Answer:
11.6g of NH₃(g) have to react
Explanation:
For the reaction:
4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g) ΔH = -905kJ
<em>4 moles of ammonia produce 905kJ</em>
Thus, if you want to produce 154kJ of energy you need:
154kJ × (4 mol NH₃ / 905kJ) = <em>0.681moles of NH₃. </em>In mass -Molar mass ammonia is 17.031g/mol-
0.681mol NH₃ × (17.031g / mol) = <em>11.6g of NH₃(g) have to react</em>
A. 1.01 is the right answer
Since
The formula is Pv= nRT
P=1 atm
V= 22.4 L
N= x
r= 0.0821
t = 273 k (bc it’s standard temperature)
So (1)(22.4)=(x)(0.0821)(273)
X= 1.001
When investigative officers get to a crime scene, they do not need to collect water samples.
<h3>What is a crime scene?</h3>
A crime scene is a place where an offence has been committed. It is usually cordoned off to enable investigation of the crime and the collection of samples.
When investigative officers get to a crime scene, they do not need to collect water samples.
Learn more about investigation:brainly.com/question/1191670
#SPJ1
<u>Answer:</u>
<u>For a:</u> The equilibrium mixture contains primarily reactants.
<u>For b:</u> The equilibrium mixture contains primarily products.
<u>Explanation:</u>
There are 3 conditions:
- When
; the reaction is product favored. - When
; the reaction is reactant favored. - When
; the reaction is in equilibrium.
For the given chemical reactions:
The chemical equation follows:
![HCN(aq.)+H_2O(l)\rightleftharpoons CN^-(aq.)+H_3O^+(aq.);K_{eq}=6.2\times 10^{-10}](https://tex.z-dn.net/?f=HCN%28aq.%29%2BH_2O%28l%29%5Crightleftharpoons%20CN%5E-%28aq.%29%2BH_3O%5E%2B%28aq.%29%3BK_%7Beq%7D%3D6.2%5Ctimes%2010%5E%7B-10%7D)
The expression of
for above reaction follows:
![K_{eq}=\frac{[CN^-][H_3O^+]}{[HCN][H_2O]}=6.2\times 10^{-10}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BCN%5E-%5D%5BH_3O%5E%2B%5D%7D%7B%5BHCN%5D%5BH_2O%5D%7D%3D6.2%5Ctimes%2010%5E%7B-10%7D)
As,
, the reaction will be favored on the reactant side.
Hence, the equilibrium mixture contains primarily reactants.
The chemical equation follows:
![H_2(g)+Cl_2(g)\rightleftharpoons 2HCl(g);K_{eq}=2.51\times 10^{4}](https://tex.z-dn.net/?f=H_2%28g%29%2BCl_2%28g%29%5Crightleftharpoons%202HCl%28g%29%3BK_%7Beq%7D%3D2.51%5Ctimes%2010%5E%7B4%7D)
The expression of
for above reaction follows:
![K_{eq}=\frac{[HCl]^2}{[H_2][Cl_2]}=2.51\times 10^{4}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BHCl%5D%5E2%7D%7B%5BH_2%5D%5BCl_2%5D%7D%3D2.51%5Ctimes%2010%5E%7B4%7D)
As,
, the reaction will be favored on the product side.
Hence, the equilibrium mixture contains primarily products.