Answer:
Average atomic mass of chlorine is 35.48 amu.
Explanation:
Given data:
Percent abundance of Cl-35 = 76%
Percent abundance of Cl-37 = 24%
Average atomic mass = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (76×35)+(24×37) /100
Average atomic mass = 2660 + 888 / 100
Average atomic mass = 3548/ 100
Average atomic mass = 35.48 amu
Average atomic mass of chlorine is 35.48 amu.
Answer:
The correct approach is Option B (Peer Review).
Explanation:
- Rather made reference to someone as a scientific peer-review, it encourages the specialist who has not been essential to the study team to analyze the study objectively and pointed out everyone's mistakes. It serves as major self-regulation for scholars and aims to make the publishing process somewhat credible. Hence, the solution to this issue is Peer Examination.
- Funding organizations rarely have the capabilities to recognize out mistakes, whereas definitive analysis is a method of study that helps to make a definitive statement. The gathering of data is simply a process of scientific study.
Other approaches do not apply to the example mentioned. Although the one mentioned is right.
I believe it was John Newlands.
Hope that helped
Answer: B
Explanation:
According to Ohm's Law, the answer is B.
Ohm's Law states that power is equal to volume x current.
If volume x current equals power, that means they are both 50% of power.
Ohm's Law:
power = voltage x current
current = voltage x power
voltage = power x current
I hope this answer helped.
Answer:
The answers are in the explanation
Explanation:
A buffer is the mixture of a weak acid with its conjugate base or vice versa. Thus:
<em>1)</em> Mixing 100.0 mL of 0.1 M HF with 100.0 mL of 0.05 M mol KF. <em>Will </em>result in a buffer because HF is a weak acid and KF is its conjugate base.
<em>2)</em> Mixing 100.0 mL of 0.1 M NH₃ with 100.0 mL of 0.1 M NH₄Br. <em>Will not </em>result in a buffer because NH₃ is a strong base.
<em>3) </em>Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.05 M KOH. <em>Will </em>result in a buffer because HCN is a weak acid and its reaction with KOH will produce CN⁻ that is its conjugate base.
<em>4)</em> Mixing 100.0 mL of 0.1 M HCl with 100.0 mL of 0.1 M KCl <em>Will not </em>result in a buffer because HCl is a strong acid.
<em>5)</em> Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.1 M KOH <em>Will not </em>result in a buffer because each HCN will react with KOH producing CN⁻, that means that you will have just CN⁻ (Conjugate base) without HCN (Weak acid).
I hope it helps!
