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Karolina [17]
3 years ago
13

The candle is lit and dilute ethanoic acid is poured down the inside of the beaker. As the acid reacts with the baking soda, bub

bles of CO2 gas form. After a few seconds
the air in the beaker is replaced by 0.20 liter of CO2 gas, causing candle flame to go out. The density of CO2 gas is 1.8 grams per liter at room temperature.
Choose the correct structural formula for the acid that was poured into the beaker.

Chemistry
1 answer:
nika2105 [10]3 years ago
5 0

Answer:

It’s answer #2

Explanation:

I only know this because i just got it wrong

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Calculate the pH of a solution with a [H 3 O + ]=5.6x10 -9 M.
Gre4nikov [31]

Answer:

8.3

Explanation:

pH is the measure of the H+ or H30 (they r the same thing) ions in a solution. it is equal to -log[H+]. [H+]=  Molar concentration of H+ ions.

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Instructions
Marysya12 [62]

Answer:

-104.7?

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Explanation:

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3 years ago
Explain how you were able to use your knowledge of how different types of blood react with anti A, anti B, and anti Rh antibodie
dybincka [34]

Depending upon the clumping reaction with anti A , anti B and anti Rh antibodies the blood types are determined.

Explanation:

Agglutination (clumping) will occur when blood that contains the particular antigen is mixed with the particular antibody.

A+ have Agglutination with Anti-A ,Anti-Rh and No agglutination with Anti-B.

A- have Agglutination with Anti-A and No agglutination with Anti-B and Anti-Rh.

B+ have Agglutination with Anti-B Anti-Rh and No agglutination with Anti-A.

B- have Agglutination with Anti-B and No agglutination with Anti-B and Anti-Rh.

Rh+ have Agglutination with Anti-A and Anti-Rh and No agglutination with Anti-B.

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3 0
3 years ago
How many moles of each element are in one mole of Be(OH)2?
REY [17]
B. 1 mole of beryllium, 2 moles of oxygen, 2 moles of hydrogen
8 0
3 years ago
Read 2 more answers
Calculate the solubility of mn(oh)2 in grams per liter when buffered at ph=7.0. assume that buffer capacity is not exhausted
Vanyuwa [196]
When PH + POH = 14 
∴ POH = 14 -7 = 7

when POH = -㏒[OH-]

          7    = -㏒ [OH-]
∴[OH-] = 10^-7

by using ICE table:

           Mn(OH)2(s) ⇄  Mn2+ (aq)  + 2OH-(aq)
initial                              0                     10^-7
change                           +X                      +2X
Equ                                 X                  (10^-7 + 2X)

when Ksp = [Mn2+][OH-]^2

when Ksp of Mn(OH)2 = 4.6 x 10^-14

by substitution:

4.6 x 10^-14 = X*(10^-7+2X)^2  by solving this equation for X

∴ X =2.3 x 10-5 M

∴ The solubility of Mn(OH)2 in grams per liter (when the molar mass of Mn(OH)2 = 88.953 g/mol
= 2.3 x10^-5 moles/L * 88.953 g/mol

= 0.002 g/ L
3 0
4 years ago
Read 2 more answers
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