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MatroZZZ [7]
3 years ago
9

The monovalent salt concentration (the predominant solute in the blood cell) for a sample of red blood cells is 0.13 moles/liter

. If one of these red blood cells were placed in pure water (at around room temperature, 300 K), and the cell comes to hydrostatic equilibrium with the water, what is the osmotic pressure of the cell (assuming it doesn't burst)
Chemistry
1 answer:
11111nata11111 [884]3 years ago
4 0

Answer:

The osmotic pressure of cell is 648.3 KPa

Explanation:

As we know the osmotic pressure is equal to

\pi = icRT

Where

i is the Van Hoff factor

c is the concentration of solution

R is the ideal gas constant

and T is the temperature.

Substituting the given values, we get -

\pi = 2 * 0.13 * 0.08206 * 300\\

\pi = 648.3 KPa

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If 2.0 mol of Zn is mixed with 3.0 mol<br> of HCl, which reactant will be limiting?
Soloha48 [4]

The following Balanced Reaction will take place:

Zn + 2HCl → ZnCl₂ + H₂

In the question, we have 2 moles of Zinc and 3 moles of HCl for this reaction

<u>Amount of HCl required to completely react with 2 moles of Zn:</u>

Since we need 2 moles of HCl for every mole of Zn, we will need 2(2) = 4 moles of HCl for every 2 moles of Zn

<u>Identifying the Limiting Reagent:</u>

But we are only given 3 moles of HCl where we need 4 moles to completely react.

So, since HCl is in less amount, it is the Limiting Reagent

8 0
3 years ago
For the following reaction, 38.3 grams of sulfuric acid are allowed to react with 33.5 grams of calcium hydroxide sulfuric acid(
Likurg_2 [28]

Answer:

What is the maximum amount of calcium sulfate that can be formed? 53.1 grams CaSO4

What is the FORMULA for the limiting reagent? H2SO4

What amount of the excess reagent remains after the reaction is complete? 4.59 grams of Ca(OH)2

Explanation:

Step 1: Data given

Mass of sulfuric acid = 38.3 grams

Molar mass of H2SO4 = 98.08 g/mol

Mass of calcium hydroxide = 33.5 grams

Molar mass of Ca(OH)2 = 74.09 g/mol

Step 2: The balanced equation

H2SO4 + Ca(OH)2 → CaSO4 + 2H2O

Step 3: Calculate moles of H2SO4

moles H2SO4 = mass H2SO4 / molar mass H2SO4

moles H2SO4 = 38.3 grams / 98.08 g/mol

moles H2SO4 = 0.390 moles

Step 4: Calculate moles of Ca(OH)2

moles Ca(OH)2 = 33.5 grams / 74.09 g/mol

moles Ca(OH)2 =0.452 moles

Step 5: Calculate limiting reactant

For 1 mol H2SO4, we need 1 mol of Ca(OH)2 to produce, 1 mol of CaSO4 and 2 mol of H2O

H2SO4 is the limiting reactant. It will completely be consumed (0.390 moles).

Ca(OH)2 is in excess. There will be consumed 0.390 moles

There will remain 0.452 - 0.390 = 0.062 moles

This is 0.062 * 74.09 g/mol = 4.59 grams

Step 6: Calculate moles of calcium sulfate

For 1 mol H2SO4, we need 1 mol of Ca(OH)2 to produce, 1 mol of CaSO4 and 2 mol of H2O

For 0.390 moles of H2SO4, there will be produced 0.390 moles of CaSO4

Step 7: Calculate mass of CaSO4

Mass CaSO4 = moles CaSO4 * molar mass CaSO4

Mass CaSO4 = 0.390 moles * 136.14 g/mol

Mass of CaSO4 = 53.1 grams

7 0
3 years ago
Calculate the number of moles found in 3.045x1024 atoms of helium.<br><br> PLS HELP
Kisachek [45]

Explanation:

so for this u have to use this equation where

Moles = number of particle/6.02×10^23

= 3.045 × 10^24/6.02×10^23

= 5.0581

write it to 3 S.F so 5.06 moles

4 0
2 years ago
g When copper(II) chloride and sodium carbonate solutions are combined, solid copper(II) carbonate precipitates, leaving a solut
Elden [556K]

Answer:

When copper(II) chloride and sodium carbonate solutions are combined, solid copper(II) carbonate precipitates, leaving a solution of sodium chloride. Write the conventional equation, total ionic equation, and net ionic equation for this reaction.

Explanation:

The word equation for the reaction is:

Copper (II) chloride(aq) + sodium carbonate (aq) ->sodium chloride (aq) +           copper carbonate(s)

The balanced chemical equation of the reaction is:

CuCl_2(aq)+Na_2CO_3(aq)->2NaCl(aq)+CuCO_3(s)

The complete ionic equation is:

Cu^2+(aq)+2Cl^-(aq)+2Na^+(aq)+CO_3^2^-(aq)->2Cl^-(aq)+2Na^+(aq)+CuCO_3(s)\\

The net ionic equation is obtained from the complete ionic equation after removing the spectator ions:

Cu^2^+(aq)+CO_3^2^-(aq)->CuCO_3(s)

5 0
3 years ago
Calculate the volume in liters of a 0.00231M copper(II) fluoride solution that contains 175.g of copper(II) fluoride CuF2. Be su
Free_Kalibri [48]

Answer:

Volume = 746 L

Explanation:

Given that:- Mass of copper(II) fluoride = 175 g

Molar mass of copper(II) fluoride = 101.543 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{175\ g}{101.543\ g/mol}

Moles_{copper(II)\ fluoride}= 1.7234\ mol

Also,

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

So,,

Volume =\frac{Moles\ of\ solute}{Molarity}

Given, Molarity = 0.00231 M

So,

Volume =\frac{1.7234}{0.00231}\ L

<u>Volume = 746 L</u>

7 0
3 years ago
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