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3 years ago

Which of the following describes scientists?

Chemistry

1 answer:

serious [3.7K]3 years ago

3 0

I believe the answer I’d D

Hope it helps

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Base A dissociates 25% in water. Base B dissociates 50%. Base C dissociates

torisob [31]

Answer:Base A is the weakest conductor electricity

Explanation:

Dissociation is a factor that affects electrical conductivity. The greater the percentage of dissociation for bases the stronger the conductivity of electricity.

Given that

Base A dissociates 25% in water

Base B dissociates 50%.

Base C dissociates 75%

We can conclude that Base A is the weakest conductor oelectricity since it has the lowest percentage of dissociation.

5 0

2 years ago

The theoretical yield for CuCO3.Cu(OH)2+2h2SO4 ®2CuSO4+2H2O+3CO2

olga nikolaevna [1]

Answer:

bxnhdjxjdjvxjsj sbdhhxhchdbd

5 0

3 years ago

What should be included in a science conclusion

QveST [7]

A science conclusion should simply have a statement identifying whether your experiment’s aim turned out to be true or false

7 0

2 years ago

Another carbon isotope has six protons and seven neutrons in its nucleus. What do you think this carbon isotope is called?

insens350 [35]

Answer:

Carbon - 13

Explanation:

For most of the elements other than that of hydrogen, the isotopes are named for the mass number.

Example : Carbon atoms with 6 neutrons have mass number of 12 ( as $6\ protons +6\ neutrons=12$ ), so they are known as carbon-12.

Given that:

Protons = 6

Neutrons = 7

Mass = 6 + 7 = 13

So the name is Carbon - 13 . The symbol is $^{13}_{6}C$

8 0

3 years ago

Read 2 more answers

Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit

Rasek [7]

Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

$0.0811 M=\frac{n}{0.0017 L}$

n = 0.0001378 mol

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

$\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol$ of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

$x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L$

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

7 0

3 years ago