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tekilochka [14]
4 years ago
15

Explain how the liquid in a thermometer changes so that it can be used to measure a temprature

Chemistry
1 answer:
mart [117]4 years ago
5 0

Answer:

The liquid that is often used in thermometers is chrome.

It is khwon for raising its volule when the temperature raises and vice-versa. ● the temperature and the volume are proprtional to each other so using Mathematics, scientists have figured out a way to benefit from it to make a thermometer.

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Calculate the [H3O+] of an aqueous solution that is 2.42 × 10−7 M in OH−.
kompoz [17]

Answer:

The (H30+) is 4,13x10-8

Explanation:

We use the formula

Kwater = (H30 +) x (OH-)

1.00e-14 = (H30 +) x 2.42e-7

(H30 +) = 1.00e-14 / 2.42e-7 = 4.13e-8M

4 0
4 years ago
What is the mass of 0.28 mole of iron?
vredina [299]

Answer:

Your answer should be 15.68 grams.

Explanation:

Seeing as 1 mole has a mass of 56 g, 56*0.28 would get you 15.68 g.

6 0
3 years ago
A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flo
atroni [7]

Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

             Weight = 0.09 \times 16.04 + 0.91 \times 29

                          = 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

                    \frac{800}{27.8336}

                     = 28.74 kmol/hr

Now, applying component balance as follows.

                0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}

                   F_{p} = 43.11 kmol/hr

                   F_{A} + F_{B} = F_{p}

                          F_{B} = 43.11 - 28.74

                                      = 14.37 kmol/hr

So, mass flow rate of pure (B), is F_{B} = 14.37 \times 29

                                                    = 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

             Molecular weight of product stream = Mol. weight \times 43.11 kmol/hr

                                  = 0.06 \times 16.04 + 0.94 \times 29

                                  = 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of O_{2} into the product stream is as follows.

                    0.21 \times 0.94 \times 43.11

                      = 8.5099 kmol/hr \times 329 g/mol

                      = 272.31 kg/hr

Therefore, calculate the mass % of O_{2} into the stream as follows.

                 \frac{272.31}{1216.67} \times 100

                     = 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of O_{2} in the product gas is 22.38%.

4 0
3 years ago
Uranium-238 had a half life of 4.5 billion years. A 100 g sample of U-238 has decayed until only 25 g remain. How long did it ta
Natasha2012 [34]

awnseris chemical hhhh Explanation:

6 0
3 years ago
Answer?????????????????????????????!!?!?!?!,!?!?!?
aniked [119]

Answer:

why cant you just say the grasshopper eats the marsh grass the shrew eats the grasshopper and the hawk eats the shrew.

Explanation:

this makes sense to me but im not sure.

8 0
3 years ago
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