Answer:
A_________________________
Based on the calculations, the approximate ductility (%el) of this brass is equal to 2.3%.
<u>Given the following data:</u>
- Yield strength = 230 mpa (33360 psi).
<h3>What is ductility?</h3>
Ductility can be defined as an important property of a material which determines its ability to become elongated due to the application of stress.
Mathematically, the ductility of a material can be expressed as percentage elongation in length:

<u>Where:</u>
is the original length.
is the final length.
is the yield strength.
For this exercise, let us assume the original length of this brass is equal to 100 meters.
Substituting the parameters into the formula, we have;

Ductility = 2.3%.
Read more on ductility here: brainly.com/question/828860
1. 1/16 x 1 000 000= 62500
2. Each half life is 5750 years long
3. The end of a half life is found by dividing the beginning of the half life by 2.
4. The number 62500 is between the 7th and the 8th half lives (refer to the number of 14C atoms on my table)
5. The 7th half life is 40250 years, and the 8th half life is 46000 years.
6. Therefore, your estimate would be a number roughly in the middle of 40250 and 46000 (for example: 43000)
Hope this helps :)
Explanation:
At 365 K temperature sulfur tetrafluoride have a density of 0.260 g/L at 0.0721 atm.
What is an ideal gas equation?
The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
First, calculate the moles of the gas using the gas law,
PV=nRT, where n is the moles and R is the gas constant. Then divide
the given mass by the number of moles to get molar mass.
Given data:
P= 0.0721 atm
n=\frac{mass}{molar \;mass}n=
molarmass
mass
R= 0.082057338 \;L \;atm \;K^{-1}mol^{-1}R=0.082057338LatmK
−1
mol
−1
T=?
Putting value in the given equation:
\frac{PV}{RT}=n
RT
PV
=n
density = \frac{2 \;atm\; X molar\; mass}{0.082057338 \;L \;atm \;K^{-1}mol^{-1} X T}density=
0.082057338LatmK
−1
mol
−1
XT
2atmXmolarmass
0.260 g/L = \frac{0.0721 \;atm\; X 108.07 g/mol}{0.082057338 \;L \;atm \;K^{-1}mol^{-1} X T}0.260g/L=
0.082057338LatmK
−1
mol
−1
XT
0.0721atmX108.07g/mol
T = 365.2158727 K= 365 K
Hence , at 365 K temperature sulfur tetrafluoride have a density of 0.260 g/L at 0.0721 atm.