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madreJ [45]
3 years ago
8

O O O A. Molecule B. Compound C. Both

Chemistry
1 answer:
ra1l [238]3 years ago
5 0
This doesn’t seem like the full question
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2) Examine the molecules below.<br> Circle the molecules that can also be classified as compounds.
sergejj [24]

Answer:

3rd

Explanation:

its the way they look

4 0
2 years ago
How would you prepare 3.5 L of a 0.9M solution of KCl?
PolarNik [594]
V=3,5L\\&#10;Cm=0,9M\\&#10;M_{KCl}=74\frac{g}{mol}\\\\&#10;C_{m}=\frac{n}{V}\\\\&#10;n=\frac{m}{M}\\\\&#10;C_{m}=\frac{m}{MV} \ \ \ \Rightarrow  \ \ \ m=C_{m}MV\\\\&#10;m=0,9\grac{mol}{L}*74\frac{g}{mol}*3,5L=233,1g

B. Add 233 g of KCl to a 3.5 L container; then add enough water to dissolve the KCl and fill the container to the 3.5 L mark. 
4 0
3 years ago
What property of ocean water affects climate the most?
weeeeeb [17]
C it's heat capacity
6 0
3 years ago
Read 2 more answers
What part of an atom is involved in a chemical reaction?
Katyanochek1 [597]

Answer:

The answer is supposed to be "Electron cloud" or "Electon".

5 0
3 years ago
A solution is made by dissolving 3.35 g of fructose in 35.0 ml of water. What is the molality of fructose in the solution? Assum
ludmilkaskok [199]

Answer:

m = 0.531 molal

Explanation:

∴ m fructose = 3.35 g

∴ V water = 35.0 mL

∴ ρ H2O = 1 g/mL

  • molality = moles solute / Kg solvent

∴ Mw fructose = 180.16 g/mol

⇒ moles fructose = 3.35 g * ( mol / 180.16 g) = 0.0186 mol fructose

⇒ m H2O = 35.0 mL * ( 1 g/mL ) * ( Kg/1000g) = 0.035 Kg H2O

⇒ molality (m) = 0.0186 mol fructose / 0.035 Kg H2O

⇒ m = 0.531 molal

6 0
3 years ago
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