Answer:
2.4 ×10^24 molecules of the herbicide.
Explanation:
We must first obtain the molar mass of the compound as follows;
C3H8NO5P= [3(12) + 8(1) + 14 +5(16) +31] = [36 + 8 + 14 + 80 + 31]= 169 gmol-1
We know that one mole of a compound contains the Avogadro's number of molecules.
Hence;
169 g of the herbicide contains 6.02×10^23 molecules
Therefore 669.1 g of the herbicide contains 669.1 × 6.02×10^23/ 169 = 2.4 ×10^24 molecules of the herbicide.
Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol
Explanation:
The given balanced chemical reaction is,

First we have to calculate the enthalpy of reaction
.

![\Delta H^o=[n_{O_2}\times \Delta H_f^0_{(O_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})+n_{O_2}\times \Delta H_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%2Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2O%29%7D%5D-%5Bn_%7BC_8H_%7B18%7D%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_8H_%7B18%7D%29%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%5D)
where,
We are given:

Putting values in above equation, we get:
![-1.0940\times 10^4=[(16\times -393.5)+(18\times -285.8)]-[(25\times 0)+(2\times \Delat H_f{C_8H_{18}(l)}]](https://tex.z-dn.net/?f=-1.0940%5Ctimes%2010%5E4%3D%5B%2816%5Ctimes%20-393.5%29%2B%2818%5Ctimes%20-285.8%29%5D-%5B%2825%5Ctimes%200%29%2B%282%5Ctimes%20%5CDelat%20H_f%7BC_8H_%7B18%7D%28l%29%7D%5D)

Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol
Answer:

Explanation:
In this reaction, we must exchange the amino group (
) for a fluorine atom (
). Also, the first step in this reaction is the addition of nitrous acid.
We must remember that the amino group in the presence of nitrous acid produces a diazonium salt. The
group is a very good leaving group and many benzene derivatives can be produced from this intermediate (see figure 1).
If what we want is to bond a fluorine atom we must use
to be able to produce m-ethylfluorobenzene (see figure 2).
I hope it helps!
Assuming we have 100 g of sample
30.45/MW of N 14g = 2.175
69.55/MW of O 16g = 4.34
4.34/2.185 = 2
for every 1 mole of N we have 2 moles of O
so the empirical formula would be NO2
without having the molecular weight of the entire molecule the molecular formula can not be determined with the information in your question
Answer:
The volume when the pressure is changed to 1.23 atm and temperature is constant will be <u><em>6.3075 L</em></u>.
Explanation:
Pressure and volume are related by Boyle's law that says:
"The volume occupied by a certain gas mass at a constant temperature is inversely proportional to the pressure"
Boyle's law is expressed mathematically as:
P * V = k
where:
- P: Pressure
- V: Volume
- k: Constant
Assuming a certain volume of gas V1 is at a pressure P1 at the beginning of the experiment. By varying the volume of gas to a new V2 value, then the pressure will change to P2, and the following will be true:
P1 * V1 = P2 * V2
In this case you have:
- P1= 754 torr= 0.9921 atm (1 atm=760 torr)
- V1= 7.82 L
- P2=1.23 atm
- V2=?
Replacing:
0.9921 atm*7.82 L=1.23 atm*V2
Resolving:

V2≅6.3075 L
<u><em>The volume when the pressure is changed to 1.23 atm and temperature is constant will be 6.3075 L.</em></u>