Answer:
Explanation:
Translation:
It is reported that an electric arc stabilized with water reached the temperature of 25600 ° F. On the absolute scale, what is the relationship between the temperature and that of an oxyacetylene flame (3500 ° C)?
Solution:
Temperature of the electric arc = 25600°F
Temperature of the oxyacetylene flame = 3500°C
The absolute temperatures are usually recorded in Kelvin. In order to compare the two given temperatures, a need to convert to Kelvin arises.
From celcius to kelvin we have:
K = 273.15 + T°C
T°C is the temperature in degree celcius
For the oxyacetylene flame:
K = 273.15 + 3500 = 3773.15K
From fahrenheit to kelvin, we have:
K =
(T °F – 32) + 273.15
T °F is the temperature in degree fahrenheit
For the electric arc:
K =
(25600 – 32) + 273.15 = 14477.59K
We can see that by expressing the two temperature on the same absolute scale, an electric arc is by far hotter than an oxyacetylene flame.
That is the element Germanium Atomic weight = 72.64
Answer:
Molarity = 0.08 M
Explanation:
Given data:
Mass of ZnCl₂ = 4.89 g
Volume of water = 500 mL (500 mL× 1L/1000 mL= 0.5 L)
Molarity of solution = ?
Solution:
Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.
Formula:
Molarity = number of moles of solute / L of solution
Number of mole ZnCl₂:
Number of moles = mass/molar mass
Number of moles = 4.89 g/ 136.3 g/mol
Number of moles = 0.04 mol
Molarity:
M = 0.04 mol / 0.5 L
M = 0.08 M
Answer:
Following are the solution to these question:
Explanation:
Calculating the mean:
![\bar{x}=\frac{175+104+164+193+131+189+155+133+151+176}{10}\\\\](https://tex.z-dn.net/?f=%5Cbar%7Bx%7D%3D%5Cfrac%7B175%2B104%2B164%2B193%2B131%2B189%2B155%2B133%2B151%2B176%7D%7B10%7D%5C%5C%5C%5C)
![=\frac{1571}{10}\\\\=157.1](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1571%7D%7B10%7D%5C%5C%5C%5C%3D157.1)
Calculating the standardn:
![\sigma=\sqrt{\frac{\Sigma(x_i-\bar{x})^2}{n-1}}\\\\](https://tex.z-dn.net/?f=%5Csigma%3D%5Csqrt%7B%5Cfrac%7B%5CSigma%28x_i-%5Cbar%7Bx%7D%29%5E2%7D%7Bn-1%7D%7D%5C%5C%5C%5C)
Please find the correct equation in the attached file.
![=28.195](https://tex.z-dn.net/?f=%3D28.195)
For point a:
![=3s+yblank \\\\=3 \times 28.195+50\\\\=84.585+50\\\\=134.585\\](https://tex.z-dn.net/?f=%3D3s%2Byblank%20%5C%5C%5C%5C%3D3%20%5Ctimes%2028.195%2B50%5C%5C%5C%5C%3D84.585%2B50%5C%5C%5C%5C%3D134.585%5C%5C)
For point b:
![=3 \ \frac{s}{m}\\\\ = \frac{(3 \times 28.195)}{1.75 \times 10^9 \ M^{-1}}\\\\= 4.833 \times 10^{-8} \ M](https://tex.z-dn.net/?f=%3D3%20%5C%20%5Cfrac%7Bs%7D%7Bm%7D%5C%5C%5C%5C%20%3D%20%5Cfrac%7B%283%20%5Ctimes%2028.195%29%7D%7B1.75%20%5Ctimes%2010%5E9%20%5C%20M%5E%7B-1%7D%7D%5C%5C%5C%5C%3D%204.833%20%5Ctimes%2010%5E%7B-8%7D%20%5C%20M)
For point c:
![= 10 \frac{s}{m} \\\\= \frac{(10 \times 28.195)}{1.75 x 10^9 \ M^{-1}}\\\\ = 1.611 \times 10^{-7}\ M](https://tex.z-dn.net/?f=%3D%2010%20%5Cfrac%7Bs%7D%7Bm%7D%20%5C%5C%5C%5C%3D%20%5Cfrac%7B%2810%20%5Ctimes%2028.195%29%7D%7B1.75%20x%2010%5E9%20%5C%20M%5E%7B-1%7D%7D%5C%5C%5C%5C%20%3D%201.611%20%5Ctimes%2010%5E%7B-7%7D%5C%20%20M)
It is calculated by using the slope value that is
. The slope value
is ambiguous.