Answer:
1.71 moles of sodium azide are needed to produce sufficient nitrogen to fill a 50.0 L air bag to a pressure of 1.25 atm at 25 C.
Explanation:
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P * V = n * R * T
In this case, you know:
- P= 1.25 atm
- V= 50 L
- n= ?
- R= 0.082
![\frac{atm*L}{mol*K}](https://tex.z-dn.net/?f=%5Cfrac%7Batm%2AL%7D%7Bmol%2AK%7D)
- T= 25 C= 298 K
Replacing:
1.25 atm* 50 L= n* 0.082
*298 K
Solving:
![n=\frac{1.25 atm* 50 L}{0.082\frac{atm*L}{mol*K}*298 K}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B1.25%20atm%2A%2050%20L%7D%7B0.082%5Cfrac%7Batm%2AL%7D%7Bmol%2AK%7D%2A298%20K%7D)
n= 2.56 moles
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:
- NaN₃: 2 moles
- Na: 2 moles
- N₂: 3 moles
Then you can apply the following rule of three: if by reaction stoichiometry 3 moles of N₂ are produced from 2 moles of NaN₃, 2.56 moles of N₂ are produced from how many moles of NaN₃?
![moles of NaN_{3} =\frac{2.56 moles of N_{2}* 2 moles of NaN_{3} }{3 moles of N_{2}}](https://tex.z-dn.net/?f=moles%20of%20NaN_%7B3%7D%20%3D%5Cfrac%7B2.56%20moles%20of%20N_%7B2%7D%2A%202%20moles%20of%20NaN_%7B3%7D%20%7D%7B3%20moles%20of%20N_%7B2%7D%7D)
moles of NaN₃= 1.71
<u><em>1.71 moles of sodium azide are needed to produce sufficient nitrogen to fill a 50.0 L air bag to a pressure of 1.25 atm at 25 C.</em></u>
Your supernova is obviously conciousness while your body is walking in reality!
Answer:
pH before addition of KOH = 4.03
pH after addition of 25 ml KOH = 7.40
pH after addition of 30 ml KOH = 7.57
pH after addition of 40 ml KOH = 8.00
pH after addition of 50 ml KOH = 10.22
pH after addition 0f 60 ml KOH = 12.3
Explanation:
pH of each case in the titration given below
(6) After addition of 60 ml KOH
Since addition of 10 ml extra KOH is added after netralisation point.
Concentration of solution after addition 60 ml KOH is calculated by
M₁V₁ = M₂V₂
or, 0.23 x 10 = (50 + 60)ml x M₂
or M₂ = 0.03 Molar
so, concentration of KOH = 0.03 molar
[OH⁻] = 0.03 molar
pOH = 0.657
pH = 14 - 0.657 = 13.34
The number of protons and neutrons in the nucleus.
Yeah, it increases the surface tension !!!