Answer:
(119 g H2O) / (18.01532 g H2O/mol) x (1 mol Pb / 2 mol H2O) x (207.21 g Pb/mol) = 684 g Pb
Explanation:
<span>Chemically speaking, rust is a base and any acid will remove it. The choice of acid is going to be the thing to consider, since acid + base = salt and water. Phosphoric acid left a residue because the salt Iron phosphate is insoluble in water. Iron's soluble salts include the chloride, the sulfate and the nitrate. Industrially speaking, you need to "pickle" your iron. Pickling is a process in which dilute sulfuric acid is used to remove any surface corrosion prior to either painting or plating an iron surface. Sulfuric acid is ordinary battery acid and the salt Iron sulfate is not toxic. Sulfuric acid is one of the most common acids used (besides hydrochloric acid). The dilute kind is not terribly corrosive but concentrated sulfuric acid is a thick, syrupy liquid which can cause some nasty chemical burns if allowed to remain on the skin. It also heats up quite a lot when water is added, so this is an "Acid to water not water to acid" situation. The other choice is Hydrochloric acid, known as muriatic acid. The 20% concentrate is available in nearly any hardware store. It isn't as corrosive as concentrated sulfuric acid, but it has a burning, acrid stench, so never use the concentrate without adequate ventilation. It is ordinarily used to remove hard water deposits (boiler scale) but does a good on on rust as well. Concentrated Iron chloride isn't entirely inert but lots of rinsing will turn it back into harmless rust/sludge, especially if the rince water is naturally hard. Nitric acid will remove corrosion from anything, but it is extremely corrosive, smells worse then Hydrochloric acid and isn't easy to get, since it can be used to create some powerful explosives</span>
Answer:
There are multiple ways to check mass but I'll tell you one. Look below
Explanation:
One easy way of checking atomic mass is by adding protons and neutrons.
For example:
We have 5 protons and 4 neutrons.
5+4=9
I hope this helps (:
Answer:
E = 0.062 V
Explanation:
(a) See the attached file for the answer
(b)
Calculating the voltage (E) using the formula;
E = - (2.303RT/nf)log Cathode/Anode
Where,
R = 8.314 J/K/mol
T = 35°C = 308 K
F- Faraday's constant = 96500 C/mol,
n = number of moles of electron = 2
Substituting, we have
E = -(2.303 * 8.314 *308/2*96500) *log (0.03/3)
= -0.031 * -2
= 0.062V
Therefore, the voltmeter will show a voltage of 0.062 V
