Answer:
Explanation:
the molecular mass of Na2C6H6O7 =236 g\mole it has a sodium that has 23 g/mole so 7.6 g of Na2C6H6O7 has x of sodium mass
236 g/mole ⇒ 23g/mole
<h2> 7.6 g ⇒ ˣ </h2>
7.6 x 23 ÷ 236 = 74.07×10-2 grams of sodium
<h2 />
3Si + 2N2 --> Si3N4 (as given)
n(Si) = m/MM = 38.25/28.085 = 1.3619 mol
n(N2) = 14.33/2*14.007 = 0.5115 mol
Therefore, N2 is limiting and Si is in excess
The molar ratio of 2N2:Si3N4 is 2:1
So, 0.0575 mol of silicon nitride is formed (dividing 0.5115 by 2)
m of silicon nitride= n*mm = 0.0575*140.283 = 8.06627... g
= 8.066g (4 significant figures)
(hopefully it is right, but double check in case i did something wrong) :)
Answer:
The correct answer is CaO > LiBr > KI.
Explanation:
Lattice energy is directly proportional to the charge and is inversely proportional to the size. The compound LiBr comprises Li+ and Br- ions, KI comprises K+ and I- ions, and CaO comprise Ca²⁺ and O²⁻ ions.
With the increase in the charge, there will be an increase in lattice energy. In the given case, the lattice energy of CaO will be the highest due to the presence of +2 and -2 ions. K⁺ ions are larger than Li⁺ ion, and I⁻ ions are larger than Br⁻ ion.
The distance between Li⁺ and Br⁻ ions in LiBr is less in comparison to the distance between K⁺ and I⁻ ions in KI. As a consequence, the lattice energy of LiBr is greater than KI. Therefore, CaO exhibits the largest lattice energy, while KI the smallest.