Decomposition of potassium chlorate (KClO3) produces potassium chloride (KCl) and pure oxygen (O2). The balanced equation for the reaction is as follows.mc017-1.jpg
What volume of oxygen gas is released at STP if 10.0 g of potassium chlorate is decomposed? (The molar mass of KClO3 is 122.55 g/mol.)
Answers
KClO₃ = 122.55 g/mol
O₂ = 32 g/mol
2 KClO₃ = 2 KCl + 3 O₂
2 x 122.55 g KClO₃ ----------> 3 x 32 g O₂
10.0 g KClO₃ ------------------> ?
Mass of O₂ = ( 10.0 x 3 x 32 ) / ( 2 x 122.55 )
Mass of O₂ = 960 / 245.1
Mass of O₂ = 3.9167 g
number of moles O₂ => 3.9167 / 32 = 0.1223 moles
1 mol -------------- 22.4 L ( at STP)
0.1223 moles ---- ?
V = 0.1223 x 22.4 / 1
V = 2.739 L
hope this helps!
Your answer is 2.74.
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What are synonyms for the word control?
Answers
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Explanation:
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Answers
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An object must overcome inertia to begin moving.
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29
Which of the following is not an example of friction?
Pouring water off a bridge and watching it turn to small drops by the time it hits the ground.
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If an object's acceleration is zero, which of these could be true?
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Both B and C could be true
Answers
C
D
C
trust me plz i am good at chemistry
(2) N2(g) + H2(g) =>NH3(g)
(3) 2NaCl(s)=>Na(s) + Cl2(g)
(4) 2KCl(s) => 2K(s) + Cl2(g)
Answers
The correct balanced chemicalequation is 2KCl(s) => 2K(s) + Cl2(g).
The correct balanced chemical equation from the options provided is (4) 2KCl(s) => 2K(s) + Cl2(g).
In this equation, two molecules of potassium chloride (KCl) react to form two molecules of potassium (K) and one molecule of chlorine gas (Cl2). This equation is balanced because the number of atoms of each element is equal on both sides of the equation.
For example, there are two atoms of potassium and two atoms of chlorine on both sides of the equation, ensuring the equation is balanced.
Learn more about Chemical Equation here:
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Answers
Ans: Volume of stock H2SO4 required = 6.94 ml
Given:
Concentration of stock H2SO4 solution M1 = 18.0 M
Concentration of the final H2SO4 solution needed M2 = 2.50 M
Final volume of H2SO4 needed, V2 = 50.0 ml
To determine:
Volume of stock needed, V1
Explanation:
Use the dilution relation:
Hello!
In a school’s laboratory, students require 50.0 mL of 2.50 M H2SO4 for an experiment, but the only available stock solution of the acid has a concentration of 18.0 M. What volume of the stock solution would they use to make the required solution?
We have the following data:
M1 (initial molarity) = 2.50 M (or mol/L)
V1 (initial volume) = 50.0 mL → 0.05 L
M2 (final molarity) = 18.0 M (or mol/L)
V2 (final volume) = ? (in mL)
Let's use the formula of dilution and molarity, so we have:
Answer:
The volume is approximately 6.94 mL
_______________________________
Answers
1 Sn 3 Sn
3 H 2 H
1 P 4 P
4 O 16 O
then you have to make the numbers the same for each side
so you get by putting the 3 infront of Sn 4 infront of the H3PO4 and 6 infront of the H2
3 Sn 3 Sn
12 H 12 H
4 P 4 P
16 O 16 O