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3 years ago

7

3. When substances cross cell membranes by simple diffusion, they

1 answer:

poizon [28]3 years ago

3 0

Answer:B

Explanation:due to osmosis and osmoregulation of the cell membrane

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The amount of energy required to raise the temperature of a swimming pool depends on the mass of water present.

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How many hydrogen bonds are between cytosine and guanine

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Three

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2 years ago

Given the unbalanced ionic equation:

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4 years ago

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In some regions of the southwest United States, the water is very hard. For example, in Las Cruces, New Mexico, the tap water co

myrzilka [38]

Answer:

a) $46.734L$ of hard water is require to produce $45L$ of pure water.

b) Same water would not work for purifying sea water as it would require so much of energy.

Explanation:

a)

Osmotic pressure of a solution is calculated by the formula,

Π= $iMRT$

Where,

Π is the osmotic pressure

$i$ is van't hoff factor

$M$ is molar concentration (mol/L)

$R$ is gas constant

$T$ is Temperature in Kelvin

Now,

$MgCO_{3}\rightarrow Mg^{2+}+ CO_{3}^{2-}$

So, according to the above equation $i=2$

Calculation concentration at which reverse osmosis will stop

$M=\frac{ Π }{iRT}$

$M=\frac{8atm}{2\times0.082L atm mol^{-1}K^{-1} \times 300K} \\M= 0.162M$

Calculating Initial concentration that is moles of $MgCO_{3}$

$n=\frac{given mass}{molar mass} \\\\n=\frac{560 \times 10^{-6}g}{84.32g/mol}\\\\N=6.642\times10^{-6}moles$

Calculating molarity

$M= 6.642 \times 10^{-6}\times 10{^3}\\M=6.642\times10^{-3}M$

Total number of moles in total volume must remain the same,

$C_{1}V_{1}=C_{2}V_{2}$

$0.006642M \times V_{1}=0.162M \times V_{2}$

$\frac{V_{1}}{V_{2}}= \frac{0.162}{0.006.642}=25.953$

$V_{1}=25.953 V_{2}$

Also, $V_{1}=45L$

$V_{2}=\frac{45}{25.953}\\ V_{2}= 1.734L$

So, net Volume,

$V=45L+1.734L=46.734L$

So, $46.734L$ is require to produce $45L$ of pure water.

b)

Using the equation for $0.60M NaCl$ as well,

$C_{1}V_{1}=C_{2}V_{2}$

$C_{1}=0.60M\\V_{1}=45L\\C_{2}=0.162M\\V_{2}=?$

$V_{2}=\frac{0.60 \times 45}{0.167}\\V_{2}= 166.66L$

$Net water=166.66+45L=211.66L$

So, the same water would not work for purifying sea water as it would require so much of energy, as we require 211.66L of water to produce 45L of water.

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