Answer:
It will take 188.06 hours for the concentration of A to decrease 10.0% of its original concentration.
Explanation:
A → B
Initial concentration of the reactant = x
Final concentration of reactant = 10% of x = 0.1 x
Time taken by the sample, t = ?
Formula used :
where,
= initial concentration of reactant
A = concentration of reactant left after the time, (t)
= half life of the first order conversion = 56.6 hour
= rate constant
Now put all the given values in this formula, we get
t = 188.06 hour
It will take 188.06 hours for the concentration of A to decrease 10.0% of its original concentration.
Answer:
None of these are correct, because there is no way to balance this equation, but I hope these steps help you figure out your answer.
Explanation:
Count out the single amounts of elements you have on both sides of the equation. To be balanced, you need to have the exact same for each element.
Before balanced Left side.
Cl-2
O-8
H-2
Before balanced right side.
H-1
Cl-1
O-3
That means we need to increase Hydrogen, Chlorine and Oxygen on the right for sure and see how that affects the equation. You can keep adding the Coefficients until the # of elements begin to match on each side.
(I tried to balance this equation, it doesn't work, there is too much on the reactants side for what the product is.)
First, we have to get how many grams of C & H & O in the compound:
- the mass of C on CO2 = mass of CO2*molar mass of C /molar mass of CO2
= 0.5213 * 12 / 44 = 0.142 g
- the mass of H atom on H2O = mass of H2O*molar mass of H / molar mass of H2O
=0.2835 * 2 / 18 = 0.0315 g
- the mass of O = the total mass - the mass of C atom - the mass of H atom
= 0.3 - 0.142 - 0.0315 = 0.1265 g
Convert the mass to mole by divided by molar mass
C(0.142/12) H(0.0315/2) O(0.1265/16)
C(0.0118) H(0.01575) O(0.0079) by dividing by the smallest value 0.0079
C1.504 H3.99 O1 by rounding to the nearst fraction
C3/2 H4/1 )1/1 multiply by 2
∴ the emprical formula C3H8O2
7.55 x 6.02 x 10²³ = 4.55 x 10²⁴ atoms
Answer:
The answer is "
"
Explanation:
Please find the complete question in the attached file.
Equation:
at 
at equilibrium 
![p= 0.47 \ \ atm\\\\SO_2=3.3-0.47 = 2.83 \ \ atm\\\\O_2= 0.74 -\frac{0.47}{2}=0.74-0.235=0.555 \ atm\\\\K_P=\frac{[PSO_3]^2}{[PSO_2]^2[PO_2]}\\\\](https://tex.z-dn.net/?f=p%3D%200.47%20%5C%20%5C%20atm%5C%5C%5C%5CSO_2%3D3.3-0.47%20%3D%202.83%20%5C%20%5C%20atm%5C%5C%5C%5CO_2%3D%200.74%20-%5Cfrac%7B0.47%7D%7B2%7D%3D0.74-0.235%3D0.555%20%5C%20atm%5C%5C%5C%5CK_P%3D%5Cfrac%7B%5BPSO_3%5D%5E2%7D%7B%5BPSO_2%5D%5E2%5BPO_2%5D%7D%5C%5C%5C%5C)
