Answer:
The correct answer is option 3.
Explanation:
The expression of 
is given by :
![K_w=[H^+][OH^-]](https://tex.z-dn.net/?f=K_w%3D%5BH%5E%2B%5D%5BOH%5E-%5D)
![K_w\propto [H^+]](https://tex.z-dn.net/?f=K_w%5Cpropto%20%5BH%5E%2B%5D)
![K_w\propto [OH^-]](https://tex.z-dn.net/?f=K_w%5Cpropto%20%5BOH%5E-%5D)
Rise in temperature will result in more dissociation of water molecules into hydrogen ions and hydroxide ions.
With an increase in concentration of hydrogen ions and hydroxide ions value of will also increase.
The correct answer is that 1.125 mol of NaOH is available, and 60.75 g of FeCl₃ can be consumed.
The mass of NaOH is 45 g
The molar mass of NaOH = 40 g/mol
The moles of NaOH = mass / molar mass
= 45 / 40
= 1.125
Thus, 1.125 mol NaOH is available
3 NaOH + FeCl₃ ⇒ Fe (OH)₃ + 3NaCl
3 mol of NaOH react with 1 mol of FeCl₃
1.125 moles of NaOH will react with x moles of FeCl₃
x = 1.125 / 3
x = 0.375 mol
0.375 mol FeCl₃ can take part in reaction
The molar mass of FeCl₃ is 162 g/mol
The mass of FeCl₃ = moles × mass
= 0.375 × 162
= 60.75 g
Thus, the amount of FeCl₃, which can be consumed is 60.75 g
Answer:
C. Chemical Energy
Explanation:
The energy that stored in the substance or in compounds is called chemical energy which released when the substance reacts.
A substance is made up of atoms and molecules and chemical energy is present in those atoms or molecules. When a chemical reaction takes place, these energy released from atoms and molecules.
Chemical energy can be released in the form of thermal energy (heat) and mechanical energy (working ability).
Hence, the correct answer is "C. Chemical Energy".
Answer:answers are in the explanation
Explanation:
(a). pH less than 7 between 1 - 3.5 are strong acid, and between 4.5-6.9 weak acid.
pH greater than 7; between 10-14 is a strong base, and between 7.1 - 9, it is weakly basic.
(b). Equation of reaction;
HBr + KOH ---------> KBr + H2O
One mole of HBr reacts with one mole of KOH to give one Mole of KBr and one mole of H2O
Calculating the mmol, we have;
mmol KOH = 28.0 ml × 0.50 M
mmol KOH= 14 mmol
mmol of HBr= 56 ml × 0.25M
mmol of HBr= 14 mmol
Both HBr and KOH are used up in the reaction, which leaves only the product,KBr and H2O.
The pH here is greater than 7
(C). [NH4^+] = 0.20 mol L^-1 × 50 ml. L^-1 ÷ 50 mL + 50mL
= 0.10 M
Ka=Kw/kb
10^-14/ 1.8× 10^-5
Ka= 5.56 ×10^-10
Therefore, ka= x^2 / 0.20
5.56e-10 = x^2/0.20
x= (0.20 × 5.56e-10)^2
x= 1.05 × 10^-5
pH = -log [H+]
pH= - log[1.05 × 10^-5]
pH = 4.98
Acidic(less than 7)
(c). 0.5 × 20/40
= 0.25 M
Ka= Kw/kb
kb= 10^-14/1.8× 10^-5
Kb = 5.56×10^-10
x= (5.56×10^-10 × 0.5)^2
x= 1.667×10^-5 M
pH will be basic
