Answer: 1123000 Joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point
Explanation:
Latent heat of vaporization is the amount of heat required to convert 1 mole of liquid to gas at atmospheric pressure.
Amount of heat required to vaporize 1 mole of lead = 177.7 kJ
Molar mass of lead = 207.2 g
Mass of lead given = 1.31 kg = 1310 g (1kg=1000g)
Heat required to vaporize 207.2 of lead = 177.7 kJ
Thus Heat required to vaporize 1310 g of lead =
Thus 1123000 Joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point
Answer:
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Explanation: Balance the reaction of KOH + H3PO4 = K3PO4 + H2O using this chemical equation balancer!
In the titration of lemon juice, the presence of ascorbic acid means the concentration of citric acid you calculated is higher.
An acid-base titration is a common way to determine the unknown concentration of an acid, given we know the concentration of the base and determine the spent volume in the titration. Let's consider the neutralization reactions that take place in a mixture of citric acid and ascorbic acid.
Citric acid titration :
3 NaOH(aq) + H₃C₆H₅O₇(aq) → Na₃C₆H₅O₇(aq) + 3 H₂O(l)
Ascorbic acid titration:
NaOH(aq) + HC₆H₇O₆(aq) → NaC₆H₇O₆(aq) + H₂O(l)
If we titrated a solution that contained only citric acid, we can relate through stoichiometry the moles and concentration of citric acid. However, if the solution also contained ascorbic acid, we would have to spend more NaOH to titrate it. Since more NaOH would react, we would conclude that there is more citric acid to react, calculating a higher concentration of the same.
In the titration of lemon juice, the presence of ascorbic acid means the concentration of citric acid you calculated is higher.
You can learn more about titration here: brainly.com/question/2728613
The
atomic mass of an element is a result of the weighted average of the
masses of its corresponding isotopes. <span>
We are given:
Ag-107: mass of 106.905 amu, 52 percent abundance
Ag-109: mass of 108.905 amu,
48 percent abundance
To be able to calculate the atomic mass of
silver, we multiply the percent abundances of the isotopes by their
respective atomic masses. Then, we add,
Atomic mass = (Atomic Mass of Ag-107) (%
Abundance) + (Atomic Mass of Ag-109) (% Abundance )
Substituting the given values, we have
Atomic mass = 106.905 (.52) + 108.905 (.48)
Atomic mass = 107.865 amu
Therefore, the atomic mass of silver would be 107.865 amu.</span>
