Answer:
(a) O = Valance Electrons (6), Inner electrons (8)
(b) Sn = Valance Electrons (2), Inner electrons (36)
(c) Ca = Valance Electrons (2), Inner electrons (20)
(d) Fe = Valance Electrons (2), Inner electrons (26)
(e) Se = Valance Electrons (6), Inner electrons (34)
Answer:
Explanation:
where,
= boiling point of solution = ?
= boiling point of solvent (X) =
= freezing point constant =
m = molality
i = Van't Hoff factor = 1 (for non-electrolyte like urea)
= mass of solute (urea) = 29.82 g
= mass of solvent (X) = 500.0 g
= molar mass of solute (urea) = 60 g/mol
Now put all the given values in the above formula, we get:
Therefore, the freezing point of solution is
The correct answer for the question that is being presented above is this one: "<span>0.3."
Here it is how to solve.
M</span><span>olecular mass of Ar = 40
</span><span>Molecular mass of Ne = 20
</span><span>Number of moles of Ar = 9.59/40 = 0.239
</span><span>Number of moles of Ne = 11.12/20= 0.556
</span><span>Mole fraction of argon = 0.239/ ( 0.239 + 0.556) = 0.3</span><span>
</span>
Answer:
15.69 dozen
Explanation:
Mass of penny = 5 g
Dozens of penny =..?
Next, we shall convert 5 g to gross. This can be obtained as follow:
3824 g = 1000 gross
Therefore,
5 g = 5 g × 1000 gross / 3824 g
5 g = 1.3075 gross
Thus, 5 g is equivalent to 1.3075 gross.
Finally, we convert 1.3075 gross to dozen. This can be obtained as follow:
1 gross = 12 dozen
Therefore,
1.3075 gross = 1.3075 gross × 12 dozen / 1 gross
1.3075 gross = 15.69 dozen
Thus, 5 g of penny is equivalent to 15.69 dozen
Answer:
Explanation:
Just saw your request regarding answering this so here it is:
All of them belong of Group 1 in periodic table and thus are highly reactive! Pattern of reactivity for Group 1 (Alkali metals) increases as you move down the group as their radius keeps increasing and thus electrons can be easily lost. Thus, to ID the lumps, Sheena should look at their reactivity and she should get the following trend:
Most reactive: Potassium (K)
Intermediate: Sodium (Na)
Least reactive: Lithium (Li)
Hope it helps!