Answer:
#1 Exposition
#2 Background information
#3 Complication
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<u>61.25 grams</u> of CO can be formed from 35 grams of oxygen.
The molecular mass of oxygen is <u>16 gmol⁻¹</u>
The molecular mass of carbon monoxide is<u> 28 gmol⁻¹</u>
Explanation:
The molar mass of carbon monoxide is molar mass of C added to that of O;
12 + 16 = 28
= 28g/mol
The molar mass of oxygen is 16 g/mol while that of oxygen gas (O₂) is 32 g/mol
Since the ration oxygen to carbon monoxide is 1: 2 moles, we begin to find out how many moles of carbon monoxide are formed by 35 g of oxygen;
35/32 * 2
= 70/32 moles
Then multiply by the molar mass of carbon monoxide;
70/32 * 28
= 61.25 g
Answer:
cinnamic acid - 150 mg
cis-stilbene - 100 μL
trans- stilbene - 100 mg
pyridinium tribromide - 200-385 mg
For this data:
moles of cinnamic acid = 0.150 g/148.16 g/mol = 0.001 mols
Theoretical mass of dibromoproduct formed = 0.001 mol x 307.97 g/mol = 0.312 g
cis-stilbene (100 ul = 0.1 ml)
moles of cis-stilbene = 0.1 ml x 1.01 g/mol/180.25 g/mol = 0.00056 mols
Theoretical mass of dibromoproduct formed = 0.00056 mol x 340.05 g/mol = 0.19 g
trans-stilbene
moles of tran-stilbene = 0.1 g/180.25 g/mol = 0.00055 mols
Theoretical mass of dibromoproduct formed = 0.00055 mol x 340.05 g/mol = 0.19 g
Explanation:
