Why is there a bond between the ions in potassium chloride
1 answer:
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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:
From which we can solve for x:
Thus, there are 5 molecules of water per each unit of CuSO₄, and the formula is:
- CuSO₄.5H₂O
If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .
will all the alcohol evaporate? or none at all?
Answer:
Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.
Explanation:
Given that:
The volume of alcohol which is placed in a small laboratory = 1.0 L
Vapor pressure of ethyl alcohol at 25 ° C = 59 mmHg
Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;
Then, we have:
=
= 0.078 atm
Temperature = 25 ° C
= ( 25 + 273 K)
= 298 K.
Density of the ethanol = 0.785 g/cm³
The volume of laboratory = l × b × h
= 3.0 m × 2.0 m × 2.5 m
= 15 m³
Converting the volume of laboratory to liter;
since 1 m³ = 100 L; Then, we have:
15 × 1000 = 15,000 L
Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:
PV = nRT
Making n the subject of the formula; we have:
n = 47. 88 mol of ethanol
Moles of ethanol in 1.0 L bottle can be calculated as follows:
Since numbers of moles =
and mass = density × vollume
Then; we can say ;
number of moles =
number of moles =
number of moles =
number of moles = 17.039 mol
Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.