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Radda [10]
4 years ago
13

Describing physical and chemical changes:

Chemistry
1 answer:
svetoff [14.1K]4 years ago
8 0
Physical: The chemist could try to bend it to find out how malleable it is. He could also try to pull it into wires to find out how ductile it is.

Chemical: The chemist could put the metal into contact with other substances to get an idea of how reactive it is, and he could try to burn it and find out how flammable it is.
You might be interested in
a solution is made by dissolving 14.57 g of sodium bromide nabr in 415 of water . what is the molality of the solution?
jenyasd209 [6]

Answer:

The answer to your question is molality = 0.34

Explanation:

Data

mass of NaBr = 14.57 g

mass of water = 415 g.

molality = ?

Process

1.- Calculate the molar mass of NaBr

NaBr = 23 + 80 = 103 g

2.- Calculate the moles of NaBr

                     103 g of NaBr ------------------ 1 mol

                      14.57 g of NaBr ---------------  x

                               x = (14.57 x 1) / 103

                               x = 14.57 / 103

                               x = 0.141 moles

3.- Calculate the molality

molality = moles / volume

-Substitution

molality = 0.141 / .415

-Result

molality = 0.34

4 0
3 years ago
Read 2 more answers
(8.6 1029) 7.4 X1029)
andrew11 [14]

Answer:

65563.914234

Explanation:

8.61029 x 7.4 x 1029

63.716146 x 1029

multiply

= 65563.914234

3 0
3 years ago
Read 2 more answers
What is temperature?
Alenkasestr [34]

Answer: Temperature is an objective measurement of how hot or cold an object is. It can be measured with a thermometer or a calorimeter. It is a means of determining the internal energy contained within a given system.

Explanation:

8 0
3 years ago
Read 2 more answers
PLZZZ HELP FASSSSSSSSSSSSTTTTTTTTTTTTT
aleksandrvk [35]

Answer:

i think multipling 30 to the 40

30 × 40 = 1200ppt

7 0
3 years ago
An unknown compound containing only C and H was burnt, yielding 10.2 g of CO2 and 6.3 g of H2O. With a molecular weight of about
Lisa [10]

Answer:

C_2H_6

Explanation:

Hello.

In this case, we can see that the mass of carbon of the unknown compound comes from the yielded mass of carbon dioxide, thus, we compute the moles of carbon as follows:

m_C=10.2gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1mol C}{1molCO_2}=0.232 molC

Moreover, the mass of hydrogen comes from the yielded water, therefore we can also compute the moles of water:

m_H=6.3gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH_2}{1molH_2O}  =0.7molH

Then, to find the subscripts in the empirical formula, we divide by the moles of carbon as the smallest:

C:\frac{0.232}{0.232}=1\\ \\H:\frac{0.7}{0.232}=3

Whose molar mass is:

M_{CH_3}=12+1*3=15g/mol

Thus, the ratio of the molecular formula to the empirical formula is:

\frac{30}{15}=2

Therefore, the molecular formula is twice the empirical formula:

C_2H_6

Which is actually ethane.

Regards.

3 0
3 years ago
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