The friction force
________________________________________________________
Answer: partial pressure of NOBr is 7792 atm
Explanation:
Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.
Equilibrium constant is given as:
![K_{p}=\frac{[p_{NOBr}]^2}{[p_{NO}]^2\times [p_{Br_2}]^1}](https://tex.z-dn.net/?f=K_%7Bp%7D%3D%5Cfrac%7B%5Bp_%7BNOBr%7D%5D%5E2%7D%7B%5Bp_%7BNO%7D%5D%5E2%5Ctimes%20%5Bp_%7BBr_2%7D%5D%5E1%7D)
![28.4=\frac{[p_{NOBr}]^2}{[(119)^2\times (151)^1}](https://tex.z-dn.net/?f=28.4%3D%5Cfrac%7B%5Bp_%7BNOBr%7D%5D%5E2%7D%7B%5B%28119%29%5E2%5Ctimes%20%28151%29%5E1%7D)
![[p_{NOBr}]=7792](https://tex.z-dn.net/?f=%5Bp_%7BNOBr%7D%5D%3D7792)
atm
Partial pressure of NOBr is 7792 atm
I'm just doing the ones that you don't have numbers already for.
2.) just leave it alone and it's correct
3.) Mg + 2AgNo3 --> Mg(No3)2 + 2Ag
5.) just leave it alone and it's correct
8.) 10C3H8O3 + 15O2 --> 30CO2 + 4H2O
10.) P4 + 6Br2 --> 4PBr3
12.) 2FeCl3 + 6NaOH --> 2Fe(OH)3 + 6NaCl
13.) 2CH3OH + 3O2 --> 2CO2 + 4H2O
14.) 2Al + 3Cu(NO3)2 --> 2Al(NO3)3 + 3Cu
15.) 3CaCl2 + 2K3AsO4 --> Ca3(AsO4)2 + 6KCl
16.) 2NH3 --> N2 + 3H2
17.) 2H3PO4 + 3Ba(OH)2 --> Ba3(PO4)2 + 6H2O
19.) Mg3N2 + 6H2O --> 3Mg(OH)2 + 2NH3
I hope this helps you!!
The correct answer is: [C]: "Br" {"bromine"} .
_______________________________________________________
<u>Answer:</u>
<em>Chemistry is the reaction between certain elements to </em><em>create a new compound</em><em> which may happen naturally but is mostly man made and hence has lot of good as well as </em><em>adverse effects.</em>
<u>Explanation:</u>
The <em>product or compound</em> that we get and the energy or outcome we receive is the ultimate result that decides whether chemical reaction that has happened is good or bad.
There are reactions that may cause pain and suffering to living beings and are also able to<em> Infuse damage and destruction</em> are are adverse nature. they can also change the course of nature hence are are <em>dangerous.
</em>
