Answer:
3.3535 g
Explanation:
Considering:
Or,
Given :
For iron(III) nitrate :
Molarity = 0.404 M
Volume = 52.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 52.0×10⁻³ L
Thus, moles of iron(III) nitrate :
Moles of iron(III) nitrate = 0.021 moles
1 mole of iron(III) nitrate forms 1 mole of iron(III) hydroxide which further forms 1 mole of iron(III) oxide.
Thus, moles of iron(III) oxide formed from 0.021 moles of iron(III) nitrate = 0.021 moles
Also, molar mass of iron(III) oxide = 159.69 g/mol
So, mass of iron(III) oxide = 0.021 moles × 159.69 g/mol = 3.3535 g
Answer:
Rate = 43 M⁻¹s⁻¹[NO₂][O₃]
Explanation:
We need to find the reaction order in
rate = (NO₂ )ᵃ (O₃ )ᵇ
given:
( NO₂ ) M ( O₃ ) M Rate M/s
0.10 0.33 1.420 (1)
0.10 0.66 2.840 (2)
0.25 0.66 7.10 (3)
When keeping the NO₂ concentration constant in the first two while doubling the concentration of O₃ , the rate doubles. Therefore it is first order with respect to O₃
Comparing (2) and (3) increasing the concentration of NO₂ by a factor of 2.5 and keeping O₃ constant , increased the rate by a factor of 2.5. Therefore the rate is first order with respect to NO₂
Then rate law is
= k (NO₂) (O₃ )
To find k take any of the three and substitute the values to find k:
1.420 M/s = k (0.10)M x (0.33)M ⇒ k = 43 /Ms
Then the answer is Rate = 43 M⁻¹s⁻¹[NO₂][O₃]
.71 because it is negative you move the decimal point to the left one time
C=m/MV
=>1.5=m/40 x 100
=> m=1.5*40*100
=6000g
