Answer: Option (b) is the correct answer.
Explanation:
According to Le Chatelier's principle, any disturbance caused in an equilibrium reaction will tend to shift the equilibrium in a direction away from the disturbance.
For example, 
Hence, expression for equilibrium constant will be as follows.
![K_{eq} = \frac{[Ca^{2+}][HCO^{-}_{3}]^{2}}{[CO_{2}][H_{2}O]}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BCa%5E%7B2%2B%7D%5D%5BHCO%5E%7B-%7D_%7B3%7D%5D%5E%7B2%7D%7D%7B%5BCO_%7B2%7D%5D%5BH_%7B2%7DO%5D%7D)
Since, the concentration for a solid substance is considered as 1 or unity. Therefore, adding or removing a solid will not affect the equilibrium.
Thus, we can conclude that according to Le Châtelier’s Principle, the amount of solid reactant or product present does not have an impact on the equilibrium because the solid does not appear in the equilibrium constant, so adding or removing solid has no effect.
There goes the ist of what you're supposed to do.
Answer:
dsfewrewr
Explanation:
https://newsengin.zendesk.com/hc/vew/community/posts/360056216292-%D9%85%D9%88%D9%82%D8%B9-%D8%A7%D9%84%D9%86%D9%88%D8%B1-%D9%85%D8%B3%D9%84%D8%B3%D9%84-%D8%A7%D9%84%D9%85%D8%A4%D8%B3%D8%B3-%D8%B9%D8%AB%D9%85%D8%A7%D9%86-%D8%A7%D9%84%D8%AD%D9%84%D9%82%D8%A9-8-%D8%A7%D9%84%D8%AB%D8%A7%D9%85%D9%86%D8%A9-%D9%83%D8%A7%D9%85%D9%84%D8%A9-%D9%85%D8%AA%D8%B1%D8%AC%D9%85%D8%A9-%D9%82%D9%8A%D8%A7%D9%85%D8%A9-%D8%B9%D8%AB%D9%85%D8%A7%D9%86-%D8%A7%D9%84%D8%AD%D9%84%D9%82%D8%A9-8-%D9%83%D8%A7%D9%85%D9%84%D8%A9-%D9%85%D8%AA%D8%B1%D8%AC%D9%85%D8%A9
https://newsengin.zendesk.com/hc/pst/community/posts/360056216252-Serial-TV-%D9%85%D8%B3%D9%84%D8%B3%D9%84-%D8%A7%D9%84%D9%85%D8%A4%D8%B3%D8%B3-%D8%B9%D8%AB%D9%85%D8%A7%D9%86-%D8%A7%D9%84%D8%AD%D9%84%D9%82%D8%A9-7-%D8%A7%D9%84%D8%B3%D8%A7%D8%A8%D8%B9%D8%A9-%D8%A7%D9%88%D9%86%D9%84%D8%A7%D9%8A%D9%86-%D8%A7%D9%84%D9%85%D8%A4%D8%B3%D8%B3-%D8%B9%D8%AB%D9%85%D8%A7%D9%86-%D8%A7%D9%84%D8%AD%D9%84%D9%82%D8%A9-7-%D9%83%D8%A7%D9%85%D9%84%D8%A9-%D9%85%D8%AA%D8%B1%D8%AC%D9%85%D8%A9-HD
https://newsengin.zendesk.com/hc/vew/community/posts/360056216292-%D9%85%D9%88%D9%82%D8%B9-%D8%A7%D9%84%D9%86%D9%88%D8%B1-%D9%85%D8%B3%D9%84%D8%B3%D9%84-%D8%A7%D9%84%D9%85%D8%A4%D8%B3%D8%B3-%D8%B9%D8%AB%D9%85%D8%A7%D9%86-%D8%A7%D9%84%D8%AD%D9%84%D9%82%D8%A9-8-%D8%A7%D9%84%D8%AB%D8%A7%D9%85%D9%86%D8%A9-%D9%83%D8%A7%D9%85%D9%84%D8%A9-%D9%85%D8%AA%D8%B1%D8%AC%D9%85%D8%A9-%D9%82%D9%8A%D8%A7%D9%85%D8%A9-%D8%B9%D8%AB%D9%85%D8%A7%D9%86-%D8%A7%D9%84%D8%AD%D9%84%D9%82%D8%A9-8-%D9%83%D8%A7%D9%85%D9%84%D8%A9-%D9%85%D8%AA%D8%B1%D8%AC%D9%85%D8%A9
Answer:
Explanation:
Two moles of magnesium (Mg) and five moles of oxygen (O2) are placed in a reaction vessel. When magnesium is ignited, it reacts with oxygen. What is the limiting reactant in this experiment?
Mg + O2 → MgO (unbalanced)
first, balance the equation
2Mg +O2-------> 2MgO
two magnesium atoms react with one diatomic oxygen molecule
there is a 1:1 ratio of magnesium to oxygen atoms
but we have 2 moles of magnesium atoms and 2X5 = 10 moles of oxygen atoms
the lesser magnesium LIMITS the amount of product we can make, so it is the LIMITING REAGENT.
