The reagent which limits the reaction is called limiting reagents.
____
For example:- N2+3H2gives 2NH3.so here nitrogen limits the reaction.
The complete table is inserted.
A table is given,
Formulas used:
pH= -log(H⁺)
pOH= -log(OH⁻)
pH+ pOH=14
Calculations:
For A: (H⁺)=2×10⁻⁸M
Using the pH formula:
pH= -log(H⁺)=-log(2×10⁻⁸)=7.69
pOH=14 - 7.69=6.3
Calculating OH concentration,
pOH= -log(OH⁻)
6.3= -log(OH⁻)
(OH⁻)=5.011×10⁻⁷M
Hence, the nature of A is basic.
Similarily,
For B,
(OH⁻)=1×10⁻⁷
Using the pH formula:
pOH= -log(OH⁻)= -log(1×10⁻⁷)=7
pH=14-7=7
Calculating H concentration,
pH= -log(H⁺)
7= -log(H⁺)
(H⁺)=1×10⁻⁷M
Hence, the nature of B is neutral.
Similarily,
For C,
pH=12.3
Using the pH formula:
pOH=14-12.3=1.7
Calculating H concentration,
pH= -log(H⁺)
12.3= -log(H⁺)
(H⁺)=5.011×10⁻¹³M
Calculating OH concentration,
pOH= -log(OH⁻)
1.7= -log(OH⁻)
(OH⁻)=1.99×10⁻²M
Hence, the nature of C is Basic.
Similarily,
For D,
pOH=6.8
Using the pH formula:
pH=14-6.8=7.2
Calculating H concentration,
pH= -log(H⁺)
7.2= -log(H⁺)
(H⁺)=6.309×10⁻⁸M
Calculating OH concentration,
pOH= -log(OH⁻)
6.8= -log(OH⁻)
(OH⁻)=1.58×10⁻⁷M
Hence, the nature of D is basic.
Learn more about the acid and bases here:
brainly.com/question/16189013
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Answer:
a) 0.100 M
b) 0.395 M
Explanation:
a) Calculate the molarity of a solution that contains 0.200 moles of NaOH (solute) in 2.00 L of solution
We will use the following expression for molarity.
[NaOH] = moles of solute / liters of solution
[NaOH] = 0.200 mol/2.00 L = 0.100 M
b) Calculate the molarity for a solution that contains 15.5 g of NaCl (solute, 58.44 g/mol) in 671 mL of solution
We will use the following expression for molarity.
[NaCl] = mass of solute / molar mass of solute × liters of solution
[NaCl] = 15.5 g / 58.44 g/mol × 0.671 L = 0.395 M
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