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:
From which we can solve for x:
Thus, there are 5 molecules of water per each unit of CuSO₄, and the formula is:
Answer:
(a) Homogeneous. 4.7 g of MgCl₂.
(b) 9.1 g
Explanation:
(a)
At 200°C, we can dissolve 54.6g of MgCl₂ in 100 g of water. The mass that we could dissolve in 38.2 g of water is:
Since we can dissolve up to 20.9 g of MgCl₂ and we added only 16.2 g, the mixture is homogeneous and we could add 20.9 g -16.2 g = 4.7 g of solute to make it saturated.
(b)
At 800°C, we can dissolve 66.1 g of MgCl₂ in 100 g of water. The mass that we could dissolve in 38.2 g of water is:
Since we can dissolve up to 25.3 g of MgCl₂ and we added only 16.2 g, we could add 25.3 g - 16.2 g = 9.1 g of solute to make it saturated.
Identify each element found in the equation. The number of atoms of each type of atom must be the same on each side of the equation once it has been balanced.
What is the net charge on each side of the equation? The net charge must be the same on each side of the equation once it has been balanced.
If possible, start with an element found in one compound on each side of the equation. Change the coefficients (the numbers in front of the compound or molecule) so that the number of atoms of the element is the same on each side of the equation. Remember, to balance an equation, you change the coefficients, not the subscripts in the formulas.
Once you have balanced one element, do the same thing with another element. Proceed until all elements have been balanced. It's easiest to leave elements found in pure form for last.
Check your work to make certain the charge on both sides of the equation is also balanced.
Answer:
Mass PbCl₂ = 50.24g
Mass AgCl = 14.84g
Explanation:
The addition of Cl⁻ ions from the KCl solution results in the precipitation of AgCl and PbCl₂ as follows:
Ag⁺ + Cl⁻ → AgCl(s)
Pb²⁺ + 2Cl⁻ → PbCl₂(s)
If we define X as mass of PbCl₂, moles of Cl⁻ from PbCl₂ are:
Xg × (1mol PbCl₂/ 278.1g) × (2moles Cl⁻ / 1 mole PbCl₂) = <em>0.00719X moles of Cl⁻ from PbCl₂</em>
<em />
And mass of AgCl will be 65.08g-X. Moles of Cl⁻ from AgCl is:
(65.08g-Xg) × (1mol AgCl/ 143.32g) × (1mole Cl⁻ / 1 mole AgCl) = <em>0.45409 - 0.00698X moles of Cl⁻ from AgCl</em>
<em />
Moles of Cl⁻ that were added in the KCl solution are:
0.242L × (1.92mol KCl / L) × (1mole Cl⁻ / 1 mole KCl) = 0.46464 moles of Cl⁻ added.
<em />
Moles Cl⁻(AgCl) + Moles Cl⁻(PbCl₂) = Moles Cl⁻(added)
0.45409 - 0.00698X moles + (0.00719X moles) = 0.46464 moles
0.45409 + 0.00021X = 0.46464
0.00021X = 0.01055
X = 0.01055 / 0.00021
X = 50.24g
As X = Mass PbCl₂
<h3>Mass PbCl₂ = 50.24g</h3>
And mass of AgCl = 65.08 - 50.24
<h3>Mass AgCl = 14.84g</h3>
