<span>In a sample of solid Ba(NO3)2 the ratio of barium ions to nitrate ions is would be one is to 2 or 1:2. Barium ion has a formal charge of positive two which means that it needs two ions which has a formal charge of negative one or 1 ion with the formal charge of negative two. However, for this case, it is bonded to a nitrate ion which has a formal charge of negative one. Therefore, it needs two nitrate ions so that for every 1 atom of barium ion, we need two ions of nitrate ions.</span>
Answer:
Percent yield = 89.1%
Explanation:
Based on the equation:
Cl₂ + 2KI → 2KCl + I₂
<em>1 mole of Cl₂ reacts with 2 moles of KI to produce to moles of KCl</em>
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To solve this quesiton we must find the moles of each reactant in order to find the limiting reactant. With the limiting reactant we can find the moles of KCl and the mass:
<em>Moles Cl₂:</em>
8x10²⁵ molecules * (1mol / 6.022x10²³ molecules) = 133 moles
<em>Moles KI -Molar mass: 166.0028g/mol-</em>
25g * (1mol / 166.0028g) = 0.15 moles
Here, clarely, the KI is the limiting reactant
As 2 moles of KI produce 2 moles of KCl, the moles of KCl produced are 0.15 moles. The theoretical mass is:
0.15 moles * (74.5513g / mol) =
11.2g KCl
Percent yield is: Actual yield (10.0g) / Theoretical yield (11.2g) * 100
<h3>Percent yield = 89.1%</h3>
No' of molecules divide by avogadro number , 6×6.023×10^23 so (2.2×10^22)÷(6.023×10^23)
= 0.03653 moles
moles × Molar mass = mass
n×Mr=m
0.03653×40 = 1.46 grams
Answer:
Option B. 10
Explanation:
If 1 mol of butanol contains 6×10²⁴ atoms of H, let's calculate the amount of H.
(number of atoms / NA)
6.02 x 10²³ atoms ___ 1 mol
6×10²⁴ atoms will occupy (6×10²⁴ / NA) = 9.96 moles
H, has 10 moles in the butano formula.
Flamability boiling point color
