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AnnZ [28]
3 years ago
9

A person with too much lactic acid in their blood, formally called lactic acidosis, is at risk for a variety of adverse health e

ffects. If the average person has approximately 4445 mL of blood, how much lactic acid (in grams) would there need to be before the healthy range is exceeded?
Chemistry
1 answer:
Akimi4 [234]3 years ago
6 0

Answer:

Lactic acidosis occurs when there is greater than 1.60 g in 4445 mL of blood

Explanation:

A normal blood lactic acid level is 0.5-1 mmol/L.

Mild to moderately elevated level of blood lactic acid known as hyperlactatemia occur when the range is between 2-4 mmol/L without metabolic acidosis.

Lactic acidosis occurs with blood lactic acid levels of > 4 mmol/L

Molar mass Lactic acid, C₃H₆O₃ = 90 g/mol

Mass 4 mmol lactic acid = 0.004 mol * 90 g/mol = 0.36 g

Mass of lactic acid in 1 litre or 1000 mL of blood = 0.36 g

Mass  of lactic acid in 4445 mL of blood = 0.36 g * (4445/1000)

<em>Mass  of lactic acid in 4445 mL of blood = 1.60 g</em>

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Why B-Cl bond is polar? ​
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Answer:

The difference in the electronegativities of chlorine and boron is 3.0 - 2.0 = 1.0 ; the difference in between chlorine and carbon is 3.0 = 2.5 = 0.5 . Consequently, the B-Cl bond is more polar ; the chlorine atom asrries the partial negative charge because it has higher electronegativity .

Explanation:

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At what temperature is the following reaction feasible: HCl(g) + NH3(g) -&gt; NH4Cl(s)?
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All four forms of energy are quantised and the quanta ‘gap’ differences increases from trans. KE ==> electronic.

Entropy (S) and energy distribution: The energy is distributed amongst the energy levels in the particles to maximise their entropy.

Entropy is a measure of both the way the particles are arranged AND the ways the quanta of energy can be arranged.

We can apply ΔSθsys/surr/tot ideas to chemical changes to test feasibility of a reaction:

ΔSθtot = ΔSθsys +  ΔSθsurr

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For example: CaCO3(s) ==> CaO(s) + CO2(g) 

ΔSθsys = ΣSθproducts – ΣSθreactants 

ΔSθsys = SθCaO(s) + SθCO2(g) – SθCaCO3(s) 

ΔSθsurr is –ΔHθ/T(K) and ΔH is very endothermic (very +ve),

Now ΔSθsys is approximately constant with temperature and at room temperature the ΔSθsurr term is too negative for ΔSθtot to be plus overall.

But, as the temperature is raised, the ΔSθsurr term becomes less negative and eventually at about 800oCΔSθtot becomes plus overall (and ΔGθ becomes negative), so the decomposition is now chemically, and 'commercially' feasible in a lime kiln.

CaCO3(s) ==> CaO(s) + CO2(g)  ΔHθ = +179 kJ mol–1  (very endothermic)

This important industrial reaction for converting limestone (calcium carbonate) to lime (calcium oxide) has to be performed at high temperatures in a specially designed limekiln – which these days, basically consists of a huge rotating angled ceramic lined steel tube in which a mixture of limestone plus coal/coke/oil/gas? is fed in at one end and lime collected at the lower end. The mixture is ignited and excess air blasted through to burn the coal/coke and maintain a high operating temperature.
ΔSθsys = ΣSθproducts – ΣSθreactants
ΔSθsys = SθCaO(s) + SθCO2(g) – SθCaCO3(s) = (40.0) + (214.0) – (92.9) = +161.0 J mol–1 K–1
ΔSθsurr is –ΔHθ/T = –(179000/T)
ΔSθtot = ΔSθsys +  ΔSθsurr
ΔSθtot = (+161) + (–179000/T) = 161 – 179000/T
If we then substitute various values of T (in Kelvin) you can calculate when the reaction becomes feasible.
For T = 298K (room temperature)

ΔSθtot = 161 – 179000/298 = –439.7 J mol–1 K–1, no good, negative entropy change

For T = 500K (fairly high temperature for an industrial process)

ΔSθtot = 161 – 179000/500 = –197.0, still no good

For T = 1200K (limekiln temperature)

ΔSθtot = 161 – 179000/1200 = +11.8 J mol–1 K–1, definitely feasible, overall positive entropy change

Now assuming ΔSθsys is approximately constant with temperature change and at room temperature the ΔSθsurr term is too negative for ΔSθtot to be plus overall. But, as the temperature is raised, the ΔSθsurr term becomes less negative and eventually at about 800–900oC ΔSθtot becomes plus overall, so the decomposition is now chemically, and 'commercially' feasible in a lime kiln.
You can approach the problem in another more efficient way by solving the total entropy expression for T at the point when the total entropy change is zero. At this point calcium carbonate, calcium oxide and carbon dioxide are at equilibrium.
ΔSθtot–equilib = 0 = 161 – 179000/T, 179000/T = 161, T = 179000/161 = 1112 K

This means that 1112 K is the minimum temperature to get an economic yield. Well at first sight anyway. In fact because the carbon dioxide is swept away in the flue gases so an equilibrium is never truly attained so limestone continues to decompose even at lower temperatures.

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