Answer:
the correct option is B
Explanation:
The correct option is b, since if we reach pH 7, it means that the acid-base reaction is neutralized, therefore the base has been neutralized by an acid or vice versa, without taking into account the proteins or the amounts of both components .
Answer:
Metals have one or two electrons in their outermost shell
C. 1-2
Explanation:
- Metals have low ionisation energy because they easily looses the outermost electrons
- They have only one- two electrons in the outer most shell.
- They loose these electron to form charged species called cation.
We can use a ratio to solve this question. Lets refer to the amount formed by the formula as a serving
1 serving needs 3/8 gal
We want to create 7/9 servings extra, so
1 + 7/9 = 16/7
1 : 3/8
16/7 : x
1/(3/8) = (16/7) / x
8 / 3 = 16x / 7
x = 7/6
He needs to use 7/6 gallons of water.
Answer:
0
Explanation:
Since HI is a strong acid, the amoung of Hydrogen ions produced by it will be the same molar as the reactant. The negative log of the concentration will reveal that the pH is 0.
The number of mole of HCl needed for the solution is 1.035×10¯³ mole
<h3>How to determine the pKa</h3>
We'll begin by calculating the pKa of the solution. This can be obtained as follow:
- Equilibrium constant (Ka) = 2.3×10¯⁵
- pKa =?
pKa = –Log Ka
pKa = –Log 2.3×10¯⁵
pKa = 4.64
<h3>How to determine the molarity of HCl </h3>
- pKa = 4.64
- pH = 6.5
- Molarity of salt [NaZ] = 0.5 M
- Molarity of HCl [HCl] =?
pH = pKa + Log[salt]/[acid]
6.5 = 4.64 + Log[0.5]/[HCl]
Collect like terms
6.5 – 4.64 = Log[0.5]/[HCl]
1.86 = Log[0.5]/[HCl]
Take the anti-log
0.5 / [HCl] = anti-log 1.86
0.5 / [HCl] = 72.44
Cross multiply
0.5 = [HCl] × 72.44
Divide both side by 72.44
[HCl] = 0.5 / 72.4
[HCl] = 0.0069 M
<h3>How to determine the mole of HCl </h3>
- Molarity of HCl = 0.0069 M
- Volume = 150 mL = 150 / 1000 = 0.15 L
Mole = Molarity x Volume
Mole of HCl = 0.0069 × 0.15
Mole of HCl = 1.035×10¯³ mole
<h3>Complete question</h3>
How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3 x 10 -5 .) Assume negligible volume of the HCl
Learn more about pH of buffer:
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