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ladessa [460]
3 years ago
10

The kinetic energy within molecules of an object _ when heat is added? Increases decreases or remains the same

Chemistry
1 answer:
kotegsom [21]3 years ago
6 0

The kinetic energy of an object increases when heat is added.

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I think the answer is a, volume, but I still might be wrong.
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Explain what is water of crystallization​
ArbitrLikvidat [17]

Explanation:

In chemistry, water(s) of crystallization or water(s) of hydration are water molecules that are present inside [crystal]s. Water is often incorporated in the formation of crystals from aqueous solutions. ... Water of crystallization can generally be removed by heating a sample but the crystalline properties are often lost

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An aluminum cylinder has a radius of 3.2 cm and a length of 8.6 cm. What is the mass of the aluminum cylinder if aluminum has a
Hatshy [7]

Answer:

750g of Aluminum

Assuming 8.6 cm lenth is the height of the cylinder. The volume of a cylinder is: V = \pi*r^2*h

V = 3.14cm x 10.24cm x 3.6cm

V = 280cm^3

Now density = mass/volume

2.7g/cm^3 = mass/280cm^3

2.7g/cm^3 x 280cm^3 = mass/<u>280cm^3</u> x <u>280cm^3</u>

= 750g of Aluminum

5 0
3 years ago
In order to prepare a 0.523 m aqueous solution of potassium iodide, how many grams of potassium iodide must be added to 2.00kg o
Lerok [7]
M = amount of the solute  / mass of the <span>solvent

0.523 = x / 2.00 

x = 0.523 * 2.00

x = 1,046  moles

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</span><span>
Mass = 1,046 * 166.0028

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hope this helps!

</span>
6 0
3 years ago
A 3.0-L sample of helium was placed in container fitted with a porous membrane. Half of the helium effused through the membrane
zhenek [66]

Explanation:

It is known that rate of effusion of gases are inversely proportional to the square root of their molar masses.

And, half of the helium (1.5 L) effused in 24 hour. So, the rate of effusion of He gas is calculated as follows.

          \frac{1.5 L}{24 hr}

            = 0.0625 L/hr

As, molar mass of He is 4 g/mol  and molar mass of O_{2} is 32 g/ mol.

Now,

   \frac{\text{Rate of He}}{\text{Rate of Oxygen}} = \sqrt{\frac{32}{4}

                               = 2.83

or, rate of O_{2} = \frac{\text{Rate of He}}{2.83}

                       = \frac{0.0625 L/hr}{2.83}

          Rate of O_{2} = 0.022 L/hr.

This means that 0.022 L of O_{2} gas effuses in 1 hr

So, time taken for the effusion of 1.5 L of O_{2} gas is calculated as follows.

         \frac{1.5 L}{0.022 L/hr}

                = 68.18 hour

Thus, we can conclude that 68.18 hours will it take for half of the oxygen to effuse through the membrane.

3 0
4 years ago
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