Answer:
Explanation:
3: Given data:
Number of moles of strontium nitrate = 3.00×10⁻³ mol
Number of atoms = ?
Solution:
There are 9 moles of atoms in 1 mole of Sr(NO₃)₂.
In 3.00×10⁻³ moles,
9 mol × 3.00×10⁻³
27.00×10⁻³ mol
Number of atoms in 3.00×10⁻³ mol of Sr(NO₃)₂:
27.00×10⁻³ mol × 6.022×10²³ atoms / 1mol
162.59×10²⁰ atoms
4)Given data:
Mass of calcium hydroxide = 4500 Kg (4500/1000 = 4.5 g)
Number of moles = ?
Solution:
Number of moles = mass in g/molar mass
by putting values,
Number of moles = 4.5 g/ 74.1 g/mol
Number of moles = 0.06 mol
5) Given data:
Number of atoms of silver nitrate = 1.06×10²³
Number of moles = ?
Solution:
1 mole of any substance contain 6.022×10²³ atoms .
1.06×10²³ atoms × 1 mol / 6.022×10²³ atoms
0.176 moles of silver nitrate
Answer: The rate of the loss of
is 0.52M/s
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate of disappearance of
=![-\frac{1d[O_3]}{2dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BO_3%5D%7D%7B2dt%7D)
Rate of formation of
=![+\frac{1d[O_2]}{3dt}](https://tex.z-dn.net/?f=%2B%5Cfrac%7B1d%5BO_2%5D%7D%7B3dt%7D)
![-\frac{1d[O_3]}{2dt}=+\frac{1d[O_2]}{3dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BO_3%5D%7D%7B2dt%7D%3D%2B%5Cfrac%7B1d%5BO_2%5D%7D%7B3dt%7D)
Rate of formation of
= 
Thus Rate of disappearance of
=![\frac{2d[O_2]}{3dt}=\frac{2}{3}\times 7.78\times 10^{-1}M/s=0.52M/s](https://tex.z-dn.net/?f=%5Cfrac%7B2d%5BO_2%5D%7D%7B3dt%7D%3D%5Cfrac%7B2%7D%7B3%7D%5Ctimes%207.78%5Ctimes%2010%5E%7B-1%7DM%2Fs%3D0.52M%2Fs)
Answer:
15.04 mL
Explanation:
Using Ideal gas equation for same mole of gas as
Given ,
V₁ = 21 L
V₂ = ?
P₁ = 9 atm
P₂ = 15 atm
T₁ = 253 K
T₂ = 302 K
Using above equation as:
Solving for V₂ , we get:
<u>V₂ = 15.04 mL</u>