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Alexxandr [17]
3 years ago
12

Biological membranes are selectively permeable, allowing certain molecules to cross the membrane, but not others. Classify the m

olecules or ions depending on how they cross a biological membrane. Note that some of these examples may also utilize active transport to traverse a membrane. However, this question is limited to passive transport processes only. Match the followings with each others
a. Simple Diffusion
b. Facilitated Diffusion

1. steroid hormones
2. K+
3. N2
4. Glucose
Chemistry
1 answer:
ddd [48]3 years ago
3 0

Answer:

Following are the solution to these question:

Explanation:

Simple diffusion implies which molecules can move across a cell membrane instead of using the incorporated protein structure channels. Its flow of molecules across these canals is enabled by diffusion. It has steroid hormones and N2.

Facilitated diffusion of molecules via cell membranes, or active transportation, which would be the random mass transit process of molecules or ions via unique cellular membranes from the transmembrane. It has K+ and Glucose.

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The sun revolved around the earth why ? help me ​
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Answer:

You are moving because the Earth and everything in our solar system is constantly moving. ... As the Earth rotates, it also moves, or revolves, around the Sun. The Earth's path around the Sun is called its orbit. It takes the Earth one year, or 365 1/4 days, to completely orbit the Sun.

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graduated cylinder is filled with water to a volume of 6.2 ML. an irregular shaped plastic object weighing 1.2 g is placed in th
Aleks04 [339]

Answer:

A. Density of object = 0.86 g/mL

B. The object will float in the water.

Explanation:

The following data were obtained from the question:

Volume of water = 6.2 mL

Mass (m) of object = 1.2 g

Volume of water + Object = 7.59 mL

Density of object =?

Density of water = 1 g/mL

Next, we shall determine the volume of the object. This can be obtained as follow:

Volume of water = 6.2 mL

Volume of water + Object = 7.59 mL

Volume of object =?

Volume of object = (Volume of water + Object) – (Volume of water)

Volume of object = 7.59 – 6.2

Volume of object = 1.39 mL

Therefore, the volume of the object is 1.39 mL

A. Determination of the density of the object.

Mass (m) of object = 1.2 g

Volume (V) of object = 1.39 mL

Density (D) of object =?

Density = mass /volume

Density = 1.2/1.39

Density of object = 0.86 g/mL

B. Determination of whether the object will float or sink.

Density of object = 0.86 g/mL

Density of water = 1 g/mL

From the above, we can see that the density of water is greater than that of the object. This implies that the object is lighter than water. Therefore, the object will float in the water.

8 0
3 years ago
In this reaction: Mg (s) + I₂ (s) → MgI₂ (s) If 2.68 moles of Mg react with 3.56 moles of I₂, and 1.76 moles of MgI₂ form, what
melomori [17]

Answer:

Y=65.7\%

Explanation:

Hello,

In this case, for the given chemical reaction, we first identify the limiting reactant by noticing that due to the 1:1 mole ratio for magnesium to iodine the reacting moles must the same, nevertheless, there are only 2.68 moles of magnesium versus 3.56 moles of iodine, for that reason, magnesium is the limiting reactant, so the theoretical turns out:

n_{MgI_2}^{theoretical}=2.68molMg*\frac{1molMgI_2}{1molMg} =2.68molMgI_2

Thus, we compute the percent yield as:

Y=\frac{n_{MgI_2}^{real}}{n_{MgI_2}^{theoretical}} *100\%=\frac{1.76mol}{2.68mol} *100\%\\\\Y=65.7\%

Best regards.

8 0
3 years ago
The thallium (present as Tl2SO4) in a 9.486-g pesticide sample was precipitated as thallium(I) iodide. Calculate the mass percen
tino4ka555 [31]

Answer: The mass percentage of Tl_2SO_4 is 5.86%

Explanation:

To calculate the mass percentage of Tl_2SO_4 in the sample it is necessary to know the mass of the solute (Tl_2SO_4 in this case), and the mass of the solution (pesticide sample, whose mass is explicit in the letter of the problem).

To calculate the mass of the solute, we must take the mass of the TlI precipitate.  We can establish a relation between the mass of TlI and Tl_2SO_4 using the stoichiometry of the compounds:

moles\ of\ TlI = \frac{0.1824 g}{331.27\frac{g}{mol} } = 5.51*10^{-4}\ mol.

Since for every mole of Tl in TlI there are two moles of Tl in Tl_2SO_4, we have:

moles\ of\ Tl_2SO_4 = 2 * moles\ of\ TlI = 1,102*10^{-3}\ mol

Using the molar mass of Tl_2SO_4 we have:

mass\ of\ Tl_2SO_4 = 1,102*10^{-3}\ mol * 504.83\ \frac{g}{mol}= 0.56\ g

Finally, we can use the mass percentage formula:

mass\ percentage = (\frac{solute\ mass}{solution\ mass} )*100 = (\frac{mass\ of\ Tl_2SO_4}{pesticide\ sample\ mass})*100 = (\frac{0.56g}{9.486g})*100 = 5.86\%

6 0
3 years ago
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