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Korvikt [17]
1 year ago
12

An individual is hospitalized and the initial blood work indicates high levels of hco3- in the blood and a ph of 7. 47. This wou

ld indicate the individual probably has?
Chemistry
1 answer:
expeople1 [14]1 year ago
5 0

An individual is hospitalized and the initial blood work indicates high levels of HCO_{3} ^{-} in the blood and a pH of 7. 47. This would indicate the individual probably has compensated respiratory acidosis.

A chronic illness usually leads to compensated respiratory acidosis because the kidneys have time to adjust to the delayed onset. Even if the PCO_{2} is elevated in a compensated respiratory acidosis, the pH is within the usual range.

The kidneys counteract a respiratory acidosis by increasing the amount of HCO_{3} that tubular cells reabsorb from the tubular fluid, the amount of H^{+} that collecting duct cells secrete while also producing HCO_{3} , and the amount of NH_{3} buffer that is formed through ammoniagenesis.

Respiratory acidosis is frequently brought on by hypoventilation as a result of: breathing depression , paralysis of the respiratory muscles, diseases of the chest wall , abnormalities of the lung parenchyma and abdominal squeezing.

Learn more about Respiratory acidosis here;

brainly.com/question/9694207

#SPJ4

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SOICHIOMETRY HOW MANY MOL OF NITROGEN DIOXIDE ARE REQUIRED TO PRODUCE 2.75 MOL NITRIC ACID? 3NO2+ H20→ 2HNO3+ NO
pishuonlain [190]

Answer: C

Explanation: Hope this help :D

4 0
3 years ago
✞What Is an Alloy?⭐⭐​
max2010maxim [7]
A alloy is a red chicken
8 0
2 years ago
What is the pH of 0.10 M NaF(aq). The Ka of HF is 6.8 x 10-4
dsp73

<u>Given information:</u>

Concentration of NaF = 0.10 M

Ka of HF = 6.8*10⁻⁴

<u>To determine:</u>

pH of 0.1 M NaF

<u>Explanation:</u>

NaF (aq) ↔ Na+ (aq) + F-(aq)

[Na+] = [F-] = 0.10 M

F- will then react with water in the solution as follows:

F- + H2O ↔ HF + OH-

Kb = [OH-][HF]/[F-]

Kw/Ka = [OH-][HF]/[F-]

At equilibrium: [OH-]=[HF] = x and [F-] = 0.1 - x

10⁻¹⁴/6.8*10⁻⁴ = x²/0.1-x

x = [OH-] = 1.21*10⁻⁶ M

pOH = -log[OH-] = -log[1.21*10⁻⁶] = 5.92

pH = 14 - pOH = 14-5.92 = 8.08

Ans: (b)

pH of 0.10 M NaF is 8.08

6 0
3 years ago
How does a chemist count the number of particles in a given number of moles of a substance?
jolli1 [7]
The chemist the count the number of particles (Atoms, Molecules or Formula Unit) in a given number of moles of a substance by using following relationship.

                              Moles  =  # of Particles / 6.022 × 10²³

Or,

                              # of Particles  =  Moles × 6.022 × 10²³

So, from above relation it is found that 1 mole of any substance contains exactly 6.022 × 10²³ particles. Greater the number of moles greater will be the number of particles.
8 0
3 years ago
If two balls have the same volume, but ball A has twice as much mass as ball B, which one will have the greater density? ball A,
rewona [7]

Density (D) is defined as the amount or mass (m) of a substance present in a unit volume(V). It can be expressed mathematically as:

Density = Mass/Volume

i.e. D = m/V -------(1)

Units: g/cm3

a)

If m1, V1 and D1 = mass, volume  and density respectively of ball A

m2, V2 and D2 = mass, volume and density respectively of ball B

It is given that: V1 = V2; but m1 = 2m2

Based on equation (1) we have:

D1/D2 = (m1/V1)* (V2/m2) = (2m2/V2)*(V2/m2) = 2

Thus, density of ball A is twice that of B.

Ans: Ball A will have a greater density than B

b)

If m1, V1 and D1 = mass, volume  and density respectively of ball C

m2, V2 and D2 = mass, volume and density respectively of ball D

It is given that:

V1 = 3V2 and m2 = 1/3(m1) i.e m1 = 3m2

Therefore,

D1/D2 = (m1/V1)* (V2/m2) = (3m2/3V2)*(V2/m2) = 1

Thus, D1 = D2

Ans: Their densities are equal

c)

If m1, V1 and D1 = mass, volume  and density respectively of ball P

m2, V2 and D2 = mass, volume and density respectively of ball Q

It is given that:

m1 = m2 but V1 = 2V2

Therefore,

D1/D2 = (m1/V1)* (V2/m2) = (m2/2V2)*(V2/m2) = 1/2

Thus, D1 = 1/2(D2)

Ans: Ball Q will have a greater density.

d)

If m1, V1 and D1 = mass, volume  and density respectively of ball X

m2, V2 and D2 = mass, volume and density respectively of ball Y

It is given that:

V1 = 2V2 and m1 = 1/2(m2)

Therefore,

D1/D2 = (m1/V1)* (V2/m2) = ((1/2(m2)/2V2)*(V2/m2) = 1/4

Thus, D1 = 1/4(D2)

Ans: Ball Y will have a greater density.



5 0
3 years ago
Read 2 more answers
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