Answer:
The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).
Explanation:
The reaction between an acid and a base is called neutralization, forming a salt and water.
Salt is an ionic compound made up of an anion (positively charged ion) from the base and a cation (negatively charged ion) from the acid.
When an acid is neutralized, the amount of base added must equal the amount of acid initially present. This base quantity is said to be the equivalent quantity. In other words, at the equivalence point the stoichiometry of the reaction is exactly fulfilled (there are no limiting or excess reagents), therefore the numbers of moles of both will be in stoichiometric relationship. So:
V acid *M acid = V base *M base
where V represents the volume of solution and M the molar concentration of said solution.
In this case:
- V acid= 13.7 mL= 0.0137 L (being 1,000 mL= 1 L)
- M acid= 0.129 M
- V base= ?
- M base= 0.135 M
Replacing:
0.0137 L* 0.129 M= V base* 0.135 M
Solving:

V base=0.0131 L = 13.1 mL
<u><em>
The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).</em></u>
Answer:
1.25 hours or 75 minutes or 1 hour and 15 minutes
Explanation:
The formula of compound is LiClO4.3H2O
<em><u>calculation</u></em>
- <em><u> </u></em>find the mole of each element
that is moles for Li,Cl,O and that of H2O
- moles = % composition/ molar mass
For Li = 4.330/ 6.94 g/mol= 0.624 moles
Cl=22.10/35.5=0.623 moles
39.89/16 g/mol =2.493 moles
H20= 33.69/18 g/mol= 1.872 moles
- find the mole ratio by dividing each moles by smallest number of mole ( 0.624 moles)
that is for Li= 0.624/0.623= 1
Cl= 0.623/0.623=1
O = 2.493/0.623 =4
H2O= 1.872/0.623=3
<h3>Therefore the formula=LiClO4.3H2O</h3><h3 />
Answer:
Your answer would be D, they are all non-renewable energy sources. Hope this helps!
Answer:
4.75 is the equilibrium constant for the reaction.
Explanation:

Equilibrium concentration of reactants :
![[CO]=0.0590 M,[H_2O]=0.00600 M](https://tex.z-dn.net/?f=%5BCO%5D%3D0.0590%20M%2C%5BH_2O%5D%3D0.00600%20M)
Equilibrium concentration of products:
![[CO_2]=0.0410 M,[H_2]=0.0410 M](https://tex.z-dn.net/?f=%5BCO_2%5D%3D0.0410%20M%2C%5BH_2%5D%3D0.0410%20M)
The expression of an equilibrium constant is given by :
![K_c=\frac{[CO_2][H_2]}{[CO][H_2O]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%7D%7B%5BCO%5D%5BH_2O%5D%7D)


4.75 is the equilibrium constant for the reaction.