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Answer:
Your answer would be a). 2.0 × 10-9
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Work:
In your question the "ph" of a 0.55 m aqueous solution of hypobromous acid temperature is at 25 degrees C, and it's "ph" is 4.48.
You would use the ph (4.48) to find the ka for "hbro"
[H+]
=
10^-4.48
=
3.31 x 10^-5 M
=
[BrO-]
or: [H+] = 10^-4.48 = 3.31 x 10^-5 M = [BrO-]
Then you would find ka:
(3.31 x 10^-5)^2/0.55 =2 x 10^-9
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<em>-Julie</em>
Answer:
A)16.3% (Hexane's formula is C6H14)
There would be an equal amounts of ELECTRONS.
In order for the atom to be neutral it would imply that the number of protons (positive particles) and the number of electrons (negative particles) are equal since the neutrons are without charge.
2C₃H₇OH + 9O₂ = 6CO₂ + 8H₂O
V(O₂)=12.0 dm³
n(C₃H₇OH)=0.1 mol
n(O₂)=12.0 dm³/22.4 dm³/mol=0.5357 mol
C₃H₇OH : O₂ 2:9 1:4.5
0.1:0.5357
oxygen in excess
V(CO₂)=3Vm*n(C₃H₇OH)
V(CO₂)=3*22.4*0.1=6.72 dm³
Answer:
0.055g/mL
Explanation:
Data obtained from the question include:
Molar Mass of the gass sample = 71g/mol
Volume of the gas sample = 1300 mL
Density =?
The density of a substance is simply mass per unit volume. It is represented mathematically as:
Density = Mass /volume.
With the above equation, we can easily obtain the density of sample of gas as illustrated below:
Density = 71g / 1300 mL
Density = 0.055g/mL
Therefore, the density of the gas sample is 0.055g/mL
