Two independent variables could change at the same time, and you would not know which variable affected the dependent variable
Answer:
%N = 25.94%
%O = 74.06%
Explanation:
Step 1: Calculate the mass of nitrogen in 1 mole of N₂O₅
We will multiply the molar mass of N by the number of N atoms in the formula of N₂O₅.
m(N): 2 × 14.01 g = 28.02 g
Step 2: Calculate the mass of oxygen in 1 mole of N₂O₅
We will multiply the molar mass of O by the number of O atoms in the formula of N₂O₅.
m(O): 5 × 16.00 g = 80.00 g
Step 3: Calculate the mass of 1 mole of N₂O₅
We will sum the masses of N and O.
m(N₂O₅) = m(N) + m(O) = 28.02 g + 80.00 g = 108.02 g
Step 4: Calculate the percent composition of N₂O₅
We will use the following expression.
%Element = m(Element)/m(Compound) × 100%
%N = m(N)/m(N₂O₅) × 100% = 28.02 g/108.02 g × 100% = 25.94%
%O = m(O)/m(N₂O₅) × 100% = 80.00 g/108.02 g × 100% = 74.06%
Answer:
2,2,3,3-tetrapropyloxirane
Explanation:
In this case, we have to know first the alkene that will react with the peroxyacid. So:
<u>What do we know about the unknown alkene? </u>
We know the product of the ozonolysis reaction (see figure 1). This reaction is an <u>oxidative rupture reaction</u>. Therefore, the double bond will be broken and we have to replace the carbons on each side of the double bond by oxygens. If
is the only product we will have a symmetric molecule in this case 4,5-dipropyloct-4-ene.
<u>What is the product with the peroxyacid?</u>
This compound in the presence of alkenes will produce <u>peroxides.</u> Therefore we have to put a peroxide group in the carbons where the double bond was placed. So, we will have as product <u>2,2,3,3-tetrapropyloxirane.</u> (see figure 2)
No it depends on the molecules strength
1. 1086.04 mmHg
2.70.213 mmHg
3. 95.954 kPa
<h3>Further explanation</h3>
Pressure (P) is the force applied per unit area
Can be formulated :
![\tt P=\dfrac{F}{A}](https://tex.z-dn.net/?f=%5Ctt%20P%3D%5Cdfrac%7BF%7D%7BA%7D)
P = pressure (SI=Pascal(Pa))
F= force applied (N)
A=area(m²)
The unit of pressure can be expressed in atm, mmHg, or Pascal
![\tt 1.\dfrac{1.429}{1}\times 760=1086.04~mmHg](https://tex.z-dn.net/?f=%5Ctt%201.%5Cdfrac%7B1.429%7D%7B1%7D%5Ctimes%20760%3D1086.04~mmHg)
![\tt 2)~9,361~Pa=9.361~kPa=\dfrac{9.361}{101.325}\times 760=70.213~mmHg](https://tex.z-dn.net/?f=%5Ctt%202%29~9%2C361~Pa%3D9.361~kPa%3D%5Cdfrac%7B9.361%7D%7B101.325%7D%5Ctimes%20760%3D70.213~mmHg)
![\tt 3)\dfrac{725}{760}\times 1=0.947~atm=\dfrac{0.947}{1}\times 101.325=95.954~kPa](https://tex.z-dn.net/?f=%5Ctt%203%29%5Cdfrac%7B725%7D%7B760%7D%5Ctimes%201%3D0.947~atm%3D%5Cdfrac%7B0.947%7D%7B1%7D%5Ctimes%20101.325%3D95.954~kPa)