Answer:104.6cm^3
Explanation:
V of a cylinder =πr2h
V = pi((1.045cm)^2)(30.48)
V = 104.57cm^3
Explanation:
Number of moles = 0.0688 moles of NaoH
volume = 0.250 L
Molarity = moles of solute / volume ( L )
M = 0.0688 / 0.250
M = 0.28 M
Answer B
Answer:
The average atomic mass is closer to Si- 28 because this isotope is present in more percentage in the sample.
Explanation:
Given data:
Atomic mass of silicon= ?
Percent abundance of Si-28 = 92.21%
Atomic mass of Si-28 = 27.98 amu
Percent abundance of Si-29 = 4.70%
Atomic mass of Si-29 = 28.98 amu
Percent abundance of Si-30 = 3.09%
Atomic mass of Si-30 = 29.97 amu
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)+(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (92.21×27.98)+(4.70×28.98)+(3.09×29.97) /100
Average atomic mass = 2580.04 +136.21+92.61 / 100
Average atomic mass = 2808.86 / 100
Average atomic mass = 28.08amu.
The average atomic mass is closer to Si- 28 because this isotope is present in more percentage in the sample.
1. True
2. True
3. False
4. True