Explanation:
b. What useful functions do oxidation numbers serve?
It is used to show oxidation and reduction (loss and gain of electrons)
b. How many molecules are in 1 mole of molecules?
1 mole = 6.022 * 10^23 molecules
c. What is the name given to the number of molecules in 1 mole?
Avogadro's Number of molecules
21. a. What is the molar mass of an element?
This is the mass of an element divided by the number of moles.
Molar mass = Mass / Number of moles
b. Write the molar mass rounded to two decimal places of carbon, neon, iron and uranium.
amu = Atomic Mass Unit
Carbon = 12.01 amu
Neon = 20.18 amu
Iron = 55.85 amu
Uranium = 238.03 amu
A. electron. The nucleus has protons and neutrons, quark is the particle which forms protons and neutrons.
Answer:
1.8 × 10² s
Explanation:
Let's consider the reduction that occurs upon the electroplating of copper.
Cu²⁺(aq) + 2 e⁻ ⇒ Cu(s)
We will establish the following relationships:
- 1 g = 1,000 mg
- The molar mass of Cu is 63.55 g/mol
- When 1 mole of Cu is deposited, 2 moles of electrons circulate.
- The charge of 1 mole of electrons is 96,486 C (Faraday's constant).
- 1 A = 1 C/s
The time that it would take for 336 mg of copper to be plated at a current of 5.6 A is:
Answer:
Glucose, found in the food animals eat, is broken down during the process of cellular respiration into an energy source called ATP. When excess ATP and glucose are present, the liver converts them into a molecule called glycogen, which is stored for later use.
Answer:
0.1g (Gallon) of chlorine
Explanation:
<u>Formula</u>
1 gallon = 3.7L; the density of water is 1.0g/ml
<u>Given</u>
2g (gallon) of chlorine to sanitize = 1,000,000g (gallon) of water
<u>Solve</u>
If 2g (gallon) chlorine = 1,000,000g (gallon)
∴, ? chlorine = 40,000
The First step; set up an equation
1000000/2 = 40000/?
The Next step; divide 1 million to 2
1000000 ÷ 2 = 500000
Then, divide the result by 40000
40000 ÷ 500000 = 0.08
In the nearest unit that is 0.1
Therefore, it will take 0.1g (gallon) of chlorine to sanitize a 40,000-gallon pool.
