First, we have to see how K2O behaves when it is dissolved in water:
K2O + H20 = 2 KOH
According to reaction K2O has base properties, so it forms a hydroxide in water.
For the reaction next relation follows:
c(KOH) : c(K2O) = 1 : 2
So,
c(KOH)= 2 x c(K2O)= 2 x 0.005 = 0.01 M = c(OH⁻)
Now we can calculate pH:
pOH= -log c(OH⁻) = -log 0.01 = 2
pH= 14-2 = 12
Answer:
2
Explanation:
Since it has to be 2H2+02→2H20 for it to be balanced.
Answer:
The correct answer is option A.
Explanation:
Equilibrium is a state when rate of forward reaction is equal to the rate of backward reaction. The concentration of reactants and products becomes constant at this state.
The ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric coefficients is termed as Equilibrium constant. It is denoted by 
.
aA + bB 
cC
![K_{eq}=\frac{[C]^c}{[A]^a[B]^b}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BC%5D%5Ec%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
Answer:
a) V air/day = 8640 L air an adult breaths / day
b) 0.0181 L CO intake a person / day
Explanation:
a) one average person has 12 breaths for min:
in each breath it take an average of 500 mL on air.
⇒ 12 breath / min * 500mL air / breath = 6000 mL air / min
the average air volume per day of a person is:
⇒ Vair/day = 6000 mL air / min * (60 min / h) * ( 24 h / day ) = 8640000 mLair / day * ( L / 1000 mL)
⇒ V air / day = 8640 L / day
b) 2.1 E-6 L CO / L air * 8640 L air / day = 0.0181 L CO / day
Answer:
A. True
Explanation:
The octane number is determined by comparing the characteristics of gasoline to isooctane (2,2,4-trimethylpentane) and heptane. The correct option is option A.
Basically, the higher the octane number, the greater the resistance of the gasoline to knocking.
