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natka813 [3]
3 years ago
9

What is one property that water and oil do not share

Chemistry
1 answer:
choli [55]3 years ago
3 0

Answer:

Molecules of oil are bigger than of water and therefore do not mix.

Hope this helps

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What is the density of ice at -10°C?
Alisiya [41]

Answer:

3.98 C my friend you welcome

6 0
3 years ago
Briefly explain in your own words why the bond angle increases as the number of electron groups decreases
tekilochka [14]

Answer:

i) The bond angle decreases due to the presence of lone pairs, which causes more repulsion on the bond pairs and as a result, the bond pairs tend to come closer. ii) The repulsion between electron pairs increases with an increase in electronegativity of the central atom and hence the bond angle increases.

Explanation:

4 0
2 years ago
A 0.150-kg sample of a metal alloy is heated at 540 Celsius an then plunged into a 0.400-kg of water at 10.0 Celsius, which is c
Zarrin [17]

Answer:

C_{alloy}=0.497\frac{J}{g\°C}

Explanation:

Hello there!

In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

Q_{allow}=-(Q_{water}+Q_{Al})

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})

Then, we solve for specific heat of the metallic alloy to obtain:

C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}

Thereby, we plug in the given data to obtain:

C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}

Regards!

3 0
2 years ago
What structural units make up network solids?
My name is Ann [436]
Non metal atoms got this from google btw
8 0
3 years ago
Read 2 more answers
Anything in red is the question
laila [671]

Answer:

  • Question 19: the three are molecular compounds.
  • Question 20: CuSO₄.5H₂O

Explanation:

<em>Question 19.</em>

  • C₂H₄
  • HF
  • H₂O₂

All of them are the combination of two kinds of different atoms in fixed proportions.

  • C₂H₄: two carbon atoms per four hydrogen atoms
  • HF: one hydrogen atom per one fluorine atom
  • H₂O₂: two hydrogen atoms per two oxygent atoms

Thus, they all meet the definition of compund: a pure substance formed by  two or more different elements with a definite composition.

Molecular compounds are formed by covalent bonds and ionic compounds are formed by ionic bonds.

Two non-metal elements, like H-F, C - C, C - H, H-O, H - H, and O - O will share electrons forming covalent bonds to complete their valence shell. Thus, the three compounds are molecular and not ionic.

<em>Question 20. </em>Formula of copper(II) sulfate hydrate with 36.0% water.

Copper(II) sulfate is CuSO₄. Its molar mass is 159.609g/mol

Water is H₂O. Its molar mass is 18.015g/mol

Calling x the number of water molecules in the hydrate, the percentage of water is:

       \dfrac{18.015x}{18.015x+159.609}=36\%\\ \\ \\ \dfrac{18.015x}{18.015x+159.609}=0.36

From which we can solve for x:

      18.015x=6.4854x+57.45924\\ \\ 11.5296x=57.45924\\ \\ x\approx4.98\approx5

Thus, there are 5 molecules of water per each unit of CuSO₄, and the formula is:

  • CuSO₄.5H₂O
4 0
3 years ago
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