Answer:
4, 3, 2
Explanation:
so I would balance this equation first by identifying the 3 2 inbalance
usually with these you find a common factor that suits other variables
eg. 6 so times the right side by 2 and the O2 by 3
now O are equal on both sides and to equalise fe times the left by 4
the equation is balanced
Answer:
Electron pairs are arranged as far apart from each other as possible.
Explanation:
Like charges repel therefore, VSEPR theory is the "...repulsions between electron pairs present in the valence shell of the central atom."
Explanation:
The given data is as follows.
Diameter = 0.1 m,
= 1000 kPa
= 500 kPa
Change in pressure
= 1000 kPa - 500 kPa = 500 kPa
Since, 1000 Pa = 1 kPa. So, 500 kPa will be equal to
.
Q = 5
=
= 0.0833 ![m^{3}/sec](https://tex.z-dn.net/?f=m%5E%7B3%7D%2Fsec)
It is known that Q = ![A \times V](https://tex.z-dn.net/?f=A%20%5Ctimes%20V)
where, A = cross sectional area
V = speed of the fluid in that section
Hence, calculate V as follows.
V = ![\frac{Q}{A}](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7BA%7D)
= ![\frac{Q \times 4}{\pi \times d^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7BQ%20%5Ctimes%204%7D%7B%5Cpi%20%5Ctimes%20d%5E%7B2%7D%7D)
= ![\frac{0.0833 \times 4}{3.14 \times (0.1)^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B0.0833%20%5Ctimes%204%7D%7B3.14%20%5Ctimes%20%280.1%29%5E%7B2%7D%7D)
= 10.61 m/sec
Also it is known that Reynold's number is as follows.
Re = ![\frac{\rho \times V \times d}{\mu}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Crho%20%5Ctimes%20V%20%5Ctimes%20d%7D%7B%5Cmu%7D)
=
= 1061032.954
As, it is given that the flow is turbulent so we cannot use the Hagen-Poiseuille equation as follows. Therefore, by using Blasius equation for turbulent flow as follows.
![\Delta P = \frac{0.241 \times \rho^{0.75} \times \mu^{0.25} \times L}{D^{4.75}} \times Q^{1.75}](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%20%5Cfrac%7B0.241%20%5Ctimes%20%5Crho%5E%7B0.75%7D%20%5Ctimes%20%5Cmu%5E%7B0.25%7D%20%5Ctimes%20L%7D%7BD%5E%7B4.75%7D%7D%20%5Ctimes%20Q%5E%7B1.75%7D)
![500 \times 10^{3} = \frac{0.241 \times (1000)^{0.75} \times (10^{-3})^{0.25} \times L}{(0.1)^{4.75}} \times (0.0833)^{1.75}](https://tex.z-dn.net/?f=500%20%5Ctimes%2010%5E%7B3%7D%20%3D%20%5Cfrac%7B0.241%20%5Ctimes%20%281000%29%5E%7B0.75%7D%20%5Ctimes%20%2810%5E%7B-3%7D%29%5E%7B0.25%7D%20%5Ctimes%20L%7D%7B%280.1%29%5E%7B4.75%7D%7D%20%5Ctimes%20%280.0833%29%5E%7B1.75%7D)
L = ![\frac{500 \times 10^{3}}{5535.36}](https://tex.z-dn.net/?f=%5Cfrac%7B500%20%5Ctimes%2010%5E%7B3%7D%7D%7B5535.36%7D)
= 90.328 m
Thus, we can conclude that 90.328 m length galvanized iron line should be used to reach the desired outlet pressure.
Answer:
the eye of the tiger in the bronx zoo tiger king meme with
Answer:
The values are:
6.55 mmHg
0.00862 atm
Explanation:
This is all about unit conversion
760 mmHg = 760 Torr
1 atm = 760 Torr
6.55 Torr . 760 mmHg / 760 Torr = 6.55 mmHg
6.55 Torr . 1 atm / 760 Torr = 0.00862 atm