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yKpoI14uk [10]
3 years ago
15

Which two philosophers challenged Democritus?

Chemistry
2 answers:
nadezda [96]3 years ago
5 0

Answer:

I don't know the second one but one of them is Aristotle

satela [25.4K]3 years ago
5 0

Answer:

Aristrotle and John Dalton

Explanation:

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Identify the following reaction:<br> deuterium<br> neutron
Darya [45]
Deuterium:also known as heavy hydrogen) is one of two stable isotopes of hydrogen (the other being protium or hydrogen-1). The nucleus of a deuterium atom,called a deuterium,contains one proton and one neutron,whereas the far more common protium has no neutron in the nucleus.
4 0
3 years ago
Essay about how a drop of water can change states from a gas to a liquid to a solid and back to a gas throught the process of ev
Crank
That will be regarding the water cycle. Write about the importance of the water cycle, each state from evaporation, condensation, precipitation, and run off. How the water cycle affects weather, animals, and the environment.
3 0
3 years ago
What is the concentration (M) of CH3OH in a solution prepared by dissolving 11.7 g of CH3OH in sufficient water to give exactly
Andrej [43]

Answer:

3.65~M

Explanation:

We have to remember the <u>molarity equation</u>:

M=\frac{mol}{L}

So, we have to calculate "mol" and "L". The total volume is 100 mL. So, we can do the <u>conversion</u>:

100~mL\frac{1~L}{1000~mL}=~0.1~L

Now we can calculate the moles. For this we have to calculate the <u>molar mass</u>:

O: 16 g/mol

H: 1 g/mol

C: 12 g/mol

(16*1)+(1*4)+(12*1)=32~g/mol

With the molar mass value we can <u>calculate the number of moles</u>:

1.7~g~of~CH_3OH\frac{1~mol~CH_3OH}{32~g~of~CH_3OH}=0.365~mol~CH_3OH

Finally, we can <u>calculate the molarity</u>:

M=\frac{0.365~mol~CH_3OH}{0.1~L}=3.65~M

I hope it helps!

7 0
3 years ago
Write the formulas for the following ionic compounds: (a) sodium oxide, (b) iron sulfide (containing the Fe2+ ion),(c) cobalt su
Inga [223]

Answer:

Explanation:

a) Na₂O

b) FeS

c)Co₂(SO₄)₃

d) BaF₂

4 0
3 years ago
Read 2 more answers
How many grams of K2CO3 are needed to prepare a 400.0 mL of a 4.25M solution?
Basile [38]

Answer:

235 g

Explanation:

From the question;

  • Volume is 400.0 mL
  • Molarity of a solution is 4.25 M

We need to determine the mass of the solute K₂CO₃,

we know that;

Molarity = Number of moles ÷ Volume

Therefore;

First we determine the number of moles of the solute;

Moles = Molarity × volume

Moles of  K₂CO₃ = 4.25 M × 0.4 L

                           = 1.7 moles

Secondly, we determine the mass of  K₂CO₃,

We know that;

Mass = Moles × Molar mass

Molar mass of  K₂CO₃, is 138.205 g/mol

Therefore;

Mass = 1.7 moles × 138.205 g/mol

         = 234.9485 g  

         = 235 g

Thus, the mass of  K₂CO₃ needed is 235 g

7 0
4 years ago
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