Answer:
1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹
2. 0.58 mol
Explanation:
1.Given ΔO₂/Δt…
2H₂O₂ ⟶ 2H₂O + O₂
-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt
d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹
d[H₂O]/dt = 2d[O₂]/dt = 2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = 6.6 × 10⁻³mol·L⁻¹s⁻¹
2. Moles of O₂
(a) Initial moles of H₂O₂
(b) Final moles of H₂O₂
The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.
(c) Moles of H₂O₂ reacted
Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol
(d) Moles of O₂ formed
Answer:
atom is the answer I think
Answer: When maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas then balanced equation is .
Explanation:
The word equation is given as maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas.
Now, in terms of chemical formulae this reaction equation will be as follows.
Here, number of atoms on reactant side are as follows.
Number of atoms on product side are as follows.
To balance this equation, multiply HCl by 4 on reactant side and multiply by 2 on product side. Therefore, the equation can be rewritten as follows.
Hence, number of atoms on reactant side are as follows.
Number of atoms on product side are as follows.
Since, this equation contains same number of atoms on both reactant and product side. Therefore, this equation is now balanced equation.
Thus, we can conclude that when maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas then balanced equation is .
MThe heat energy required to raise the temperature of 0.36Kg of copper from 22 c to 60 c is calculate using the following formula
MC delta T
m(mass)= 0.360kg in grams = 0.360 x1000 = 360 g
c(specific heat energy) = 0.0920 cal/g.c
delta T = 60- 23 = 37 c
heat energy is therefore= 360g x0.0920 cal/g.c x 37 c= 1225.44 cal
Basicity of an acid is the number of hydrogen ions which can be produced by one molecule of an acid