The correct description that holds true for the above question would be B. Atomic number - 29 and Mass Number - 63.
Answer:
3M
Explanation:
moles ÷ liters = molarity
4.8 ÷ 1.6 = 3M
Adopting the number of avogrado 6.02 * 10²³ / mol
<span>Sodium chloride (table salt)</span> Molar Mass = 58.44 g / mol
We will first have to find the number of moles in 35 grams of the element, like this:
1 mol ----------------- 58.44 g
X ---------------------- 35 g
58.44 * x = 35 * 1
58.44x = 35

X = 0.598904...
X ≈ 0.60<span> mol </span>
Now we will find how many atoms there are in 0.60 mol of this element, like this:
1 mol -------------------- 6.02 * 10²³ atoms
0.60 mol ----------------- X
X = 0.60 * 6.02 * 10²³
Answer:
185.05 g.
Explanation
Firstly, It is considered as a stichiometry problem.
From the balanced equation: 2LiCl → 2Li + Cl₂
It is clear that the stichiometry shows that 2.0 moles of LiCl is decomposed to give 2.0 moles of Li metal and 1.0 moles of Cl₂, which means that the molar ratio of LiCl : Li is (1.0 : 1.0) ratio.
We must convert the grams of Li metal (30.3 g) to moles (n = mass/atomic mass), atomic mass of Li = 6.941 g/mole.
n = (30.3 g) / (6.941 g/mole) = 4.365 moles.
Now, we can get the number of moles of LiCl that is needed to produce 4.365 moles of Li metal.
Using cross multiplication:
2.0 moles of LiCl → 2.0 moles of Li, from the stichiometry of the balanced equation.
??? moles of LiCl → 4.365 moles of Li.
The number of moles of LiCl that will produce 4.365 moles of Li (30.3 g) is (2.0 x 4.365 / 2.0) = 4.365 moles.
Finally, we should convert the number of moles of LiCl into grams (n = mass/molar mass).
Molar mass of LiCl = 42.394 g/mole.
mass = n x molar mass = (4.365 x 42.394) = 185.05 g.