Answer:
- Question 19: the three are molecular compounds.
Explanation:
<em>Question 19.</em>
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](https://tex.z-dn.net/?f=%5Cdfrac%7B18.015x%7D%7B18.015x%2B159.609%7D%3D36%5C%25%5C%5C%20%5C%5C%20%5C%5C%20%5Cdfrac%7B18.015x%7D%7B18.015x%2B159.609%7D%3D0.36)
From which we can solve for x:
![18.015x=6.4854x+57.45924\\ \\ 11.5296x=57.45924\\ \\ x\approx4.98\approx5](https://tex.z-dn.net/?f=18.015x%3D6.4854x%2B57.45924%5C%5C%20%5C%5C%2011.5296x%3D57.45924%5C%5C%20%5C%5C%20x%5Capprox4.98%5Capprox5)
Thus, there are 5 molecules of water per each unit of CuSO₄, and the formula is:
Answer:
36.92 mg of oxygen required for bio-degradation.
Explanation:
![5C_6H_6+15O_2\rightarrow 12CO_2+15H_2O](https://tex.z-dn.net/?f=5C_6H_6%2B15O_2%5Crightarrow%2012CO_2%2B15H_2O)
Mass of benzene = 30 mg = 0.03 g (1000 mg = 1 g )
Moles benzene =![\frac{0.03 g}{78 g/mol}=0.0003846 mol](https://tex.z-dn.net/?f=%5Cfrac%7B0.03%20g%7D%7B78%20g%2Fmol%7D%3D0.0003846%20mol)
According to reaction 5 moles of benzene reacts with 15 moles of oxygen gas.
Then 0.0003846 mol of benzene will react with:
of oxygen gas
Mass of 0.0011538 moles of oxygen gas:
0.0011538 mol × 32 g/mol = 0.03692 g = 36.92 mg
36.92 mg of oxygen required for bio-degradation.
Answer: Option B. 76.83L
Explanation:
1 mole of a gas occupy 22.4L at stp. This implies that 1mole of Radon also occupy 22.4L at stp.
If 1 mole of Radon = 22.4L
Therefore, 3.43 moles of Radon = 3.43 x 22.4 = 76.83L