Answer:
Explanation:
When calculating an empirical formula from percentages, assume you have a 100g sample. This allows you to convert the percentages directly to grams, because X % of 100g is X grams.
So:
24.42 % = 24.42 g Ca, 17.07% = 17.07g N, 58.5% = 58.5g O
The next step is to divide each mass by their molar mass to convert your grams to moles.
24.42/40.08 = 0.6092 mol
17.07/14.01 = 1.218 mol
58.85/15.99 = 3.680 mol
Then you will divide all of your mol values by the SMALLEST number of moles. This gives you whole numbers that are the mole ratio (subcripts) of the empircal formula.
0.6092 mol/0.6092 mol = 1
1.218 mol/0.6092 mol = 2
3.680 mol/0.6092 mol = 6
So the empirical formula is
1. A soluble salt can be prepared by reacting an acid with a suitable insoluble reactant including:
a metal
a metal oxide
a carbonate
3. I don’t know this one
4. A term base or glossary is a database containing single words or expressions related to a specific subject.
5. Strong acid is an acid that ionizes completely in aqueous solution. It always loses a proton (H+) when dissolved in water. Weak acid is an acid that ionizes partially in a solution. ... Because the rate of reaction depends upon the degree of dissociation αand strong acids have higher degrees of dissociation.
im not sure of the rest
Answer:
T2 = 260 K
Explanation:
<em>Given data:</em>
P1 = 150.0 k Pa
T1 = (-23+ 273.15) K = 250.15 K
V1 = 1.75 L
P2 = 210.0 kPa
V2 = 1.30 L
<em>To find:</em>
T2 = ?
<em>Formula:</em>
<em>Calculation:</em>
T2 = (210.0 kPa) x (1.30 L) x (250.15 K) / (150.0 kPa) x (1.75 L)
T2 = 260 K
Answer:
Faraday's constant will be smaller than it is supposed to be.
Explanation:
If the copper anode was not completely dry when its mass was measured, mass of the copper must be heavier than it should have been. Hence, the calculated Faraday’s constant would be smaller than it is supposed to be since when calculating Faraday’s Constant, the charge transferred is divided by the moles of electrons.
