Answer:
6.14
Explanation:
If the pH falls as temperature increases, this does not mean that water becomes more acidic at higher temperatures. A solution is acidic if there is an excess of hydrogen ions over hydroxide ions (i.e., pH < pOH). In the case of pure water, there are always the same concentration of hydrogen ions and hydroxide ions and hence, the water is still neutral (pH = pOH) - even if its pH changes.
The problem is that we are all familiar with 7 being the pH of pure water, that anything else feels really strange. Remember that to calculate the neutral value of pH from Kw . If that changes, then the neutral value for pH changes as well. At 100°C, the pH of pure water is 6.14, which is "neutral" on the pH scale at this higher temperature. A solution with a pH of 7 at this temperature is slightly alkaline because its pH is a bit higher than the neutral value of 6.14.
For a 568B crossover cable that already has wht-org and org on pins 1 and 2 of the connector at one end, the connector at the OTHER end should have wht-grn and grn on pins 1 and 2 respectively.
The wht-org and org at that end should drop to pins 3 and 6 respectively.
You're welcome, and good luck.
Answer:
9.965 nF
Explanation:
The capacitance of the axon C = εA/d where ε = dielectric constant = 24.78 × 10⁻¹² F/m, A = surface area of axon = 2πrL where r = radius of axon = 8 μm = 8 × 10⁻⁶ m and L = length of axon = 8 cm = 8 × 10⁻² m and d = thickness of membrane = 0.01 μm = 0.01 × 10⁻⁶ m
So, C = εA/d
C = ε2πrL/d
Substituting the of the values variables into the equation, we have
C = ε2πrL/d
C = 24.78 × 10⁻¹² F/m × 2π × 8 × 10⁻⁶ m × 8 × 10⁻² m/0.01 × 10⁻⁶ m
C = 9964.63 × 10⁻²⁰ Fm/0.01 × 10⁻⁶ m
C = 996463 × 10⁻¹⁴ F
C = 9.96463 × 10⁻⁹ F
C = 9.96463 nF
C ≅ 9.965 nF
A wave created by shaking a rope up and down is called a:
(According to me and certified experts) Transverse wave.
If this helps, then please give Brainliest!
Answer: 8*10^-15 N
Explanation: In order to calculate the force applied on an electron in the middle of the two planes at 500 V we know that, F=q*E
The electric field between the plates is given by:
E = ΔV/d = 500 V/0.01 m=5*10^3 N/C
the force applied to the electron is: F=e*E=8*10^-15 N
