C6H12O6
That is the molecular formula of glucose
<u>Answer:</u> The additional information that is helpful in calculating the mole percent of XCl(s) and ZCl(s) is the molar masses of Z and X
<u>Explanation:</u>
To calculate the mole percent of a substance, we use the equation:

Mass percent means that the mass of a substance is present in 100 grams of mixture
To calculate the number of moles, we use the equation:

We require the molar masses of Z and X to calculate the mole percent of Z and X respectively
Hence, the additional information that is helpful in calculating the mole percent of XCl(s) and ZCl(s) is the molar masses of Z and X
1 mol of any gas or mix of gases at STP conditions will have a volume of 22.4 L. Since the problem doesn’t said what are the conditions I will asume that are STP condition and the volume of one mole of the mix will have a volume of 22.4 L.
You may know that density is
D=m/v
In one mole of air I will have 80% of Nitrogen (N2) and 20% oxygen (O2).
So the mass of one mole of air will be
14 x2x0.80+16x2x0.20 = 22.4 g + 6.4 g = 28.8 g
D= 28.8/22.4 = 1.28 g/L
Of course if the temperature is higher the density will be smaller because the volume of one mole will be bigger and viceversa if the temperature decrease. Also if the pressure is different than one atm the volume of a mol will change.
In general, we have this rate law express.:
![\mathrm{Rate} = k \cdot [A]^x [B]^y](https://tex.z-dn.net/?f=%5Cmathrm%7BRate%7D%20%3D%20k%20%5Ccdot%20%5BA%5D%5Ex%20%5BB%5D%5Ey)
we need to find x and y
ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).
then we go to compare two experiments in which only one concentration is changed
compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4) by the smaller [B] (experiment 1) and call it Δ[B]
Δ[B]= 0.3 / 0.1 = 3
now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:
ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...
solve for y in the equation
![\Delta \mathrm{Rate} = \Delta [B]^y](https://tex.z-dn.net/?f=%5CDelta%20%5Cmathrm%7BRate%7D%20%3D%20%5CDelta%20%5BB%5D%5Ey)

To this point,
![\mathrm{Rate} = k \cdot [A]^x [B]^1](https://tex.z-dn.net/?f=%5Cmathrm%7BRate%7D%20%3D%20k%20%5Ccdot%20%5BA%5D%5Ex%20%5BB%5D%5E1%20)
do the same to find x.
choose two experiments in which only the concentration of B is unchanged:
Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4
solve for x for
![\Delta \mathrm{Rate} = \Delta [A]^x](https://tex.z-dn.net/?f=%5CDelta%20%5Cmathrm%7BRate%7D%20%3D%20%5CDelta%20%5BA%5D%5Ex)

the rate law is
Rate = k·[A]²[B]