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beks73 [17]
3 years ago
15

Carbon dioxide and water react to form bicarbonate ion and hydronium ion. Hyperventilation (rapid breathing) causes more carbon

dioxide to be exhaled than normal. How will hyperventilation affect the pH of blood?
this is about a unit on shifting equilibrium and the common-ion effect so try to connect the explanation to those ideas please!
Chemistry
1 answer:
Juliette [100K]3 years ago
4 0
Hyperventilation<span> will blow off carbon dioxide which reduces the alkaline component of the </span>blood<span>, resulting in acidosis and a lower </span>blood pH<span>. (See Andromeda Strain). This is why re-breathing into a bag to increase carbon dioxide in your </span>blood will make you feel less faint. Basically <span>It increases the pH ... you can look at CO2 as an acid ... and when reduced ... alkalinization of the pH occurs</span> 
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Stacy reads that a 50:50 mixture of methanol and water is best for keeping a car’s radiator from freezing when the temperature g
salantis [7]
1. The hypothesis for this is experiment is that the 50:50 of methanol-water mixture will not turn to solid when the temperature reaches to -40°C.

2. The procedure for this is measuring equal volumes of water and methanol using the graduated cylinder. You can measure 100 mL of water and 100 mL of methanol using the graduated cylinder. Then, mix them in the beaker. Next, measure 200 mL of water, and another 200 mL of methanol. Don't mix them. Also, make a 60:40 mixture by measuring 120 mL of water and 80 mL of methanol, then mix them together. Place them all in the refrigerator at the same time. Record the time when they would freeze to solid.

3. The controls for this experiment are the 200 mL water alone, and the 200 mL methanol alone.

4. The independent variable in here is the time, while the dependent variable is the temperature of the mixtures.

5. If the hypothesis turns out to be true, then all the mixtures prepared should freeze and become solid after a certain period of time, with the exception of the 50:50 mixture. The 50:50 mixture should still remain as a liquid even when left overnight.
5 0
2 years ago
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What is the molar concentration of 35 mL of H2SO4 that neutralizes 25 mL of 0.320M NaOH
stira [4]
V ( H2SO4) = 35 mL / 1000 => 0.035 L

M ( H2SO4) = ?

V ( NaOH ) = 25 mL / 1000 => 0.025 L 

M ( NaOH ) = 0.320 M

number of moles NaOH:

n = M x V

n = 0.025 x 0.320 => 0.008 moles of NaOH

Mole ratio:

<span>2 NaOH + H2SO4 = Na2SO4 + 2 H2O
</span>
2 moles NaOH ---------------------- 1 mole H2SO4
0.008 moles moles NaOH ---------- ??

0.008 x 1 / 2 => 0.004 moles of H2SO4 :

Therefore:

M ( H2SO4) = n / V

M = 0.004 /  0.035

= 0.114 M

hope this helps!



6 0
3 years ago
How many molecules of co2,h2o,c3h8, and o2 will be present if the reaction goes to completion?
Nadya [2.5K]

Let initially there are 10 molecules of O2 and 3 molecules of C3H8 present

The reaction will be

C3H8(g)  + 5O2(g)  ----> 3CO2(g)   + 4H2O

so here oxygen molecules are limiting as for 3 molecules of C3H8 we need 15 molecules of O2

now the given 10 molecules of O2 will react with only 2 molecules of C3H8 and they will form six molecules of CO2 and 8 molecules of H2O

Hence answer is

molecules of CO2 formed = 6

Molecules of H2O formed = 8

molecules of C3H8 left = 1

molecules of O2 left = 0


7 0
3 years ago
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