5% carbon, 9.1% hydrogen and 36.4% oxygen. Its molar mass is 88.0 g/mol.
What is its molecular formula?

Answers

Answer:

Question 7 to 12 are given in attached file because character limit is only 5000

Explanation:

1. A 15.0 gram sample of a compound is found to contain 8.83 grams of sodium and 6.17 grams of sulfur. Calculate the empirical formula of this compound.

Given data:

Mass of sample = 15 g

Mass of sodium = 8.83 g

Mass of sulfur = 6.17 g

Empirical formula = ?

Solution:

Number of gram atoms of Na = 8.83 / 23 = 0.4

Number of gram atoms of S = 6.17 / 32 = 0.2

Atomic ratio:

Na : S

0.4/0.2 : 0.2/0.2

2 : 1

Na : S = 2 : 1

Empirical formula is Na₂S.

2. Analysis of a 10.150 gram sample of a compound known to contain only phosphorus and oxygen indicates a phosphorus content of 4.433 grams. What is the empirical formula of this compound?

Given data:

Mass of phosphorus = 4.433 g

Mass of oxygen = 10.150 g - 4.433 g = 5.717 g

Empirical formula = ?

Solution:

Number of gram atoms of P = 4.433 / 30.9738 = 0.1431

Number of gram atoms of O = 5.717/ 15.999 = 0.3573

Atomic ratio:

P : O

0.1431/0.1431 : 0.3573/0.1431

1 : 2.5

P : O = 2(1 : 2.5)

Empirical formula is P₂O₅.

3. A compound is found to contain 36.48%Na, 25.41% S, and 38.11% O. Find its empirical formula.

Given data:

Percentage of sodium = 36.48%

Percentage of sulfur = 25.41%

Percentage of oxygen = 38.11%

Empirical formula = ?

Solution:

Number of gram atoms of Na = 36.48 / 23 = 1.6

Number of gram atoms of S = 25.41/ 32 = 0.8

Number of gram atoms of O = 38.11/ 16 = 2.4

Atomic ratio:

Na : S : O

1.6/0.8 : 0.8/0.8 : 2.4/0.8

2 : 1 : 3

Na: S : O = 2 : 1 : 3

Empirical formula is Na₂SO₃.

4. A compound is found to contain 63.52% iron and 36.48% sulfur. Find its empirical formula.

Given data:

Percentage of iron = 63.52%

Percentage of sulfur = 36.48%

Empirical formula = ?

Solution:

Number of gram atoms of Fe = 63.52 / 55.845 = 1.14

Number of gram atoms of S = 36.48 / 32 = 1.14

Atomic ratio:

Fe : S

1.14/1.14 : 1.14/1.14

1 : 1

Fe : S = 1 : 1

Empirical formula is FeS.

5. Qualitative analysis shows that a compound contains 32.38%sodium, 22.65% sulfur and 44.99 % oxygen. Find the empirical formula of this compound.

Given data:

Percentage of sodium = 32.38%

Percentage of sulfur = 22.65%

Percentage of oxygen = 44.99%

Empirical formula = ?

Solution:

Number of gram atoms of Na = 32.38 / 23 = 1.4

Number of gram atoms of S = 22.65/ 32 = 0.7

Number of gram atoms of O = 44.99/ 16 = 2.8

Atomic ratio:

Na : S : O

1.4/0.7 : 0.7/0.7 : 2.8/0.7

2 : 1 : 4

Na: S : O = 2 : 1 : 4

Empirical formula is Na₂SO₄.

6. Analysis of a 20.0 gram sample of a compound containing only calcium and bromine indicates that 4.00 grams of calcium are present. What is the empirical formula of the compound formed?

Given data:

Mass of sample = 20g

Mass of bromine = 20 g - 4 g = 16 g

Mass of calcium = 4 g

Empirical formula = ?

Solution:

Number of gram atoms of bromine = 16 / 80= 0.2

Number of gram atoms of calcium = 4/ 40= 0.1

Atomic ratio:

Ca : Br

0.1/0.1 : 0.2/0.1

1 : 2

Ca: Br = 1 : 2

Empirical formula is CaBr₂.

