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sattari [20]
3 years ago
12

How do you expect the water level to change when the apparatus is placed in warm water? In cool water?

Chemistry
1 answer:
wlad13 [49]3 years ago
6 0
I am assuming that the apparatus you're referring to would be a container wherein there is water placed inside. To answer your question, I expect the water level to rise when it's placed in warm water and I expect it to fall when placed in cool water. This is because the water molecules get excited when placed in water molecule, thus, moving with more kinetic energy. The opposite goes with cool water. 
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How many ATOMS of boron are present in 3.61 grams of boron trifluoride ?
Delvig [45]

Hi there!

\large\boxed{3.203 *  10^{22} atoms}

We can use the following conversions to solve:

Total mass --> amount of mols --> amount of atoms (Avogadro's number)

Begin by calculating the amount of boron trifluoride in 3.61 grams:

3.61 g * (1 mol BF₃ / 67.8 g) ≈ 0.0532 mol BF₃

Use avogadro's number to convert:

0.0532 mol * 6.02× 10²³atoms / 1 mol = 3.203 × 10²² atoms

3 0
3 years ago
By using the following reactions, calculate the heat of combustion of pentane:
Sergio [31]
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6 0
3 years ago
The Olympian swam 200 meters in 20 seconds. <br> 1) velocity<br> 2) acceleration<br> 3) speed
irakobra [83]

Answer: velocity

Explanation:

3 0
2 years ago
Read 2 more answers
A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate o
Ostrovityanka [42]

Answer:

19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }

where,

Diffusion rate of unknown gas = 155 mL/s

CO_2 diffusion rate = 102 mL/s

CO_2 molar mass = 44 g mol^-1

Unknown gas molercualr mass = M_unknown

Now placing these values to the above formula

\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

4 0
3 years ago
Which statement is true when considering color in a fiber?
Arisa [49]
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3 years ago
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