Answer : The expression for reaction quotient will be :
(1) ![Q_c=\frac{[SO_2][HF]^4}{[SF_4]}](https://tex.z-dn.net/?f=Q_c%3D%5Cfrac%7B%5BSO_2%5D%5BHF%5D%5E4%7D%7B%5BSF_4%5D%7D)
(2) ![Q_c=\frac{[O_2]^2[Xe]}{[XeF_2]}](https://tex.z-dn.net/?f=Q_c%3D%5Cfrac%7B%5BO_2%5D%5E2%5BXe%5D%7D%7B%5BXeF_2%5D%7D)
Explanation :
Reaction quotient
: It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.
(1) The given balanced chemical reaction is,

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted. So, the expression for reaction quotient will be :
![Q_c=\frac{[SO_2][HF]^4}{[SF_4]}](https://tex.z-dn.net/?f=Q_c%3D%5Cfrac%7B%5BSO_2%5D%5BHF%5D%5E4%7D%7B%5BSF_4%5D%7D)
(2) The given balanced chemical reaction is,
![2MoO_2(s)+XeF_2(g)\rightarrow 2MoF(l)+Xe(g)+2O_2(g)[/texIn this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted. So, the expression for reaction quotient will be :[tex]Q_c=\frac{[O_2]^2[Xe]}{[XeF_2]}](https://tex.z-dn.net/?f=2MoO_2%28s%29%2BXeF_2%28g%29%5Crightarrow%202MoF%28l%29%2BXe%28g%29%2B2O_2%28g%29%5B%2Ftex%3C%2Fp%3E%3Cp%3EIn%20this%20expression%2C%20only%20gaseous%20or%20aqueous%20states%20are%20includes%20and%20pure%20liquid%20or%20solid%20states%20are%20omitted.%20%20So%2C%20the%20expression%20for%20reaction%20quotient%20will%20be%20%3A%3C%2Fp%3E%3Cp%3E%5Btex%5DQ_c%3D%5Cfrac%7B%5BO_2%5D%5E2%5BXe%5D%7D%7B%5BXeF_2%5D%7D)
The amount of oxygen bound to hemoglobin is 98.5%
Answer:
3.43×10¹ mol
Explanation:
Given data:
Initial number of moles = 12.4 mol
Initial volume = 122.8 L
Final number of moles = ?
Final volume = 339.2 L
Solution:
The number of moles and volume are directly proportional to each other at same temperature and pressure.
V₁/n₁ = V₂/n₂
122.8 L/ 12.4 mol = 339.2 L / n₂
n₂ = 339.2 L× 12.4 mol / 122.8 L
n₂ = 4206.08 L.mol /122.8 L
n₂ = 34.3mol
In scientific notation:
3.43×10¹ mol
It’s sulfur because it shows little reactivity.
Answer: m= 3.15x10-3 g NaHCO3
Explanation: To find the mass of NaHCO3 we will use the relationship between moles and molar mass. The molar mass of NaHCO3 is 84 g.
3.75x10-5 moles NaHCO3 x 84 g NaHCO3 / 1 mole NaHCO3
= 3.15x10-3 g NaHCO3