Answer:
3 is the slope
Step-by-step explanation:
The number of mole of pure 11C (carbon-11) initially present is 0.2727 mole
<h3>Description of mole </h3>
The mole of a substance is related to it's mass and molar mass according to the following equation
Mole = mass / molar mass
With the above formula, we can obtain the number of mole of pure 11C. Detail below:
<h3>How to determine the mole of 11C</h3>
From the question given, the following data were obtained:
- Molar mass of 11C = 11 g/mol
- Mass of 11C = 3 g
- Mole of 11C =?
Mole = mass / molar mass
Mole = 3 / 11
Mole = 0.2727 mole
Therefore, the number of mole of pure 11C initially present is 0.2727 mole
Learn more about mole:
brainly.com/question/13314627
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Area inside the semi-circle and outside the triangle is (91.125π - 120) in²
Solution:
Base of the triangle = 10 in
Height of the triangle = 24 in
Area of the triangle = 

Area of the triangle = 120 in²
Using Pythagoras theorem,




Taking square root on both sides, we get
Hypotenuse = 23 inch = diameter
Radius = 23 ÷ 2 = 11.5 in
Area of the semi-circle = 

Area of the semi-circle = 91.125π in²
Area of the shaded portion = (91.125π - 120) in²
Area inside the semi-circle and outside the triangle is (91.125π - 120) in².