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4 years ago

What is the condition in which an odorless gas combines rapidly with hemoglobin and crowds out oxygen?

Chemistry

2 answers:

slavikrds [6]4 years ago

8 0

Answer:

The correct answer is - carbon monoxide poisoning.

Explanation:

Carbon monoxide poisoning is the condition which is involves the combining rapidly with hemoglobin with a odorless gas and crowds out or reduction of the oxygen.

Carbon monoxide or CO is a odorless gas can be present in various fumes exhaust from the tobacco smoke or cars. it joins with the hemoglobin like oxygen not let go.

Thus, the correct answer is - carbon monoxide poisoning.

Nesterboy [21]4 years ago

3 0

Answer: Carbon monoxide poisoning

Explanation:

Carbon monoxide poisoning is the condition in which the hemoglobin in the body gets attached with the carbon monoxide gas and there is a overall reduction in the oxygen level.

Carbon monoxide is basically released from the burning of fossil fuels and is very suffocating in nature.

It can be understood in the way the beside oxygen the carbon monoxide is attached to the hemoglobin molecule.

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Using the van der waals equation, the pressure in a 22.4 l vessel containing 1.50 mol of chlorine gas at 0.00 °c is ________ atm

Lana71 [14]

Answer is: the pressure in a vessel is 1.48 atm.

V(Cl₂) = 22.4 L; pressure of chlorine gas.

n(Cl₂) = 1.50 mol; amount of chlorine gas.

T = 0.00°C = 273.15 K; temperature.

a = 6.49 L²·atm/mol²; the constant a provides a correction for the intermolecular forces.

b = 0.0562 L/mol; value is the volume of one mole of the chlorine gas.

R = 0.08206 L·atm/mol·K, universal gas constant.

Van de Waals equation: (P + an² / V²)(V - nb) = nRT.

(P + 6.49 L²·atm/mol² · (1.5 mol)² / (22.4 L)²) · (22.4 L - 1.5 mol·0.0562 L/mol) = 1.5 mol · 0.08206 L·atm/mol·K · 273.15 K.

(P + 6.49 L²·atm/mol² · (1.5 mol)² / (22.4 L)²) = (1.5 mol · 0.08206 L·atm/mol·K · 273.15 K) ÷ (22.4 L - 1.5 mol · 0.0562 L/mol).

P + 0.029 atm = 33.62 L·atm ÷ 22.31 L.

P = 1.507 atm - 0.029 atm.

P = 1.48 atm; the pressure.

7 0

3 years ago

For the wild type (unmutated) enzyme, you measure a rate of p-nitrophenol release by the change in absorbance at 405 nm (for the

Aleks04 [339]

Answer:

1.2x10⁻⁵M = Concentration of the product released

Explanation:

Lambert-Beer's law states the absorbance of a solution is directly proportional to its concentration. The equation is:

A = E*b*C

<em>Where A is the absotbance of the solution: 0.216</em>

<em>E is the extinction coefficient = 18000M⁻¹cm⁻¹</em>

<em>b is patelength = 1cm</em>

<em>C is concentration of the solution</em>

<em />

Replacing:

0.216 = 18000M⁻¹cm⁻¹*1cm*C

<h3>1.2x10⁻⁵M = Concentration of the product released</h3>

4 0

3 years ago

In terms of drug development, rank the following microbes from most difficult (1) to least difficult (3):

iVinArrow [24]

Answer:

See below ↓↓↓

Explanation:

Helminths = 1 Most Difficult

Fungi = 2 Moderately Difficult [Option not given, but 2 is right answer]

Bacteria = 3 Least Difficult

6 0

2 years ago

ANSWER FOR BRAINLYEST

Olenka [21]

Is there something else that goes along with this?

6 0

3 years ago

Read 2 more answers

A 0.020 M solution of niacin has a pH of 3.26. (a) What percentage of the acid is ionized in this solution

mariarad [96]

Answer: 2.75%

Explanation:

$pH=-log [H+]$

$3.26 = -log [H+]$

$[H+] = 5.495\times 10^{-4} M$

$HA\rightleftharpoons H^++A^-$

initial 0.020 0 0

eqm 0.020 -x x x

$K_a=\frac{[H+][A-]}{[HA]}$

$K_a=\frac{[x][x]}{[0.020-x]}$

$x=5.495\times 10^{-4}$

$K_a=\frac{[5.495\times 10^{-4}]^2}{[0.020-5.495\times 10^{-4}]}$

$K_a =1.553\times 10^{-5}$

percent dissociation = $\frac{[H^+_eqm]}{[Acid_{initial}]}\times 100$

percent dissociation= $\frac{5.495\times 10^{-4}}{0.020}\times 100$

Thus percent dissociation= 2.75 %

5 0

3 years ago