Answer:
E: the point where the acid and base have been added in proper stoichiometric amounts
Explanation:
Equivalence point in titration is simply the point where the amounts of acid and base used just sufficiently reacts chemically to cause neutralization whereas the endpoint is the point where the indicator of the titration changes colour.
The Equivalence point occurs before the endpoint.
Thus, option E is correct.
In calorimetry, energy is measured through heat transfer from one substance to
another. Which of the following is NOT a method of heat transfer?
Answer:
Refraction
Explanation:
What is the limiting reactant in the following equation? How much Fe2O3 will be produced if 2.1 g of Fe reacts with 2.1 g of O2?
4 Fe + 3O2 —> 2Fe2O3
Answer:
Fe is limiting reactant and 3.00g of Fe2O3 will be produced
Explanation:
To solve this question we must convert the mass of each reactant to moles and, using the reaction we can find limiting reactant. With moles of limiting reactant we can find moles of Fe2O3 and its mass as follows:
Moles Fe -Molar mass: 55.845g/mol-
2.1g * (1mol / 55.845g) = 0.0376 moles
Moles O2 -Molar mass: 32g/mol-
2.1g * (1mol / 32g) = 0.0656 moles
For a complete reaction of 0.0656 moles of O2 are needed:
0.0656moles O2 * (4mol Fe / 3 mol O2) = 0.0875 moles Fe
As there are just 0.0376 moles,
Fe is limiting reactant
The mass of Fe2O3 is:
Moles:
0.0376 moles Fe* (2mol Fe2O3 / 4mol Fe) = 0.0188 moles Fe2O3
Mass:
0.0188 moles Fe2O3 * (159.69g / mol) =
3.00g of Fe2O3 will be produced
PLEASE HELP!! ORGANIC CHEMISTRY
A sample of a diatonic gas is loaded into an evacuated bottle at STP. The 0.25 L bottle contains 1.76 grams of the unidentified gas. Calculate the molar mass of the gas. What is the identity of the diatomic gas?
Answer:
(a) 157.7 g
(b) 7.04 g/dm³
Explanation:
(a) From the question,
According to Avogadro's Law,
1 mole of every gas at STP occupies a volume of 22.4 dm³
But mass of 1 mole of the diatomic gas = molar mass of the gas.
This Implies that,
The molar mass of the gas at STP occupies a volume of 22.4 dm³
From the question,
If,
0.25 L bottle contain 1.76 g of the gas,
Therefore,
Molar mass of the gas = (1.76×22.4)/0.25
Molar mass of the gas = 157.7 g.
(b) Density of the gas = mass/volume
D = m/v
Given: m = 1.76 g, v = 0.25 L = 0.25 dm³
Therefore,
D = 1.76/0,25
D = 7.04 g/dm³