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Nonamiya [84]
2 years ago
10

Identify the true statements about surface tension. Molecules along the surface of a liquid behave differently than those in the

bulk liquid. Water forming a droplet as it falls from a faucet is a primary example of surface tension. Surface tension increases as the temperature of the liquid rises. Cohesive forces attract the molecules of the liquid to one another. Mercury has a lower surface tension than water.
Chemistry
1 answer:
Alexxandr [17]2 years ago
4 0

Answer:

Molecules along the surface of a liquid behave differently than those in the bulk liquid.

Cohesive forces attract the molecules of the liquid to one another.

Water forming a droplet as it falls from a faucet is a primary example of surface tension.

Explanation:

Surface tension is the force that stretches the liquid surface. This force acts normal to the surface. It is the downward force that acts on the surface of the liquids which is due to the cohesive forces of the liquids.

The water molecules are bonded by a strong hydrogen bond force which is between hydrogen atom and the electronegative oxygen atom. At the surface the water molecules are attracted strongly by other water molecules which lies below the surface and are stretched at the surface. Thus the water molecules at the surface acts differently than in the bulk liquid.

Mercury have a strong cohesive force than the water and have a higher surface tension force than the water.

Surface water acquires minimum surface area, hence acquiring spherical shape of water.

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Pesticide concentrations in the Rhine River between Germany and France between 1969 and 1975 averaged 0.55 mg/L of hexachloroben
sertanlavr [38]

Answer:

1.93×10⁻³ mmoles/L of C₆Cl₆; 1.58×10⁻⁴ mmoles/L of C₁₂H₈Cl₆O; 3.51×10⁻³ mmoles/L of C₆H₆Cl₆

Explanation:

We have to find out the molar mass of each pesticide to calculate the moles, and then the milimoles

C₆Cl₆ → 12. 6 + 35.45 .6 = 284.7 g/m

C₁₂H₈Cl₆O →  12 . 12 + 8.1 + 35.45 .6 + 16 = 380.7 g/m

C₆H₆Cl₆ → 12.6 + 6.1 + 35.45 .6 = 290,7 g/m

Let's convert mg to g (/1000)

0.55 mg / 1000 = 5.5×10⁻⁴ g

0.060 mg / 1000 = 6×10⁻⁵ g

1.02 mg / 1000 = 1.02×10⁻³ g

Now we can know the moles (mass / molar mass)

5.5×10⁻⁴ g / 284.7 g/m = 1.93×10⁻⁶ moles of C₆Cl₆

6×10⁻⁵ g / 380.7 g/m = 1.58×10⁻⁷ moles of C₁₂H₈Cl₆O

1.02×10⁻³ g / 290,7 g/m = 3.51×10⁻⁶ moles of C₆H₆Cl₆

Milimoles = Mol . 1000

1.93×10⁻⁶ . 1000 = 1.93×10⁻³ mmoles of C₆Cl₆

1.58×10⁻⁷ . 1000 = 1.58×10⁻⁴ mmoles of C₁₂H₈Cl₆O

3.51×10⁻⁶ . 1000 = 3.51×10⁻³ mmoles of C₆H₆Cl₆

6 0
2 years ago
A collection of coins contains 14 pennies, 16 dimes, and 7 quarters. What is the percentage of pennies in the collection?A colle
marin [14]

Answer:

=37.83783784

Explanation:

Find the total sum of all coins,

which is 37, take the number of pennies and the total of all coins put in parenthesis( 14/37) like so and than * times them by 100

you equation should look like this

(14/37)* 100= and than the answer shown above should be the one you received. I have checked this with multiple calculators, it should be accurate.  

8 0
2 years ago
Read 2 more answers
Calculate the pH of a buffer solution prepared by dissolving 0.20 mole of cyanic acid (HCNO) and 0.80 mole of sodium cyanate (Na
Afina-wow [57]

The pH of a buffer solution : 4.3

<h3>Further explanation</h3>

Given

0.2 mole HCNO

0.8 mole NaCNO

1 L solution

Required

pH buffer

Solution

Acid buffer solutions consist of weak acids HCNO and their salts NaCNO.

\tt \displaystyle [H^+]=Ka\times\frac{mole\:weak\:acid}{mole\:salt\times valence}

valence according to the amount of salt anion  

Input the value :

\tt \displaystyle [H^+]=2.10^{-4}\times\frac{0.2}{0.8\times 1}\\\\(H^+]=5\times 10^{-5}\\\\pH=5-log~5\\\\pH=4.3

7 0
2 years ago
Determine the velocity of a marble (m = 8.66 g) with a wavelength of 3.46 × 10-33m.a.45.2 m/sb.11.3 m/sc.22.1 m/sd.38.8 m/se.52.
Damm [24]

Answer:

Velocity = 22.11 m/s

Explanation:

Using the equation;

λ = h / mv;

where  

λ = wavelength ( 3.46 x 10^-33 )  

m = mass of the marble ( 8.66 g )  

v = velocity ( we gotta find out )  

h = planck's constant ( 6.626 * 10^-34 )  

Making v the subject of the formula;

V = h /mλ

  =(6.626 x 10^-34 ) / 0.00866 ( 3.46 x 10^-33 )  

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≈ 22.11 m/s

6 0
2 years ago
50g nitrous oxide combines with 50g oxygen form dinitrogen tetroxide according to the balanced equation below.
photoshop1234 [79]

Limiting reactant : O₂

Mass of  N₂O₄ produced = 95.83 g

<h3>Further explanation</h3>

Given

50g nitrous oxide

50g oxygen

Reaction

2N20 + 302 - 2N204

Required

Limiting reactant

mass of N204 produced

Solution

mol N₂O :

\tt =\dfrac{50}{44}=1.136

mol O₂ :

\tt =\dfrac{50}{32}=1.5625

2N₂O+3O₂⇒ 2N₂O₄

ICE method

1.136    1.5625

1.0416  1.5625    1.0416

0.0944    0          1.0416

Limiting reactant : Oxygen-O₂

Mass N₂O₄(MW=92 g/mol) :

\tt =mol\times MW=1.0416\times 92=95.83~g

7 0
2 years ago
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