Answer:
1) There are 0.0128 moles of BBr3 present in 3.20 grams of this compound
2) There are 909.35 grams of BBr3 present in 3.63 moles of this compound
Explanation:
<u>Step 1: </u>Given data
Boron tribromide = BBr3
Molar mass of Boron = 10.81 g/mole
Molar mass of Bromide = 79.9 g/mole
Molar mass of Boron tribromide = 10.81 + 3*79.9 = 250.51 g/mole
<u>Step 2:</u> Calculating number of moles
Number of moles = mass / molar mass
Number of moles of BBr3 = 3.20 grams / 250.51 g/mole
Number of moles of BBr3 = 0.0128 moles
<u>Step 3:</u> Calculating mass
If we have 3.63 moles of boron tribromide (= BBr3)
Mass of BBr3 = number of moles of BBr3 * Molar mass of BBr3
Mass of BBr3 = 3.63 moles * 250.51 g/mole
Mass of BBr3 = 909.35 grams
Is would be heat or light
Answer:
Explanation:
Heat transfer occurs by three main mechanisms: conduction, where rigorously vibrating molecules transfer their energy to other molecules with lower energy; convection, in which the bulk movement of a fluid causes currents and eddies that promote mixing and the distribution of thermal energy
Answer 1)
a=1° = 0.01745329252 radiansExplanation : According to the conversion chart, to convert 1° this formula is used;
<span>radians = degrees × π / 180°
</span>
while solving for, One degree which is equal 0.01745329252 radians:
1° = π/180° = 0.005555556π = 0.01745329252 rad
Answer 2) b = 1' = = 0.0002907 radians
Explanation : A minute of arc is π/10800 of radian; which is a measurement unit of angular which is equal to 1/60 of 1° degree.
So substituting the value of 1° as and on solving we get,
0.01745/60 = 0.0002907 radians.
A minute of arc is <span>π/<span>10800</span></span><span> of a </span>radian.
Answer 3) C= 1'' = 4.8481
Explanation - When calculating second of arc which is considered to be 1/60 of an arc minute, and 1/3600 of a degree;
So all these are inter-related units which can be calculated if any one value is given;
<span>on solving we get, </span>π/648000 (about 1/206265) of a radian =
4.8481 X
Answer:
0.0258 mol <em>Answer</em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>