A caption is a positively charged ion. They are formed when an ion loses one or more electrons. Typical it is the loss Of their valence electrons.
Since its volume you do what it says 55 cm x 100 cm x 80cm and do 100 x 50 which is 5,000. then 5,000 x 80 which is 40,000. sorry i cant show work on here.
Answer:this is all i found out
Oxygen is known as a diatomic molecule. Oxygen atom has atomic number as 8 so it has its K shell completely filled with 2 electrons. While it has only 6 electrons in L shell which requires 8 electrons to get inert gas configuration. So oxygen atom shares 2 electrons with another oxygen atom to form a diatomic molecule.
Explanation:
In nature there are many more variations amino acids than the simple 20 found in humans. However, when analyzing the human genome sequence, there is a code for all 64 permutations (4^3), only some of them share amino acids. This is a safe-guard against mutations of one or two nucleotides. For example, the amino acid Alanine is coded with four different nucleotide sequences: GCA, GCC, GCG, GCU. Also some amino acids code the same like UUU &UUC
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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