It is called the periodic table
Answer:
0.550
Explanation:
The absorbance (A) of a substance depends on its concentration (c) according to Beer-Lambert law.
A = ε . <em>l</em> . c
where,
ε: absorptivity of the species
<em>l</em>: optical path length
A 45 mM phosphate solution (solution A) had an absorbance of 1.012.
A = ε . <em>l</em> . c
1.012 = ε . <em>l</em> . 45 mM
ε . <em>l</em> = 0.022 mM⁻¹
We can find the concentration of the second solution using the dilution rule.
C₁ . V₁ = C₂ . V₂
45mM . 11mL = C₂ . 20.0 mL
C₂ = 25 mM
The absorbance of the second solution is:
A = (ε . <em>l</em> ). c
A = (0.022 mM⁻¹) . 25 mM = 0.55 (rounding off to 3 significant figures = 0.550)
According to the equation, the ratio of the reactant Cu and product CuO is 2:2, thus 1:1. Therefore to produce 2.44 mol CuO, 2.44 mol Cu is required. The molecular weight of Cu is 64. So the mass of Cu that is required to produce 2.44mol CuO is 2.44mol * 64 g/mol = 156.16 g.
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
Answer:
4.07L of a 0.110M NaF are needed
Explanation:
Based on the reaction:
SrCl₂(aq)+2NaF(aq)⟶SrF₂(s)+2NaCl(aq)
<em>1 mole of strontium chloride react with 2 moles of NaF</em>
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361mL of 0.620M SrCl₂ solution has:
0.361L ₓ (0.620mol / L) = 0.22382 moles SrCl₂.
Moles of NaF for a complete reaction must be:
0.22382 moles SrCl₂ ₓ (2 mol NaF / 1 mol SrCl₂) = <em>0.44764 moles of NaF</em>
If you have a solution of 0.110M NaF, the moles of NaF needed are:
0.44764 moles of NaF ₓ (1L / 0.110mol NaF) = <em>4.07L of a 0.110M NaF are needed</em>
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