Zero is the value of water potential of pure water at atmospheric pressure.
Answer:
Density, d = 1.779 g/cm³
Explanation:
The density of a material is given by its mass per unit volume.
Here, height of a piece of magnesium cylinder, h = 5.62 cm
Its diameter, d = 1.34 cm
Radius = 0.67 cm
Volume of he cylinder,


So, the density of the sample is 1.779 g/cm³.
Answer:
Zn =⇒ Zn+2(0.10) + 2e- (anode)
Zn+2(?M) + 2e- === Zn(s) (cathode)
Zn + Zn+2(?M) ===⇒ Zn+2(0.10) + Zn
E = E^o -0.0592 log Q; in this case E^o is zero.
E = - 0.0592 /n logQ where n is the number of electrons transferred, in this
case n = 2
23 mV x 1 volt/1000mv = 0.023 Volts
0.023 = -0.0592 / 2 log(0.10) / [Zn+2]
0.023 = -0.0296 { log 0.10 – log [Zn+2] }
0.023 = -0.0296{ -1 - log[Zn+2] }
0.023 = +0.0296 + 0.0296log[Zn+2]
-0.0066 = 0.0296log[Zn+2]
-0.22= log[Zn+2]
[Zn+2] = 10^-0.22 = 0.603 Molar
If you were to engineer an everyday solution for conserving water which activity do you think would be the most impactful ?
Answer:
38.75 L
Explanation:
From the question,
Applying Boyles Law,
PV = P'V'....................... Equation 1
Where P = Original pressure of the Argon gas, V = Original Volume of Argon gas, P' = Final pressure of Argon gas, V' = Final Volume of Argon gas.
make V the subject of the equation
V = P'V'/P.................... Equation 2
Given: P = 34.6 atm, V' = 456 L, P' = 2.94 atm.
Substitute these values into equation 2
V = (456×2.94)/34.6
V = 38.75 L