Explanation:
In order to be able to calculate the volume of oxygen gas produced by this reaction, you need to know the conditions for pressure and temperature.
Since no mention of those conditions was made, I'll assume that the reaction takes place at STP, Standard Temperature and Pressure.
STP conditions are defined as a pressure of
100 kPa
and a temperature of
0
∘
C
. Under these conditions for pressure and temperature, one mole of any ideal gas occupies
22.7 L
- this is known as the molar volume of a gas at STP.
So, in order to find the volume of oxygen gas at STP, you need to know how many moles of oxygen are produced by this reaction.
The balanced chemical equation for this decomposition reaction looks like this
2
KClO
3(s]
heat
×
−−−→
2
KCl
(s]
+
3
O
2(g]
↑
⏐
⏐
Notice that you have a
2
:
3
mole ratio between potassium chlorate and oxygen gas.
This tells you that the reaction will always produce
3
2
times more moles of oxygen gas than the number of moles of potassium chlorate that underwent decomposition.
Use potassium chlorate's molar mass to determine how many moles you have in that
231-g
sample
231
g
⋅
1 mole KClO
3
122.55
g
=
1.885 moles KClO
3
Use the aforementioned mole ratio to determine how many moles of oxygen would be produced from this many moles of potassium chlorate
1.885
moles KClO
3
⋅
3
moles O
2
2
moles KClO
3
=
2.8275 moles O
2
So, what volume would this many moles occupy at STP?
2.8275
moles
⋅
22.7 L
1
mol
=
64.2 L
Al(NO3)3(aq) + 3NaOH(s) --> Al(OH)3 (s) + 3NaNO3 (aq)
The precipitate here is Al(OH)3 (s), since the solid reactant is the precipitate in the aqueous solution. Usually, it is okay to assume in basic chemistry that the transition metal is going to be part of the compound that is the precipitate, especially in an acidic salt and a strong base reaction that we have here.
The answer <span>is <span>8.9 g/mL</span>.</span>
The density (D) is <span>equal to mass (m) divided by volume (V): D = m/V
Let's find the mass of the object:
m = 156 g - 105.5 g = 50.5 g
Let's find the volume of the volume:
V = 30.7 mL - 25 mL = 5.7 mL
The density is:
D = m/V = 50.5 g / 5.7 mL = 8.9 g/mL</span>
Explanation:
We are given: entropy of Fe2O3 = 90J/K.mol
: entropy of C = 5.7J/K.mol
: entropy of Fe = 27.2J/K.mol
: entropy of CO = 198J/K.mol
Answer:
The correct answer is C.
The average weight of an atom of an element, formerly based on the
weight of one hydrogen atom taken as a unit or on 1/16 (0.0625) the
weight of an oxygen atom, but after 1961 based on 1/12 the weight of the
carbon-12 atom.
