Answer:
pH = 12.52
Explanation:
Given that,
The [H+] concentration is 
.
We need to find its pH.
We know that, the definition of pH is as follows :
![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
Put all the values,
![pH=-log[3\times 10^{-13}]\\\\pH=12.52](https://tex.z-dn.net/?f=pH%3D-log%5B3%5Ctimes%2010%5E%7B-13%7D%5D%5C%5C%5C%5CpH%3D12.52)
So, the pH is 12.52.
Answer:
Mg(NO4)2 is 180.3 g/mol
Explanation:
First find the substance formula.
Magnesium Nitrate.
Magnesium is a +2 charge.
Nitrate is a -1 charge.
So to balance the chemical formula,
We need 1 magnesium atom for every nitrate atom.
2(1) + 1(-2) = 0
So the substance formula is Mg(NO4)2.
Now find the molar mass of Mg(NO4)2.
Mg = 24.3 amu
N = 14.0 amu
O = 16.0 amu
They are three nitrogen and twelve oxygen atoms.
So you do this: 24.3 + 14.0(2) + 16.0(8) = 180.3 g/mol
So the molar is mass is 180.3 g/mol.
The final answer is Mg(NO4)2 is 180.3 g/mol
Hope it helped!
Answer:
Formation. When the solar system settled into its current layout about 4.5 billion years ago, Earth formed when gravity pulled swirling gas and dust in to become the third planet from the Sun. Like its fellow terrestrial planets, Earth has a central core, a rocky mantle and a solid crust.
I think the correct answer from the choices listed above is option D. Outer planets are mostly made up of gases and are huge in size. These gases are hydrogen and helium. <span>These outer planets are Saturn, Jupiter, Uranus, and Neptune. Hope this answers the question.</span>
Answer:
A. Cu^+2(aq)cathode ---> Cu^+2(aq)anode
Explanation:
Electrolysis is the process in which current is passed through a solution thereby causing a chemical change at the anode and cathode. Copper is being purified using electrolysis by using impure copper at the anode and pure copper at the cathode. This pure and impure copper are placed in a copper(ii)sulfate electrolyte solution and dc current is made to pass through it. The resulting changes at the anode and cathode are given by the equation:
cathode: Cu²⁺ + 2e⁻ ⇒ Cu
anode: Cu ⇒ Cu²⁺ + 2e⁻
