The reaction: 2H2(g) + O2(g) → 2H2O(g), can be interpreted as: a. 2 moles of hydrogen gas reacts with 1 mole of oxygen gas to produce 2 moles of water.
Answer:
V₂ = 0.6 V.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n is constant, and have different values of P, V and T:
<em>(P₁V₁T₂) = (P₂V₂T₁).</em>
<em></em>
V₁ = V, P₁ = P, T₁ = T.
V₂ = ??? V, P₂ = 1.25 P, T₂ = 0.75 T.
<em>∴ V₂ = (P₁V₁T₂)/(P₂T₁) =</em> (P)(V)(0.75 T)/(1.25 P)(T)<em> = 0.6 V.</em>
<u>Answer:</u> The concentration of 
required will be 0.285 M.
<u>Explanation:</u>
To calculate the molarity of 
, we use the equation:
Moles of = 0.016 moles
Volume of solution = 1 L
Putting values in above equation, we get:
For the given chemical equations:
![Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9](https://tex.z-dn.net/?f=Ni%5E%7B2%2B%7D%28aq.%29%2B6NH_3%28aq.%29%5Crightleftharpoons%20%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%2BC_2O_4%5E%7B2-%7D%28aq.%29%3BK_f%3D1.2%5Ctimes%2010%5E9)
Net equation: ![NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?](https://tex.z-dn.net/?f=NiC_2O_4%28s%29%2B6NH_3%28aq.%29%5Crightleftharpoons%20%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%2BC_2O_4%5E%7B2-%7D%28aq.%29%3BK%3D%3F)
To calculate the equilibrium constant, K for above equation, we get:
The expression for equilibrium constant of above equation is:
![K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BC_2O_4%5E%7B2-%7D%5D%5B%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%5D%7D%7B%5BNiC_2O_4%5D%5BNH_3%5D%5E6%7D)
As, is a solid, so its activity is taken as 1 and so for 
We are given:
![[[Ni(NH_3)_6]^{2+}]=0.016M](https://tex.z-dn.net/?f=%5B%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%5D%3D0.016M)
Putting values in above equations, we get:
![0.48=\frac{0.016}{[NH_3]^6}}](https://tex.z-dn.net/?f=0.48%3D%5Cfrac%7B0.016%7D%7B%5BNH_3%5D%5E6%7D%7D)
![[NH_3]=0.285M](https://tex.z-dn.net/?f=%5BNH_3%5D%3D0.285M)
Hence, the concentration of required will be 0.285 M.
Answer is: 1 megamole is equal to 1000000 (one million) mole or 10⁶<span> moles</span>.
The SI base unit for amount of substance is the mole. The SI prefix "mega" represents a factor of 10⁶, or in exponential notation, 1E6.
<span>The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 12 grams of carbon C-12.</span>
