Answer:
Strong acids. hope this helps :)
Answer:
NaS2 it is the most stable
Answer:
-3.82ºC is the freezing point of solution
Explanation:
We work with the Freezing point depression to solve the problem
ΔT = m . Kf . i
ΔT = Freezing point of pure solvent - freezing point of solution
Let's find out m, molality (moles of solute in 1kg of solvent)
15 g / 58.45 g/mol = 0.257 moles of NaCl
NaCl(s) → Na⁺ (aq) + Cl⁻(aq)
i = 2 (Van't Hoff factor, numbers of ions dissolved)
m = mol /kg → 0.257 mol / 0.250kg = 1.03 m
Kf = Cryoscopic constant → 1.86 ºC/m (pure, for water)
0ºC - Tºf = 1.03m . 1.86ºC/m . 2
Tºf = -3.82ºC
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
It is known that Q =
where, A = cross sectional area
V = speed of the fluid in that section
Hence, calculate V as follows.
V =
=
=
= 10.61 m/sec
Also it is known that Reynold's number is as follows.
Re =
=
= 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.
L =
= 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:
3 rd period 2 electrons mean it should be Calcium
Explanation: