Answer:

Explanation:
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In this case, since the decomposition of potassium chlorate is:

We can see a 2:3 mole ratio between potassium chlorate and oxygen (molar mass 32.0 g/mol), thus, via stoichiometry, we compute the mass of oxygen that are produced by the decomposition of 2.50 moles of this reactant:

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Answer:
V = 5.17L
Explanation:
Mass of gas = 8.7g
T = 23°C = (23 + 273.15)K = 296.15K
P = 1.15 atm
V = ?
R = 0.082atm.L / mol.K
From ideal gas equation
PV = nRT
P = pressure of the gas
V = volume of the gas
n = no. Of moles
R = ideal gas constant
T = temperature of the gas
no of moles = mass / molar mass
Molar mass of Chlorine = 35.5g / mol
No. Of moles = 8.7 / 35.5
No. Of moles = 0.245 moles
PV = nRT
V = nRT / P
V = (0.245 * 0.082 * 296.15) / 1.15
V = 5.9496 / 1.15
V = 5.17L
The volume of the gas is 5.17L
Answer:
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Answer:
2K + 2H2O → H2 + 2KOH
Explanation:
Find how many atoms you have on both sides then add 2 to both sides.
Reactant: Products:
K: 1+1=2 K: 1+1=2
H: 2+2=4 H: 3+1=4
O: 1+1=2 O: 1+1=2
Therefore it is balanced. Hope this helps
The answer for the following problem is mentioned below.
- <u><em>Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule </em></u>
Explanation:
Given:
no of moles of the oxygen gas = 0.692
Also given:
2 HgO → 2 Hg + 
where,
HgO represents mercuric oxide
Hg represents mercury
represents oxygen
To calculate:
Molar mass of HgO:
Molar mass of HgO = 216 grams
molar mass of mercury (Hg) = 200 grams
molar mass of oxygen (O) =16 grams
HgO = 200 +16 = 216 grams
We know;
2×216 grams of HgO → 1 mole of oxygen molecule
? → 0.692 moles of oxygen molecule
= 
= 298.944 grams of HgO
<u><em>Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule </em></u>
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