The molarity of a solution is 0.909 M.
m(CaCl₂) = 82 g; mass of calcium chloride
M(CaCl₂) = 111 g/mol; molar mass of calcium chloride
n(CaCl₂) = m(CaCl₂) ÷ M(CaCl₂)
n(CaCl₂) = 82 g ÷ 111 g/mol
n(CaCl₂) = 0.74 mol; amount of calcium chloride
V(CaCl₂) = 812 mL = 0.812 L; volume of calcium chloride
c(CaCl₂) = n(CaCl₂) ÷ V(CaCl₂)
c(CaCl₂) = 0.74 mol ÷ 0.812 L
c(CaCl₂) = 0.909 mol/L = 0.909 M; molarity of calcium chloride
Calcium chloride is a inorganic salt with ionic bonds between calcium and chlorine. It is a colorless crystalline solid at room temperature, highly soluble in water.
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Answer:
1.The electonic configuration of elements and their position in the periodic table are related to each other, From the electronic configuration of the elements, we can determine the period and the group to which the element belongs
Let's consider, sodium with atomic number 11 and k, l, and M shells have 2,8,and 1 electrons. since, there are 3 principal energy levels so we concluded sodium belongs to third period M Shell(valance shell) has only 1 electrons. so sodium belongs to group 1.
2. Entire D-block elements are known as Transition Elements.
3. Group 17 is the halogen group.
4. Main group of elements are...... 1,2, and 13 through 18.
5. Group 18 are the noble gas elements .
12. a). Smaller
b). Increases
c). More reactive
d). Softer
7. a). k › Ca › Ge › Br › Kr
b). Ra › Ba › Sr › Ca › Mg › Be
9. a). Ca(calcium) ion is smaller.
b). Cl(chlorine) atom is smaller.
c). Mg(magnesium) atom is smaller.
10. a). F(fluorine)
b). Sr(strontium)
c). Pb(lead)
d). At(Astatine)
For both of them, used the balanced equation and it’s mole ratio to convert whatever you need to into moles. See the attacked work.
1) D 5 mols
2) A 0.55 mols
First let us determine the electronic configuration of
Bromine (Br). This is written as:
Br = [Ar] 3d10 4s2 4p5
Then we must recall that the greatest effective nuclear
charge (also referred to as shielding) greatly increases as distance of the
orbital to the nucleus also increases. So therefore the electron in the
farthest shell will experience the greatest nuclear charge hence the answer is:
<span>4p orbital</span>
