Answer:
One mole of cadmium (6multiply1023 atoms) has a mass of 112 grams, as shown in the periodic table on the inside front cover of the textbook. The density of cadmium is 8.65 grams/cm3.
Explanation:
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>
<em />
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>
<span>Boyle's Law is k = PV so
Initial k = 13.0 L x 4.0 atm = 52 L atm
Final kf = 6.5 L x 8 atm = 52 L atm
The gas obeys Boyle's Law
The answer with two significant figures separated by a comma is k = 52, kf = 52.</span>
Answer:
The balanced chemical equation: 2 Al + 3Cl2→ 2 AlCl3
Mole-mole relationship: 2 moles Al + 3 moles Cl2→ 2 moles AlCl3
Given: 0.600 moleCl2; 0.500 mole Al
Required: Excess reactant___; Number of moles of AlCl3 produced__
Solution: Use dimensional analysis using the mole-mole rel
0.600 mole Cl2 * 2 moles Al/3 moles Cl2 = 0.4 mole Al
0.5 mole Al* 3 moles Cl2/2 moles Al = 0.75 mole Cl2
Based on the given:
0.6mole Cl2 + 0.4 mole Al ( this is possible based on the given)
0.5mole Al + 0.75 mole Cl2 (this is not possible because the given is only 0.600 mole of Cl 2)
Answer: Excess reactant is Al; Limiting reactant is Cl2
The amount of AlCl3 produced = 0.6 mole Cl2 + 0.4 mole Al = 1.0 mole AlCl3
