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3 years ago

Why does the structure of the plasma membrane make this type of transport necessary for fluids

1 answer:

8 0

Answer: because the phospholipid bilayer formed by these interactions makes a good barrier between the interior and exterior of the cell, because water and other polar or charged substances cannot easily cross the hydrophobic core of the membrane.

Explanation:

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1.33 dm3 of water at 70°C are saturated by 2.25

astraxan [27]

Given that 4.50 dm³ of Pb(NO₃)₂ is cooled from 70 °C to 18 °C, the

amount amount of solute that will be deposited is 1,927.413 grams.

<h3>How can the amount of solute deposited be found?</h3>

The volume of water 1.33 dm³ of water 70 °C.

The number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water at 70 °C = 2.25 moles

At 18 °C, the number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water = 0.53 moles

Therefore;

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C is therefore;

1.33 dm³ contains 2.25 moles.

$Number \ of \ moles \ in \ 4.50 \ dm^3 = \dfrac{2.25}{1.33} \times 4.50 \approx \mathbf{7.613 \, moles}$

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C ≈ 7.613 moles

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C is therefore;

1.33 dm³ contains 0.53 moles

$Number \ of \ moles \ in \ 4.50 \ dm^3 = \dfrac{0.53}{1.33} \times 4.50 \approx \mathbf{1.79 \, moles}$

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C ≈ 1.79 moles

The number of moles that precipitate out = The amount of solute deposited

Which gives;

Amount of solute deposited = 7.613 moles - 1.79 moles = 5.823 moles

The molar mass of Pb(NO₃)₂ = 207 g + 2 × (14 g + 3 × 16 g) = 331 g

The molar mass of Pb(NO₃)₂ = 331 g/mol

The amount of solute deposited = Number of moles × Molar mass

Which gives;

The amount of solute deposited = 5.823 moles × 331 g/mol = 1,927.413 g

Learn more about saturated solutions here:

brainly.com/question/2624685

5 0

3 years ago

Suppose of potassium sulfate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of potassium

dimulka [17.4K]

Answer:

This question is incomplete, here's the complete question:

"Suppose 0.0842g of potassium sulfate is dissolved in 50.mL of a 52.0mM aqueous solution of sodium chromate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium sulfate is dissolved in it. Round your answer to 2 significant digits."

Explanation:

Reaction :-

K2SO4 + Na2CrO4 ------> K2CrO4 + Na2SO4

Mass of K2SO4 = 0.0842 g, Molar mass of K2SO4 = 174.26 g/mol

Number of moles of K2SO4 = 0.0842 g / 174.26 g/mol = 0.000483 mol

Concentration of Na2CrO4 = 52.0 mM = 52.0 * 10^-3 M = 0.052 mol/L

Volume of Na2CrO4 solution = 50.0 ml = 50 L / 1000 = 0.05 L

Number of moles of Na2CrO4 = 0.05 L * 0.052 mol/L = 0.0026 mol

Since number of moles of K2SO4 is smaller than number of moles Na2CrO4, so 0.000483 mol of K2SO4 will react with 0.000483 mol of Na2CrO4 will produce 0.000483 mol of K2CrO4.

0.000483 mol of K2CrO4 will dissociate into 2* 0.000483 mol of K^+

Final concentration of potassium cation

= (2*0.000483 mol) / 0.05 L = 0.02 mol/L = 0.02 M

8 0

3 years ago

Calculate the molarity of a solution of sodium hydroxide, naoh, if 23.64 ml of this solution is needed to neutralize 0.5632 g of

GuDViN [60]

The molarity of NaOH needed is calculated as follows
calculate the moles of KhC8h4O4

that is moles = mass/molar mass of KhC8h4O4(204.22 g/mol)

=0.5632g /204.22g/mol= 2.76 x10^-3 moles

write the equation for reaction

khc8h4O4 + NaOH ---> KNaC8h4O4 + H2O

from the equation above the reacting ratio of KhC8h4O4 to NaOh is 1:1 therefore the moles of Naoh is also 2.76 x10^-3 moles

molarity of NaOh = (moles of NaOh / volume ) x 1000

that is { (2.76 x10^-3) / 23.64} x100 =0.117 M

8 0

3 years ago

How many grams of lead sulfide from when 10.0 g of lead are heated with 3.0 g of sulfur

zaharov [31]

The reaction forms 11.5 g PbS.

We have the masses of two reactants, so this is a limiting reactant problem.

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.

Step 1. Gather all the information in one place with molar masses above the formulas and everything else below them.

M_r: 207.2 32.06 239.28

Pb + S → PbS

Mass/g: 10.0 3.0

Step 2. Calculate the moles of each reactant

Moles of Pb = 10.0 g Pb × (1 mol Pb /207.2 g Pb) = 0.048 26 mol Pb

Moles of S = 30 g S × (1 mol S/32.06 g S) = 0.9357 mol S

Step 3. Identify the limiting reactant

Calculate the moles of PbS we can obtain from each reactant.

From Pb: Moles of PbS = 0.048 26 mol Pb × (1 mol PbS/1 mol Pb )

= 0.048 26 mol PbS

From S: Moles of S = 0.9357 mol S × (1 mol PbS/1 mol S) = 0.9357 mol PbS

Pb is the limiting reactant because it gives the smaller amount of PbS.

Step 4. Calculate the mass of PbS.

Mass = 0.048 26 mol PbS × (239.28 g PbS/1 mol PbS) = 11.5 g PbS

The reaction produces 11.5 g PbS.

4 0

3 years ago

If yttrium loses three electrons to be similar to Krypton (to try and become Royal), what type of ion does it become.

zhenek [66]

Answer:

Explanation:

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5 0

3 years ago