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KengaRu [80]
3 years ago
7

How does increased exhalation of CO2 at the lungs (pulmonary regulation) restore the correct pH range?

Chemistry
1 answer:
PilotLPTM [1.2K]3 years ago
3 0

Answer:

When CO2 levels become excessive, a condition known as acidosis occurs. This is defined as the pH of the

blood becoming less than 7.35. The body maintains the balance mainly by using bicarbonate ions in the

blood. As the body responds to neutralize this condition, an electrolyte imbalance – an increase of plasma

chloride, potassium, calcium and sodium, can occur.

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If a sample of hydrogen gas was collected over water. If the total pressure 7 points
ludmilkaskok [199]

Answer:

53.3

Explanation:

6 0
3 years ago
A decorative "ice" sculpture is carved from dry ice (solid CO2) and held at its sublimation point of –78.5°C. Consider the proce
Leno4ka [110]

Answer:

The answers to the questions are;

a. The entropy of sublimation for carbon dioxide (the system) is  

134.07 J/Kmol.

b. The entropy of the universe for this reversible process is 376 J/K.

Explanation:

Entropy of sublimation is the entropy change experienced following the transformation of a mole of solid to vapor at  the temperature where the sublimation is taking place

a. We note that the mass of the solid CO₂ = 389 g

Molar mass of CO₂ = 44.01 g/mol

Number of moles of CO₂ in the sculpture = Mass/(Molar mass)

= (389 g)/(44.01 g/mol) = 8.84 Moles

Entropy of sublimation is given by

ΔS_{sublimation} = S_{vapor} - S_{solid} = \frac{\Delta H_{sublimation}}{T}

Where:

ΔH_{sublimation}  = 26.1 KJ/mol

T = Temperature = –78.5°C = ‪194.65‬ K

Therefore the amount of heat required to cause the 389 g of dry ice to sublime =    26.1 KJ/mol  × 8.84 Moles = 230.695 KJ

Therefore the entropy of sublimation = ΔS_{sublimation} = \frac{230.695 KJ}{194.65 K}

= 1.185 KJ/K

= 1185 J/K = 1185/8.84 J/Kmol = 134.07 J/Kmol

b. The entropy of the universe is given by;

ΔS_{universe} = \Delta S_{system} + ΔS_{surrounding}  

If the heat absorbed by the system is the same as the heat given off by the surrounding, then we have;

ΔS_{universe} = \frac{Q}{ T_{system}}  -\frac{Q}{T_{surrounding}}  

                =1.185 KJ/K - -\frac{230.695 KJ}{285.15K} = 1.185 KJ/K - 0.809 KJ/K = 0.376 KJ/K

= 376 J/K.

7 0
3 years ago
What is the mass of H2O in 3 moles?
ss7ja [257]

The answer is 18.02 grams


4 0
4 years ago
if burning wood causes the energy of the surroundings to increase by 12,000J what is the change in energy of the wood
Hitman42 [59]
Due to the law of conservation of energy we know that the energy that heated the surrounding had to have come from the reaction which means that the wood (with oxygen since this is combustion) lost 12000J. The change in energy for the wood is negative. I hope this helps. Let me know if anything is unclear.
5 0
3 years ago
Are normal salts acidic, basic or neutral salt?
lorasvet [3.4K]

Answer:

neutral idk

Explanation:

6 0
4 years ago
Read 2 more answers
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