Answer:
The final volume is 990.8 L
Explanation:
Let calculate the moles of gas in the first situation:
P . V = n . R . T
1.20 atm . 2.90 L = n . 0.082 . 293K
(1.20 atm . 2.90 L) / (0.082 . 293K) = 0.145 moles
This are the same moles in the second situation:
P . V = n . R . T
0.003atm . V = 0.145 moles . 0.082 . 250K
V = (0.145 moles . 0.082 . 250K) / 0.003atm
V = 990.8 L
Answer:
Because sodium and potassium are very reactive metals so they react explosively with HCL and H2SO4 evolving a large amount of heat.
Explanation:
For NaBr(aq) hydrogen gas is formed at the cathode. Hydrogen in water is reduced to hydrogen gas. For NaBr(aq), bromine, Br2(l), will be produced instead of oxygen gas at the anode.
<span>For sodium sulfate hydrogen gas is formed at the cathode. Hydrogen in water is reduced to hydrogen gas. Oxygen gas will be produced at the anode. </span>
<span>Should someone suggest that sodium metal is formed at the cathode, rest assured that that can't happen. Sodium metal reacts with water to make Na+. </span>
<span>Lead(II) iodide is insoluble in water. Therefore, not much will happen since there is no electrolyte.</span>
A) Since the plot 1/[AB] vs time gives straight line, the order of the reaction with respect to A is second order:
rate constant, K = slope = 5.5 x 10⁻² M⁻¹S⁻¹
b) Rate law : Rate = k[AB]²
c) half life period of the 2nd order is inversely proportional to the initial concentration of the reactants
t 1/2 =

.

t 1/2 =

d) k = 5.5 x 10⁻² M⁻¹s⁻¹
Initial concentration of AB, [A₀] = 0.250 M
concentration of AB after 75 s = [A]
k =
![\frac{1}{t} [ \frac{1}{[A]} - \frac{1}{[Ao]} ]](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7Bt%7D%20%5B%20%5Cfrac%7B1%7D%7B%5BA%5D%7D%20-%20%20%5Cfrac%7B1%7D%7B%5BAo%5D%7D%20%5D)
[A] = 0.123 M
Equation: AB → A + B
concentration of AB after 75 s = 0.123 M
Amount of AB dissociated = 0.25 - 0.123 = 0.127 M
concentration of [A] produced = concentration of [B] produced = Amount of AB reacted = 0.127 M
It could be source for the battery, switch or interrupter for the lever, charge for the bulb and conductor for the wire.