Answer:

1)Na2S

2)P2O5

3)Na2SO3

4)FeS

5)Na2SO4

6)CaBr2

7)C8H20Pb

8)P4O10

9)C3Cl3N3O3

10)C2H4O2

11)C4H10O

12)C4H8O2

Explanation:

1. A 15.0 gram sample of a compound is found to contain 8.83 grams of sodium and 6.17 grams of sulfur. Calculate the empirical formula of this compound.

Moles Na = 8.83 grams / 22.98 g/mol = 0.384 moles

Moles S = 6.17 grams / 32.065 g/mol = 0.192 moles

To find the mol ratio we divide by the smallest amount of moles

Na: 0.384 / 0.192 = 2

S: 0.192 /0.192 = 1

For each mol S we have 2 mol Na

The empirical formula is Na2S

2. Analysis of a 10.150 gram sample of a compound known to contain only phosphorus and oxygen indicates a phosphorus content of 4.433 grams. What is the empirical formula of this compound?

Mass of oxygen = 10.150 - 4.433 = 5.717 grams

Moles P = 4.433 grams / 30.97 g/mol

Moles P = 0.143 moles

Moles O = 5.717 grams / 16.0 g/mol = 0.357 moles

To find the mol ratio we divide by the smallest amount of moles

P: 0.143 moles / 0.143 moles = 1

O = 0. 357 moles / 0.143 moles = 2.5

For each P atom we have 2.5 O atoms

The empirical formula is P2O5

3. A compound is found to contain 36.48%Na, 25.41% S, and 38.11% O. Find its empirical formula.

Suppose the mass of the compound = 100 grams

Moles Na = 36.48 grams / 22.98 g/mol = 1.587 moles

Moles S = 25.41 grams / 32.065 g/mol = 0.792 moles

Moles O = 38.11 grams / 16.0 g/mol = 2.382 moles

To find the mol ratio we divide by the smallest amount of moles

Na: 1.587 moles / 0.792 moles = 2

S: 0.792 moles / 0.792 moles = 1

O: 2.382 moles / 0.792 = 3

The empirical formula = Na2SO3

4. A compound is found to contain 63.52% iron and 36.48% sulfur. Find its empirical formula.

Suppose the mass of the compound = 100 grams

Moles Fe = 63.52 grams / 55.845 g/mol = 1.137 moles

Moles S = 36.48 grams / 32.065 g/mol = 1.137 moles

The empirical formula is FeS

Which compounds are classified as Arrhenius acids?(1) HCl and NaOH
(2) HNO3 and NaCl
(3) NH3 and H2CO3
(4) HBr and H2SO4

Answers

Answer:The correct answer is option 4.

Explanation:

Arrhenius acids are those compounds which gives $H^+$ ions when dissolved in their aqueous solution.

$HA(aq)\rightarrow H^++A^-$

Arrhenius bases are those compounds which gives $OH^-$ ions when dissolved in their aqueous solution.

$BOH(aq)\rightarrow OH^-+B^+$

$HBr \& H_2SO_4$ are Arrhenius acids because they form $H^+$ ions in their respective aqueous solution.

$HBr(aq)\rightarrow H^++Br^-$

$H_2SO_4(aq)\rightarrow 2H^++SO_(4)^(2-)$

Hence, the correct answer is option 4.

(4) HBr i H2SO4...........

Help I don't get it XD

Answers

I think it takes about 2 weeks to go from H to B on the diagram since H is the new moon and B is the full moon so going from H to B represents half of the lunar cycle and one lunar cycle takes about 28 days (which is about 4 weeks).

How do machines transfer and convert energy

Answers

Energy transformation or energy conversion is the process of changing one form of energy to another. In physics, the term energy describes the capacity to produce certain changes within a system, without regards to limitations in transformation imposed.

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A. What is the Potential energy (PE) of a 2 kg (mass) hammer which is 0.4 meter up? (use g = 9.8) Show your work. b. What is the PE of an object with a mass of 10 kg, and 2 meters up?c. Finish this sentence to describe what happens to potential energy: As the height of an object increases…d. Which has more potential energy, a bowling ball on the floor or a grape on the floor? Explain.