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Answer:
1) 0.3g Mg
2)0.5g MgO
3)0.2g O
4)0.01mol Mg & 0.01mol O
5)0.01mol MgO
6) Empirical formula MgO
Explanation:
The mass og Mg is obtained by substracting 24.36g from 24.66g:
24.66 - 24.36 = 0.3g Mg
The ignition of Mg means that it's reacting with oxygen to form an oxide. The increase in the crucible mass after the Mg ignition is due to the addition of oxygen. However, the addition of few drops of water produces a new compound: a hydroxide. According to the oxidation state og Mg (2+), the only magnesium oxide possible is MgO. It happens because the oxidation state of oxygen in oxides is 2-. Which means that just one oxygen atom is required to electrically neutralize one magnesium atom.
We can use a conversion factor to know how much MgO is made from from 0.3 g of Mg:
*
= 0.2g O
Thereby the mass of the oxide is 0.2g O + 0.3g Mg = 0.5g MgO
We convert the mass of oxygen and magnesium to the respective amounts in moles by using conversion factors:
*
= 0.01mol O
*
= 0.01mol Mg
The moles of MgO can be obtained from:
*
= 0.01mol MgO
To obtain the empirical formula, the amount fo moles of each elements must be divided by the smallest one, in this case, 0.01.
The result for both number of Mg atoms and O atoms is 1. This can be interpreted to mean that there is a Mg atom for each O atom forming the formula unit of the compound.
The step when water is added to the compound resulting after heating does not affect the calculations necessary for the magnesium oxide.
Answer:
4.1x10⁻⁵
Explanation:
The dissociation of an acid is a reversible reaction, and, because of that, it has an equilibrium constant, Ka. For a generic acid (HA), the dissociation happens by:
HA ⇄ H⁺ + A⁻
So, if x moles of the acid dissociates, x moles of H⁺ and x moles of A⁻ is formed. the percent of dissociation of the acid is:
% = (dissociated/total)*100%
4.4% = (x/[HA])*100%
But x = [A⁻], so:
[A⁻]/[HA] = 0.044
The pH of the acid can be calcualted by the Handersson-Halsebach equation:
pH = pKa + log[A⁻]/[HA]
3.03 = pKa + log 0.044
pKa = 3.03 - log 0.044
pKa = 4.39
pKa = -logKa
logKa = -pKa
Ka =
Ka =
Ka = 4.1x10⁻⁵