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Answer = B = Neutrons and Mass Number
Isotopes are defined as those atoms which have same atomic number but different atomic masses.
Atomic mass is basically the number of protons and neutrons present in an atom.
Atomic number is the number of protons present in an atom.
So, in isotopes the number of protons are same but the number of neutrons vary due to which atomic masses also vary.
In given three isotopes, all have same number of protons but different number of neutrons.
i.e.
H-1 = 1 P + 0 N = 1 u (Proton)
H-2 = 1 P + 1 N = 2 u (Deuterium)
H-3 = 1 P + 2 N = 3 u (Tritium)
Hence, it is clear that the number after H shows a change in number of neutrons and mass number.
Isotopes are defined as those atoms which have same atomic number but different atomic masses.
Atomic mass is basically the number of protons and neutrons present in an atom.
Atomic number is the number of protons present in an atom.
So, in isotopes the number of protons are same but the number of neutrons vary due to which atomic masses also vary.
In given three isotopes, all have same number of protons but different number of neutrons.
i.e.
H-1 = 1 P + 0 N = 1 u (Proton)
H-2 = 1 P + 1 N = 2 u (Deuterium)
H-3 = 1 P + 2 N = 3 u (Tritium)
Hence, it is clear that the number after H shows a change in number of neutrons and mass number.
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Answer:
(a) pH = 12.73
(b) pH = 10.52
(c) pH = 1.93
Explanation:
The net balanced reaction equation is:
KOH + HBr ⇒ H₂O + KBr
The amount of KOH present is:
n = CV = (0.1000 molL⁻¹)(30.00 mL) = 3.000 mmol
(a) The amount of HBr added in 9.00 mL of 0.1000 M HBr is:
(0.1000 molL⁻¹)(9.00 mL) = 0.900 mmol
This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:
(3.000 mmol) - (0.900 mmol) = 2.100 mmol KOH
After the addition of HBr, the volume of the KOH solution is 39.00 mL. The concentration of KOH is calculated as follows:
C = n/V = (2.100 mmol) / (39.00 mL) = 0.0538461 M KOH
The pOH and pH of the solution can then be calculated:
pOH = -log[OH⁻] = -log(0.0538461) = 1.2688
pH = 14 - pOH = 14 - 1.2688 = 12.73
(b) The amount of HBr added in 29.80 mL of 0.1000 M HBr is:
(0.1000 molL⁻¹)(29.80 mL) = 2.980 mmol
This amount of HBr will neutralize an equivalent amount of KOH, leaving the following amount of KOH:
(3.000 mmol) - (2.980 mmol) = 0.0200 mmol KOH
After the addition of HBr, the volume of the KOH solution is 59.80 mL. The concentration of KOH is calculated as follows:
C = n/V = (0.0200 mmol) / (59.80 mL) = 0.0003344 M KOH
The pOH and pH of the solution can then be calculated:
pOH = -log[OH⁻] = -log(0.0003344) = 3.476
pH = 14 - pOH = 14 - 3.476 = 10.52
(c) The amount of HBr added in 38.00 mL of 0.1000 M HBr is:
(0.1000 molL⁻¹)(38.00 mL) = 3.800 mmol
This amount of HBr will neutralize all of the KOH present. The amount of HBr in excess is:
(3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr
After the addition of HBr, the volume of the analyte solution is 68.00 mL. The concentration of HBr is calculated as follows:
C = n/V = (0.800 mmol) / (68.00 mL) = 0.01176 M HBr
The pH of the solution can then be calculated:
pH = -log[H⁺] = -log(0.01176) = 1.93
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<h2>Answer: 13985.4 g</h2>
Explanation:
Mass = volume × density
Mass = 652 cm³ × 21.45 g/cm³
= 13985.4 g
Explanation:
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