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Alchen [17]
3 years ago
9

Which of the following molecules may diffuse freely through a phospholipid bilayer? (select two answers) asparagine, an amino ac

id with a large, hydrophilic side chain alanine, an amino acid with a small, hydrophobic side-chain testosterone, a steroid hormone fructose, a monosaccharide molecular oxygen (O2)
Chemistry
1 answer:
lisabon 2012 [21]3 years ago
3 0

Answer:

testosterone, a steroid hormone

molecular oxygen (O₂)

Explanation:

Phospholipid Bi layer is permeable to some molecules and it will allow them to pass through it freely. Molecules which are soluble in water (polar molecules or hydrophillic molecules) are less permeable through phospholipid Bi layer . In given options Molecular Oxygen and Testosterone( steroid Hormone) are permeable through the layer because of its lipid solubility and size of the molecule .

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Answer:

c?

Explanation:

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Alcohols are polar molecules. true false
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All alcohols are polar. True
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An increase in blood co2 levels leads to __________.
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Respiratory Acidosis

Explanation:

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In which reaction are electrons transferred from one reactant to another reactant?
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The answer is (1). The electrons transfer form one reactant to another means that there is the change of valence. For other choices, the valence of reactant does not change.
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12. In a gas mixture consisting of carbon dioxide (CO2) and nitrogen (N2), the carbon dioxide molecules have a root-mean-square
zaharov [31]

Answer:

The correct answer is option e.

Explanation:

The formula used for root mean square speed is:

\nu_{rms}=\sqrt{\frac{3kN_AT}{M}}

where,

\nu_{rms} = root mean square speed

k = Boltzmann’s constant = 1.38\times 10^{-23}J/K

T = temperature

M = Molar mass

N_A = Avogadro’s number = 6.02\times 10^{23}mol^{-1}

Root mean square speed of  carbon dioxide molecule:

\nu_{rms}= 550 m/s

Temperature of the mixture = T =?

Molar mass of carbon dioxide = 44 g/mol = 0.044 kg/mol

\nu_{rms}=550 m/s=\sqrt{\frac{3\times 1.38\times 10^{-23}J/K\times 6.022\times 10^{23}mol^{-1}T}{0.044 kg/mol}}

T = 533.87 K

Root mean square speed of nitrogen  molecule:

\nu'_{rms}= ?s

Molar mass of nitrogen = 28 g/mol = 0.028 kg/mol

\nu'_{rms}=\sqrt{\frac{3\times 1.38\times 10^{-23}J/K\times 6.022\times 10^{23}mol^{-1}\times 533.87 K}{0.028 kg/mol}}

\nu'_{rms}=689.46 m/s\approx 689 m/s

689 m/s is the root-mean-square speed of the nitrogen molecules in the sample.

7 0
3 years ago
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