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nikitadnepr [17]
3 years ago
15

This is something easy can someone please finish this one. i'll give brainliest.

Chemistry
2 answers:
Ainat [17]3 years ago
4 0

Answer:

B B C A C B A A C C

Explanation:

jk it was a big guess

Flauer [41]3 years ago
4 0

Answer:

1. C maybe

2. C

3. B

4. A

5. A

6. B

7. C

8. A

9. C

10. A

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When a solution of barium nitrate and a solution of copper (II) sulfate are mixed, a chemical reaction produces solid barium sul
Artemon [7]
The total mass of the products is 10.76 g + 204.44 g = 215.20 g.
The masses of all the reactants but one are known so,

215.20 g - 120.00 g - 8.15 g - 75.00 g = 12.05 g

12.05 g is the mass of the unweighed barium nitrate. 
8 0
3 years ago
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A log has a density of .8 g/cm³. What will happen to this log in freshwater, which has a density of 1.0 g/cm³?
Inga [223]

Answer:

The log will float on the water because its density is lower than the liquid, so it will stay at the top due to Archimedes' principle.

6 0
2 years ago
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Calculate the number of atoms of sodium in a 4.5-gram sample
allochka39001 [22]

Answer:

1.18×10²³ atoms.

Explanation:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ atoms.

From the above concept, 1 mole of sodium also contains 6.02×10²³ atoms.

1 mole of sodium = 23 g.

Thus,

23 g of sodium contains 6.02×10²³ atoms.

Therefore, 4.5 g of sodium will contain = (4.5 × 6.02×10²³)/23 = 1.18×10²³ atoms.

From the above calculation,

4.5 g of sodium contains 1.18×10²³ atoms.

7 0
3 years ago
When 136g of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezin
loris [4]

The given question is incomplete. The complete question is:

When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

Answer: The vant hoff factor for sodium chloride in X is 1.9

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=8.2^0C = Depression in freezing point

K_f = freezing point constant

i = vant hoff factor = 1 ( for non electrolyte)

m= molality =\frac{136g\times 1000}{950g\times 75.07g/mol}=1.9

8.2^0C=1\times K_f\times 1.9

K_f=4.32^0C/m

Now Depression in freezing point for sodium chloride is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=20.0^0C = Depression in freezing point

K_f = freezing point constant  

m= molality = \frac{136g\times 1000}{950g\times 58.5g/mol}=2.45

20.0^0C=i\times 4.32^0C\times 2.45

i=1.9

Thus vant hoff factor for sodium chloride in X is 1.9

3 0
2 years ago
What do the individual spheres represent in the model!
Trava [24]

Answer:

the individual atom in the molecule

Explanation:

In chemistry, the ball-and-stick model is a molecular model of a chemical substance.  Invidual spheres there represent atoms in the molecule. The bigger atomic number the atom has, the larger diameter of the spheres this atom has in this model.

I hope this answer will help you. Have a nice day !

5 0
2 years ago
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