Answer:
Rubidium-85=61.2
Rubidium-87=24.36
Atomic Mass=85.56 amu
Explanation:
To find the atomic mass, we must multiply the masses of the isotope by the percent abundance, then add.
<u>Rubidium-85 </u>
This isotope has an abundance of 72%.
Convert 72% to a decimal. Divide by 100 or move the decimal two places to the left.
- 72/100= 0.72 or 72.0 --> 7.2 ---> 0.72
Multiply the mass of the isotope, which is 85, by the abundance as a decimal.
- mass * decimal abundance= 85* 0.72= 61.2
Rubidium-85=61.2
<u>Rubidium-87</u>
This isotope has an abundance of 28%.
Convert 28% to a decimal. Divide by 100 or move the decimal two places to the left.
- 28/100= 0.28 or 28.0 --> 2.8 ---> 0.28
Multiply the mass of the isotope, which is 87, by the abundance as a decimal.
- mass * decimal abundance= 87* 0.28= 24.36
Rubidium-87=24.36
<u>Atomic Mass of Rubidium:</u>
Add the two numbers together.
- Rb-85 (61.2) and Rb-87 (24.36)
Answer:
Using the coarse adjustment knob of the microscope in high power may lead to the breaking of the slide if adjusted and raised the slide too much which can damage the sample as well as the high power lens.
In this case, I would recommend using the fine adjustment knob and moving away from the end of the viewing area of the microscope so there would no collision take place. The fine adjustment will help to get a clear image.
The volume of copper : 3.24 ml
<h3>Further explanation
</h3>
Density is a quantity derived from the mass and volume
Density is the ratio of mass per unit volume
With the same mass, the volume of objects that have a high density will be smaller than objects with a smaller type of mass
The unit of density can be expressed in g/cm³ or kg/m³
Density formula:
ρ = density
, g/cm³ or kg/m³
m = mass
, g or kg
v = volume
, cm³ or m³
A common example is the water density of 1 gr / cm³
The density of copper : 8.96 gr/ml
mass of copper : 29 g
then the volume :
After subtracting the volume needed from the volume dispensed, we got a remainder of 35ml
<h3>Subtraction of Numbers</h3>
Given Data
- Volume of Hexane dispensed = 40ml
Let us compute the amount of excess hexane/ the volume that will remain
Remainder = The difference in volume dispensed and the volume needed
Remainder = 40-5
Remainder = 35 ml
The remainder is 35ml
Learn more about subtraction of numbers here:
brainly.com/question/4721701
Answer: 2 mol
Explanation:
- According to the ideal gas law, One mole of an ideal gas at STP (standard temperature and normal pressure) occupies 22.4 liters.
- Using cross multiplication,
1 mol of (O2) → 22.4 L
? → 43.9 L
Therefore, the number of moles of oxygen in 43.9 L = (43.9 × 1)/ 22.4 = 1.96 mol≈ 2 mol..
