Free expansion refers to uninhibited or provoked expansion, i.e. it’s expanding on its own, by itself. The only situation in which this occurs is B, as the gas is allowed to expand by itself. A involves a reaction that forces the expansion, C involves the gas performing work, and D involves thermal expansion.
Hope this helps!
Answer:
FeCl₂
Explanation:
Empirical formula is the simplest formula of a compound.
Mass percentage of Chlorine = 56%
Mass percentage Iron = 100 - 56 = 44%
Elements
Fe Cl
Percentage mass 44 56
moles 44/55.85 56/35.5
0.79 1.56
Divide by the smallest
number 0.79/0.79 1.56/0.79
1 2
The empirical formula of the compound is FeCl₂
Answer:
The moles of O₂ which are present in the sample are 0.433. (<u>option D</u>)
Explanation:
We should apply the Ideal Gases Law, to solve this:
P . V = n. R. T
1.38 atm . 7.51L = n . 0.082 . 292K
(1.38 atm . 7.51L) / (0.082 . 292K) = n
0.433 moles = n
Answer:
91.4%
Explanation:
Potassium hydrogen tartrate, KHT, reacts with KOH as follows.
KHT + KOH → H₂O + K₂T
<em>where 1 mole of acid (KHT) reacts per mole o base (KOH), -That is acid/base ratio 1:1</em>
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The endpoint of a titration is the point in which moles of KOH = moles of KHT, you can see this endpoint with an indicator or doing a potentiometric titration.
As the endpoint requires 21.58mL = 0.02158L of a 0.1125M KOH, moles of KOH = moles of KHT are:
0.02158L × (0.1125mol / L) = 2.428x10⁻³ moles of KOH = moles of KHT
To convert these moles to grams you use molar mass of KHT (188.177g/mol):
2.428x10⁻³ moles of KHT × (188.177g / mol) = 0.457g of KHT are in the sample.
As you add 0.500g of sample, percentage of KHT in the sample is:
(0.457g / 0.500g) × 100 =
<h3>
91.4%</h3>
-That is the purity of the sample-
Answer:
4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium
Explanation:
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) you can see that the following amounts in moles of each compound react and are produced:
- HCl: 2 moles
- Na: 1 mole
- NaCl: 2 moles
- H₂: 1 mole
You know the following masses of each element:
- H: 1 g/mole
- Cl: 35.45 g/mole
- Na: 23 g/mole
So, the molar mass of each compound participating in the reaction is:
- HCl: 1 g/mole + 35.45 g/mole= 36.45 g/mole
- Na: 23 g/mole
- NaCl: 23 g/mole + 35.45 g/mole= 58.45 g/mole
- H₂: 2* 1 g/mole= 2 g/mole
Then, by stoichiometry of the reaction, the following amounts in grams of each of the compounds participating in the reaction react and are produced:
- HCl: 2 moles* 36.45 g/mole= 72.9 g
- Na: 1 mole* 23 g/mole= 23 g
- NaCl: 2 moles* 58.45 g/mole= 116.9 g
- H₂: 1 mole* 2 g/mole= 2 g
So, a rule of three applies as follows: if by stoichiometry, when reacting 72.9 grams of HCl 2 grams of H₂ are formed, when reacting 175 grams of HCl how much mass of H₂ will be formed?
mass of H₂= 4.8 g
<u><em>4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium</em></u>
