Silver and lead are special elements, where silver is insoluble in all halogen anions (like AgCl, AgI, AgBr)
and lead is insoluble in sulphates and halogen anions ( PbSO4, PbCl2, etc.)
Mercury is special because it is the only metal that is a liquid at room temperature
hope this helps!!
Answer:
H₂O is the limiting reactant
Theoretical yield of 240 g Al₂O₃ and 14 g H₂
Explanation:
Find how many moles of one reactant is needed to completely react with the other.
6.5 mol Al × (3 mol H₂O / 2 mol Al) = 9.75 mol H₂O
We need 9.75 mol of H₂O to completely react with 6.5 mol of Al. But we only have 7.2 mol of H₂O. Therefore, H₂O is the limiting reactant.
Now find the theoretical yield:
7.2 mol H₂O × (1 mol Al₂O₃ / 3 mol H₂O) × (102 g Al₂O₃ / mol Al₂O₃) ≈ 240 g Al₂O₃
7.2 mol H₂O × (3 mol H₂ / 3 mol H₂O) × (2 g H₂ / mol H₂) ≈ 14 g H₂
Since the data was given to two significant figures, we must round our answer to two significant figures as well.
Mass number = protons + neutrons. 12 + 11 = 23.
Answer:
The attractive force is negative and MgO has a higher melting point
Explanation:
From Couloumb's law:
Energy of interaction, E = k 
where q1 and q2 are the charges of the ions, k is Coulomb's constant and r is the distance between both ions, i.e the atomic radii of the ions.
If you look at Coulomb's law, you note that in the force is negative (because q1 is negative while q2 is positive).
In addition to that, the compounds MgO and NaF have similar combined ionic radii, then we can determine the melting point trend from the amount of energy gotten
The melting point of ionic compounds is determined by 1. charge on the ions 2. size of ions. while NaF has smaller charges (+1 and -1), MgO (+2 and -2) has larger charges and greater combined atomic radii. This implies that the compound with greater force would have a higher melting point.
Hence the compound MgO would have a higher melting point than NaF.
Answer:
0.295 L
Explanation:
It seems your question lacks the final concentration value. But an internet search tells me this might be the complete question:
" A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to 24.0 mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits. "
Keep in mind that if your value is different, the answer will be different as well. However the methodology will remain the same.
To solve this problem we can<u> use the formula</u> C₁V₁=C₂V₂
Where the subscript 1 refers to the concentrated solution and the subscript 2 to the diluted one.
- 47.2 mL * 150 mM = 24.0 mM * V₂
And <u>converting into L </u>becomes:
- 295 mL *
= 0.295 L