A monobromination reaction of an alkane involves an alkane and bromine. The position of the hydrogen atom that will be substituted by the bromine free radical will depend on the order of the alkane. The bromine will attach to the carbon that has the most substituents.
Answer:
0.209 mol/L
Explanation:
Given data
- Mass of copper(lI) sulfate (solute): 11.7 g
- Volume of solution: 350 mL = 0.350 L
The molar mass of copper(Il) sulfate is 159.61 g/mol. The moles corresponding to 11.7 grams are:
11.7 g × (1 mol/159.61 g) = 0.0733 mol
The molarity of copper(Il) sulfate is:
M = moles of solute / liters of solution
M = 0.0733 mol / 0.350 L
M = 0.209 mol/L
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Answer:
9.0 g/cm³
Explanation:
Density can be computed with the formula:
Where:
D = Density
M = Mass
V = Volume
In your problem we are given:
84 cm³ = volume
760 g = mass
So we just plug in our given into the formula:
The answer would then be:
9.0 g/cm³
Answer:
5
Explanation:
Sorry, don't have one
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