Answer
is: concentration of hydrogen cations in lemon juice is 4.00 × 10−3 m.
pH
= 2.4.
pH
= -log[H₃O+].
[H₃O+] = 10∧(-pH).
[H₃O+] = 10∧(-2.4).
[H₃O+] = 0.004 mol/L.
pH value (potential of hydrogen - [H+]) is a logarithmic scale that
specify the acidity or basicity of an aqueous solution. When pH is greater than
seven, aqueous solution is basic, below seven is acidic and when pH is equal
seven, solution is neutral.
ΔH = -15.78 KJ when 25.0 g of S₈ reacts.
<h3>What is Balanced Chemical Equation ?</h3>
The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.
The given reaction balanced equation is:
1/8 S₈(s) + H₂(g) → H₂S(g) ΔH = -20.22kj
Now,
When 1 mol will react, then ΔH = -20.2 × 8
= -161.6 KJ
When 1 mol of S8 = 8 × 32 g
= 256 g
When 256g of S8 react ΔH = -161.6 KJ
When 1 g reacts, ΔH =
=- 0.63125 KJ
When 25 g reacts, ΔH = - 0.63125 × 25
= -15.78 KJ
Thus from the above conclusion we can say that ΔH = -15.78 KJ when 25.0 g of S₈ reacts.
Learn more about the Balanced chemical equation here: brainly.com/question/26694427
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Answer:
![[CO]=[Cl_2]=0.01436M](https://tex.z-dn.net/?f=%5BCO%5D%3D%5BCl_2%5D%3D0.01436M)
![[COCl_2]=0.00064M](https://tex.z-dn.net/?f=%5BCOCl_2%5D%3D0.00064M)
Explanation:
Hello there!
In this case, according to the given chemical reaction at equilibrium, we can set up the equilibrium expression as follows:
![K=\frac{[CO][Cl_2]}{[COCl_2]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BCO%5D%5BCl_2%5D%7D%7B%5BCOCl_2%5D%7D)
Which can be written in terms of x, according to the ICE table:

Thus, we solve for x to obtain that it has a value of 0.01436 M and therefore, the concentrations at equilibrium turn out to be:
![[CO]=[Cl_2]=0.01436M](https://tex.z-dn.net/?f=%5BCO%5D%3D%5BCl_2%5D%3D0.01436M)
![[COCl_2]=0.015M-0.01436M=0.00064M](https://tex.z-dn.net/?f=%5BCOCl_2%5D%3D0.015M-0.01436M%3D0.00064M)
Regards!
In acidic solutions you have H+ but in basic solutions you have OH-.
You need know that for to balance the reaction.