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WITCHER [35]
3 years ago
6

Which statement about the density of cold water is most likely correct?

Chemistry
2 answers:
Tema [17]3 years ago
5 0

the answer is A) Cold water is more dense than warm water

podryga [215]3 years ago
3 0

Answer: Option (A) is the correct answer.

Explanation:

Density is defined as the mass of a substance divided by the volume of the substance.

Mathematically,    Density = \frac{mass}{volume}

In cold water, molecules are more closer to each other due to which there occurs increase in mass of the substance. As density is proportional to mass of the substance. So, density of cold water will be more.

On the other hand, in warm water molecules have more kinetic energy so, there will be more number of collisions. As a result, molecules will move away from each other due to which there will be decrease in density.

Thus, we can conclude that the statement cold water is more dense than warm water, is true about the density of cold water.

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List three examples of waves that can transfer energy without going through a medium
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Answer:

Explanation:

Light waves.

GAMA waves.

Radio waves.

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2 years ago
How do electrons transition between energy levels?
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The electron with its extra packet of energy becomes excited, and promptly moves out of its lower energy level and takes up a position in a higher energy level.
7 0
3 years ago
PLEASE HELP MEEE!!!!
Assoli18 [71]
It would be 35.8 Calories or calories. Not sure about that part. Hope this helps though.
3 0
3 years ago
600 s after initiation of a first order reaction 48.5% of the initial reactant concentration remains present. What is the rate c
Ludmilka [50]

Answer:

k=1.20x10^{-3} s^{-1}

Explanation:

For a first order reaction the rate law is:

v=\frac{-d[A]}{[A]}=k[A]

Integranting both sides of the equation we get:

\int\limits^a_b {\frac{d[A]}{[A]}} \, dx =-k\int\limits^t_0 {} \, dt

where "a" stands for [A] (molar concentration of a given reagent) and "b" is {A]0 (initial molar concentration of a given reagent), "t" is the time in seconds.

From that integral we get the integrated rate law:

ln\frac{[A]}{[A]_{0} } =-kt

[A]=[A]_{0}e^{-kt}

ln[A]=ln[A]_{0} -kt

k=\frac{ln[A]_{0}-ln[A]}{t}

therefore k is

k=\frac{ln1-ln0,485}{600}=1,20x10^{-3}

8 0
3 years ago
An aqueous solution of a soluble compound (a nonelectrolyte) is prepared by dissolving 33.2 g of the compound in sufficient wate
stiv31 [10]

Answer:

2710.2g/mol

Explanation:

Step 1:

Data obtained from the question. This include the following:

van 't Hoff factor (i) = 1 (since the compound is non-electrolyte)

Mass of compound = 33.2g

Volume = 250mL

Osmotic pressure (Π) = 1.2 atm

Temperature (T) = 25ºC = 25ºC + 273 = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Molar mass of compound =.?

Step 2:

Determination of the molarity of the compound.

The molarity, M of the compound can be obtained as follow:

Π = iMRT

1.2 = 1 x M x 0.0821 x 298

Divide both side by 0.0821 x 298

M = 1.2 / (0.0821 x 298)

M = 0.049mol/L

Step 3:

Determination of the number of mole of compound in the solution. This can be obtain as follow:

Molarity = 0.049mol/L

Volume = 250mL = 250/1000 = 0.25L

Mole of compound =..?

Molarity = mole /Volume

0.049 = mole / 0.25

Cross multiply

Mole = 0.049 x 0.25

Mole of compound = 0.01225 mole.

Step 4:

Determination of the molar mass of the compound. This is illustrated below:

Mole of the compound = 0.01225 mole.

Mass of the compound = 33.2g

Molar mass of the compound =.?

Mole = Mass /Molar Mass

0.01225 = 33.2/Molar Mass

Cross multiply

0.01225 x molar mass = 33.2

Divide both side by 0.01225

Molar mass = 33.2/0.01225

Molar mass of the compound = 2710.2g/mol

Therefore, the molar mass of the compound is 2710.2g/mol

6 0
3 years ago
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