Answer: 15.8 g of
will be required to produce 1.60 grams of 
Explanation:
To calculate the moles :

According to stoichiometry :
As 1 mole of
is given by = 2 moles of 
Thus 0.05 moles of
is given by =
of 
Mass of 
Thus 15.8 g of
will be required to produce 1.60 grams of 
When 32.0 g of F₂ react with enough N₂, 39.86g of NF₃ are formed.
<h3>What is stoichiometry?</h3>
Stoichiometry is the study of the quantitative relationships or ratios between two or more substances undergoing a physical change or chemical change.
Let's consider the following balanced equation.
N₂ + 3 F₂ ⇒ 2 NF₃
To calculate the mass of NF₃ formed from 32.0 g of F₂, we need to consider the following relationships.
- The molar mass of F₂ is 38.00 g/mol.
- The molar ratio of F₂ to NF₃ is 3:2.
- The molar mass of NF₃ is 71.00 g/mol.

When 32.0 g of F₂ react with enough N₂, 39.86g of NF₃ are formed.
Learn more about stoichiometry here: brainly.com/question/16060199
![\\ \tt\leadsto \dfrac{d[NH_3]}{dt}=1.50\times 10^{-6}](https://tex.z-dn.net/?f=%5C%5C%20%5Ctt%5Cleadsto%20%5Cdfrac%7Bd%5BNH_3%5D%7D%7Bdt%7D%3D1.50%5Ctimes%2010%5E%7B-6%7D)
- dt remains same for reaction
![\\ \tt\leadsto \dfrac{d[H_2]}{dt}=\dfrac{3}{2}\dfrac{d[NH_3]}{dt}](https://tex.z-dn.net/?f=%5C%5C%20%5Ctt%5Cleadsto%20%5Cdfrac%7Bd%5BH_2%5D%7D%7Bdt%7D%3D%5Cdfrac%7B3%7D%7B2%7D%5Cdfrac%7Bd%5BNH_3%5D%7D%7Bdt%7D)
![\\ \tt\leadsto \dfrac{d[H_2]}{dt}=\dfrac{3}{2}(1.5\times 10^{-6})](https://tex.z-dn.net/?f=%5C%5C%20%5Ctt%5Cleadsto%20%5Cdfrac%7Bd%5BH_2%5D%7D%7Bdt%7D%3D%5Cdfrac%7B3%7D%7B2%7D%281.5%5Ctimes%2010%5E%7B-6%7D%29)
![\\ \tt\leadsto \dfrac{d[H_2]}{dt}=2.25\times 10^{-6}Ms^{-1}](https://tex.z-dn.net/?f=%5C%5C%20%5Ctt%5Cleadsto%20%5Cdfrac%7Bd%5BH_2%5D%7D%7Bdt%7D%3D2.25%5Ctimes%2010%5E%7B-6%7DMs%5E%7B-1%7D)
M is molarity here not metre
Explanation:
As the given reaction will be 
Since, it is given that volume of reaction vessel is doubled. Hence, moles of species involved will also increase.
Therefore, the reaction equation will become as follows.

As it is given that partial pressure of
is 0.670 bar,
is 0.438 bar and
is 0.228 bar.
Expression to calculate new equilibrium pressure if the volume of reaction vessel is double is as follows.
![K_{p} = \frac{[P_{SbCl_{5}}]^{2}}{[P_{SbCl_{3}}]^{2}[P_{Cl_{2}}]^{2}}](https://tex.z-dn.net/?f=K_%7Bp%7D%20%3D%20%5Cfrac%7B%5BP_%7BSbCl_%7B5%7D%7D%5D%5E%7B2%7D%7D%7B%5BP_%7BSbCl_%7B3%7D%7D%5D%5E%7B2%7D%5BP_%7BCl_%7B2%7D%7D%5D%5E%7B2%7D%7D)
= 
= 
Thus, we can conclude that new equilibrium pressures if the volume of the reaction vessel is doubled is
.
Answer:
The obvious difference between substitution and elimination reactions is that substitution reactions replace one substituent with another while elimination reactions simply remove the substituent. ... The major difference between the groups is the number of steps the reaction takes place over.
substitution and elimination can be seenis by looking at the bond or the H+ . When substitution occurs, the H+ will be subtituted with other atom. As for elimination, when this reaction occur which mostly involve alkane, new bondwill appear.
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