Answer:
![5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}](https://tex.z-dn.net/?f=5.31%2A10%5E%7B-10%7D%20%3D%20%5Cfrac%7B%5B%5DH_%7B2%7D%5D%5E%7B2%7D%5BO_%7B2%7D%5D%7D%7B%5BH_%7B2%7DO%5D%5E%7B2%7D%7D)
Explanation:
For a chemical reaction, equilibrium is a state at which the rate of the forward reaction equals that of the reverse reaction. The equilibrium constant Keq is a parameter characteristic of this state which is expressed as a ratio of the concentration of the products to that of the reactants.
For a hypothetical reaction:
xA + yB ⇄ zC
The equilibrium constant is :
![Keq = \frac{[A]^{x}[B]^{y}}{[C]^{z} }](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BA%5D%5E%7Bx%7D%5BB%5D%5E%7By%7D%7D%7B%5BC%5D%5E%7Bz%7D%20%7D)
The given reaction involves the decomposition of H2O into H2 and O2
The equilibrium constant is expressed as :
![Keq = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5B%5DH_%7B2%7D%5D%5E%7B2%7D%5BO_%7B2%7D%5D%7D%7B%5BH_%7B2%7DO%5D%5E%7B2%7D%7D)
Since Keq = 5.31*10^-10
Answer:
the pressure will decrease
Explanation:
After 25 days, it remains radon 5.9x10^5 atoms.
Half-life is the time required for a quantity (in this example number of radioactive radon) to reduce to half its initial value.
N(Ra) = 5.7×10^7; initial number of radon atoms
t1/2(Ra) = 3.8 days; the half-life of the radon is 3.8 days
n = 25 days / 3.8 days
n = 6.58; number of half-lifes of radon
N1(Ra) = N(Ra) x (1/2)^n
N1(Ra) = 5.7×10^7 x (1/2)^6.58
N1(Ra) = 5.9x10^5; number of radon atoms after 25 days
The half-life is independent of initial concentration (size of the sample).
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Number 9 adding oil lubricates the chain making it easier to pedal. Also the oil prevents rusting
The moles of HCl to neutralize the sodium hydroxide produced is<u> 0.0135 mole. </u>
Neutralization or neutralization is a chemical response wherein an acid and a base react quantitatively with each other. In a reaction in water, neutralization outcomes in there being no excess of hydrogen or hydroxide ions gift in the answer.
<u>calculation:-</u>
<u />
2Na + 2H₂O -----> 2NaOH + H₂
2 mol or 46g of Na produces 80 grams of NaOH
∴ 0.31 g of Na will produce = 80/46 × 0.31
= 0.54 gram of NaOH.
mol of NaOH = 0.54/40
= 0.0135
Since both Hcl and NaOH have the same valance factor,
1 mole NaOH is needed to neutralize 1 mol HCl
∴ 0.0135 mole of NaOH will need = 0.0135 mole of HCl
mass = 0.0135 × 36.5
=<u> 0.493 grams of HCL.</u>
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