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svetlana [45]
3 years ago
15

Surface tension ________________________.

Chemistry
1 answer:
Alex73 [517]3 years ago
6 0
C) is higher for nonpolar substances than for polar ones

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Order these clouds from highest (1) to lowest (3):
tatiyna
Stratus clouds reside below the 6,000ft range.
Altocumulus clouds reside in the 6,500-20,000ft range.
Cirrus clouds reside above the 20,000ft range.

Ordered from highest to lowest, the sequence of clouds will be:
Cirrus
Altocumulus
Stratus
5 0
3 years ago
Identify the functional groups present in each molecule.
Colt1911 [192]

Answer:

Carbon 3 is double bonded to an oxygen and attached to carbon 2 and carbon 4. :

Answer: Carbonyl group ( Ketone or aldehyde)

Carbon 17 is attached to an oxygen, which is attached to a hydrogen. :

Answer: Carboxyl group (Carboxylic acid)

A central carbon is attached to an amine, two hydrogens, and a carbon that is double bonded to an oxygen and single bonded to an oxygen attached to a hydrogen. :

Answer: Amide group

An amide group contains both amine and carboxyl

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5 0
3 years ago
One atom of silicon can properly be combined in a compound with
Over [174]

your answer is c. two atoms of oxygen.

7 0
3 years ago
Read 2 more answers
Calculate the freezing point of a solution containing 5. 0 grams of kcl and 550. 0 grams of water. the molal-freezing-point-depr
lutik1710 [3]

The freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is  - 0.45°C

Using the equation,

ΔT_{f} = iK_{f}m

where:

ΔT_{f} = change in freezing point (unknown)

i = Van't Hoff factor

K_{f} = freezing point depression constant

m = molal concentration of the solution

Molality is expressed as the number of moles of the solute per kilogram of the solvent.

Molal concentration is as follows;

MM KCl = 74.55 g/mol

molal concentration = \frac{5.0g*\frac{1mol}{74.55g} }{550.0g*\frac{1kg}{1000g} }

molal concentration = 0.1219m

Now, putting in the values to the equtaion ΔT_{f} = iK_{f}m we get,

ΔT_{f} = 2 × 1.86 × 0.1219

ΔT_{f} = 0.4536°C

So, ΔT_{f} of solution is,

ΔT_{f_{solution} } = 0.00°C - 0.45°C

ΔT_{f_{solution} } =  - 0.45°C

Therefore,freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is  - 0.45°C

Learn more about freezing point here;

brainly.com/question/3121416

#SPJ4

7 0
2 years ago
The half life of 226/88 Ra is 1620 years. How much of a 12 g sample of 226/88 Ra will be left after 8 half lives?
Aloiza [94]

Answer:

0.0468 g.

Explanation:

  • The decay of radioactive elements obeys first-order kinetics.
  • For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 1620 years).

∴ k = ln2/(t1/2) = 0.693/(1620 years) = 4.28 x 10⁻⁴ year⁻¹.

  • For first-order reaction: <em>kt = lna/(a-x).</em>

where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).

t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).

a is the initial concentration (a = 12.0 g).

(a-x) is the remaining concentration.

∴ kt = lna/(a-x)

(4.28 x 10⁻⁴ year⁻¹)(12960 year) = ln(12)/(a-x).

5.54688 = ln(12)/(a-x).

Taking e for the both sides:

256.34 = (12)/(a-x).

<em>∴ (a-x) = 12/256.34 = 0.0468 g.</em>

8 0
3 years ago
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