Grams of Chromium(III) nitrate produced : 268.95 g
<h3>Further explanation</h3>Given
0.85 moles of Lead(IV) nitrate
Required
grams of Chromium(III) nitrate
Solution
Reaction
Balanced equation :
<em>2Cr₂ + 3Pb(NO₃)₄ ⇒ 4Cr(NO₃)₃ + 3Pb </em>
From the equation, mol ratio of Pb(NO₃)₄ : Cr(NO₃)₃ = 3 : 4, so mol Cr(NO₃)₃
mol Cr(NO₃)₃ =
Mass of Chromium(III) nitrate (MW=238.0108 g/mol) :
mass = mol x MW
mass = 1.13 x 238.0108
mass = 268.95 g
Answer : The number of moles of is,
moles.
Solution :
As we know that,
1 mole contains number of molecules
As, number of molecules of
are present in 1 mole of
So, number of molecules of
are present in
moles of
Therefore, the number of moles of is,
moles.
Answer : The angular acceleration is
Solution : Given,
The given angular velocity equation is,
At t = 0,
At t = t,
Angular acceleration : It is defined as the rate of change of angular velocity with respect to time.
Formula used for angular acceleration :
where,
= angular acceleration
= angular velocity at time 't'
= angular velocity at time '0'
t = time
Now put all the given values in this formula, we get the angular acceleration.
Therefore, the angular acceleration is