We are given the resistance and voltage of this lamp and we are asked to find the current; the equation that relates these together is Ohm’s Law, V = IR. Simply plug and solve:
V = IR
(220 V) = I(484 Ohms)
I = 0.4545 Amps
The lamp has a current of 0.4545 Amps passing through it under these conditions.
Hope this helps!
Answer:
![p[H+] = 10.042](https://tex.z-dn.net/?f=p%5BH%2B%5D%20%3D%2010.042)
Explanation:
As we know that
......eq (1)
we will calculate the pH of OH- and then we will calculate the pH of H+
So p[OH-] ![= - log [1.10 * 10^{-4}]](https://tex.z-dn.net/?f=%3D%20-%20log%20%5B1.10%20%2A%2010%5E%7B-4%7D%5D)
Solving the right side of the equation, we get
p[OH-]
![= - [-3.958]\\= 3.958](https://tex.z-dn.net/?f=%3D%20-%20%5B-3.958%5D%5C%5C%3D%203.958)
Now we know that

Substituting the value of pOH in the above equation, we get -
![14.0 = p[H+] + 3.958\\p[H+] = 14 - 3.958\\p[H+] = 10.042](https://tex.z-dn.net/?f=14.0%20%3D%20p%5BH%2B%5D%20%2B%203.958%5C%5Cp%5BH%2B%5D%20%3D%2014%20-%203.958%5C%5Cp%5BH%2B%5D%20%3D%2010.042)
Chromium sulfate.
Cr is chromium, and SO4 is sulfate.
Explanation:
Moles of phosphorus pentachloride present initially = 2.5 mol
Moles of phosphorus trichloride at equilibrium = 0.338 mol

Initially
2.5 mol 0 0
At equilibrium:
(2.5 - x) mol x x
So, from above, the moles of phosphorus trichloride at equilibrium , x= 0.338 mol
Mass of 0.338 moles of phosphorus trichloride at equilibrium:
= 0.338 mol × 137.5 g/mol = 46.475 g
Moles of phosphorus pentachloride present at equilibrium :
= (2.5 - 0.338) mol = 2.162 mol
Mass of 2.162 moles of phosphorus pentachloride at equilibrium:
= 2.162 mol × 208.5 g/mol = 450.777 g
Moles of chloride gas present at equilibrium : 0.338 mol
Mass of 0.338 moles of chloride gas at equilibrium:
= 0.338 mol × 71 g/mol = 23.998 g