What is the attraction between water molecules called?.

NO LINKS PLEASE! Why does dye dropped into a cup of water eventually spread throughout the water?

Nezavi [6.7K]

Answer:

C

Explanation:

okay, you need to look at the structures of the particles of matter in the solid, liquid and gas.

particles in a solid are in fixed positions, where they can only vibrate in those positions ( take a look at ice, or rather, a brick)
liquids have very small or rather, no spaces between them, but they can slide or rub against each other, like people in a really tight crowd I guess
gas particles have very large spaces between them and they move randomly. these exibit what's called brownian motion.

since water particles (and all other liquid particles) have negligible spacings and limited movement, that allows the dye particles to move from a region of high concentration to that of a low concentration. the aim for this is for the mixture/solution to reach an equilibrium, that is the mixture must get to a point where all regions have the same concentration of the dye.

you can refer to your coursebooks :)

correct where wrong please:)

5 0

2 years ago

2C₂H6 + 702 —>4C02 + 6H₂O

navik [9.2K]

Answer:

975.56×10²³ molecules

Explanation:

Given data:

Number of molecules of C₂H₆ = 4.88×10²⁵

Number of molecules of CO₂ produced = ?

Solution:

Chemical equation:

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

Number of moles of C₂H₆:

1 mole = 6.022×10²³ molecules

4.88×10²⁵ molecules×1mol/6.022×10²³ molecules

0.81×10² mol

81 mol

Now we will compare the moles of C₂H₆ with CO₂.

C₂H₆ : CO₂

2 : 4

81 : 4/2×81 = 162 mol

Number of molecules of CO₂:

1 mole = 6.022×10²³ molecules

162 mol ×6.022×10²³ molecules / 1 mol

975.56×10²³ molecules

8 0

3 years ago

How do you calculate mass using density and volume?

skad [1K]

Example:

Mass = ?

Density = 25 g/mL

Volume = 5 mL

therefore:

d = m / V

25 = m / 5

m = 25 x 5

m = 125 g

hope this helps!

7 0

3 years ago

For a particular isomer of C8H18, the combustion reaction produces 5104.1 kJ of heat per mole of C8H18(g) consumed, under standa

Vladimir79 [104]

Answer:

Explanation:

Hello,

Considering the chemical reaction, the enthalpy of reaction is given by:

ΔH°rxn=ΔfHCO2+ΔfHH2O-ΔfHC8H18

(ΔfHO2=0)

Taking into account that the reaction produces energy, ΔH°rxn is negative. No, solving for ΔfHC8H18:

ΔfHC8H18=-ΔH°rxn+8*ΔfHCO2+9*ΔfHH2O

ΔfHC8H18=-(-5104.1 kJ/mol)+9*(-292.74kJ/mol)+8*(-393.5 kJ/mol)

ΔfHC8H18=-678.56 kJ/mol

Best regards.

3 0

3 years ago

(a) Write the balanced neutralization reaction that occurs between H2SO4 and KOH in aqueous solution. Phases are optional. (b) S

Sunny_sXe [5.5K]

These are two questions and two answers

Answer:

Question 1:

H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l)

Question 2:

0.201 M

Explanation:

Question 1:

The neutralization reaction that occurs between H₂SO₄ and KOH is an acid-base reaction.

The products of an acid-base reaction are salt and water.

This is the sketch of such neutralization reaction:

1) Word equation:

sulfuric acid + potassium hydroxide → potassium sulfate + water

↑ ↑ ↑ ↑

acid base salt water

2) Skeleton equation (unbalanced)

H₂SO₄ + KOH → K₂SO₄ + H₂O

#) Balanced chemical equation (including phases)

H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l) ← answer

Question 2:

1) Mol ratio:

Using the stoichiometric coefficients of the balanced chemical equation you get the mol ratio:

1 mol H₂SO₄ (aq) : 2 mol KOH (aq) : 1 mol K₂SO₄ (aq) : mol 2H₂O (l)

2) Moles of H₂SO₄:

V = 0.750 liter
M = 0.480 mol/liter
M = n/V ⇒ n = M×V = 0.480 mol/liter × 0.750 liter = 0.360 mol

3) Moles of KOH:

V = 0.700 liter
M = 0.290 mol/liter
M = n/V ⇒ n = M × V = 0.290 mol/liter × 0.700 liter = 0.203 mol

4) Determine the limiting reagent:

a) Stoichiometric ratio:

1 mol H₂SO₄ / 2 mol NaOH = 0.500 mol H₂SO4 / mol NaOH

b) Actual ratio:

0.360 mol H₂SO4 / 0.203 mol NaOH = 1.77 mol H₂SO₄ / mol NaOH

Since hte actual ratio of H₂SO₄ is greater than the stoichiometric ratio, you conclude that H₂SO₄ is in excess.

5) Amount of H₂SO₄ that reacts:

Since, KOH is the limiting reactant, using 0.203 mol KOH and the stoichiometryc ratio 1 mol H₂SO₄ / 2 mol KOH, you get:

x / 0.203 mol KOH = 1 mol H₂SO₄ / 2 mol KOH ⇒

x = 0.203 / 2 = 0.0677 mol of H₂SO₄

6) Concentration of H₂SO₄ remaining:

Initial amount - amount that reacted = 0.360 mol - 0.0677 mol = 0.292 mol

Total volume = 0.700 liter + 0.750 liter = 1.450 liter

Concetration = M

M = n / V = 0.292 mol / 1.450 liter = 0.201 M ← answer

6 0

3 years ago