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Travka [436]
3 years ago
10

Which of the solutions have greatest osmotic pressure 30% sucrose 60% sucrose or 30% magnesium sulfate?

Chemistry
1 answer:
timofeeve [1]3 years ago
8 0

Answer:Osmotic pressure is the minimum amount of pressure a solution must exert in order to prevent from crossing a barrier by osmosis. Solute molecules have difficulty crossing semipermeable membranes, so the more solutes that are in a solution, the higher the osmotic pressure will be. Between 30% sucrose and 60% sucrose, 60% sucrose will have a greater osmotic pressure than 30% because it has a higher percentage of solutes. However, since sucrose has a higher potential to cross semipermeable membranes and is more absorbable than magnesium sulfate, magnesium sulfate would have a higher osmotic pressure than 60% sucrose even though 60% sucrose has higher molecules.

Explanation:

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What is the wavelength of a wave having a frequency of 2500 Hz
Lemur [1.5K]
We will assume that the wave is moving in free space, thus, the velocity of the wave would be equal to the speed of light = 3 * 10^8 meters/sec

The velocity of the wave can be calculated using the following rule:
velocity = frequency * wavelength where:
velocity = 3 * 10^8 m/sec
frequency = 2500 Hz
wavelength is the unknown that we want to calculate.

Substitute with the givens in the above equation to get the wavelength as follows:
3 * 10^8 = 2500*wavelength
wavelength = 120 * 10^3 meters
5 0
2 years ago
Read 2 more answers
How many grams of sodium phosphate monobasic would we add to a liter and how many grams of sodium phosphate dibasic would we add
Nostrana [21]

Answer :

The correct answer   for Mass of Na₂HPO₄ = 4.457 g and mass of  NaH₂PO₄  = 8.23 g

Given :  pH = 6.86

Total concentration of Phosphate buffer = 0.1 M

Asked : Mass of  Sodium phosphate monobasic (NaH₂PO₄) = ?

Mass of  Sodium phosphate dibasic(Na₂HPO₄)= ?

Following steps can be done to find the masses of NaH₂PO₄ and Na₂HPO₄ :

(In phosphate buffer , Na+ ion from  NaH₂PO₄ and Na₂HPO₄ acts as spectator ion , so only H₂PO₄⁻ and HPO₄²⁻ will be considered )

<u>Step 1 : To find pka </u>

H₂PO₄⁻  <=> HPO₄²⁻  

The above reaction has pka = 7.2 ( from image shown )

<u>Step 2 : Plug values in Hasselbalch- Henderson equation </u>.

Hasselbalch -Henderson equation is to find pH  for buffer solution which is as follows :

pH = pka + log\frac{[A^-]}{[HA]}

pH = 6.86         pKa = 7.2

6.86 = 7.2 + log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}

Subtracting  both side by 7.2

6.86-7.2 = 7.2 -7.2+ log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}

-0.34 =  log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}

Removing log

10^-^0^.^3^4 =   \frac{[HPO_4^2^-]}{[H_2PO_4^-]}

\frac{[HPO_4^2^-]}{[ H_2PO_4^-]} = 0.457 ---------------- equation (1)

<u>Step 3 : To find  molarity of H₂PO₄⁻ and HPO₄²⁻</u>

Total concentration of buffer = [H₂PO₄⁻] + [HPO₄²⁻] = 0.1 M

Hence,  [H₂PO₄⁻ ] + [ HPO₄²⁻ ] =  0.1 M

Assume [H₂PO₄⁻ ] = x

So ,  [x ] + [ HPO₄²⁻ ] =  0.1 M

[ HPO₄²⁻ ] =  0.1 - x

Step 4 : Plugging value of [H₂PO₄⁻ ]  and  [ HPO₄²⁻ ]

[H₂PO₄⁻ ]  = x

 [ HPO₄²⁻ ] = 0.1 - x

Equation (1) = >\frac{[HPO_4^2^-]}{[ H_2PO_4^-]} = 0.457

Plug value of [H₂PO₄⁻ ]  and  [ HPO₄²⁻ ] ( from step 3 ) into equation (1)  as :

\frac{[0.1 - x ]}{[ x]} = 0.457

Cross multiplying

0.1 - x  = 0.457 x

Adding x on both side

0.1 -x + x = 0.457 x + x

0.1  = 1.457 x

Dividing both side by 1.457

\frac{0.1}{1.457} = \frac{1.457 x }{1.457}

x = 0.0686 M

Hence , [H₂PO₄⁻ ]  = x  = 0.0686 M

 [ HPO₄²⁻ ] = 0.1 - x

 [ HPO₄²⁻ ]  =   0.1 - 0.0686  

[ HPO₄²⁻ ] = 0.0314 M

Step 5 : To find moles of  H₂PO₄⁻ ( NaH₂PO₄) and HPO₄²⁻ (Na₂HPO₄ ) .

Molarity is defined as mole of solute per 1 L volume of solution .

Molarity of NaH₂PO₄ = 0.0686 M  or 0.0686 mole per 1 L

Molarity of Na₂HPO₄ = 0.0314 M  or 0.0314 mole per 1 L

Since  that volume of buffer solution  is 1 L , so Molarity  = mole

Hence Mole of NaH₂PO₄  = 0.0686 mol

Mole of Na₂HPO₄ = 0.0314 mol

<u>Step 6 : To find mass  of Na₂HPO₄  and NaH₂PO₄ </u>

Moles of  Na₂HPO₄  and NaH₂PO₄  can be converted to their masses using molar mass as follows :

Molar mass of  Na₂HPO₄  = 141.96 \frac{g}{mol}

Molar mass of NaH₂PO₄ = 119.98 \frac{g}{mol}

Mass (g) = mole (mol)* molar mass(\frac{g}{mol})

Mass of Na_2HPO_4 = 0.0314 mol * 141.96 \frac{g}{mol}

Mass of Na₂HPO₄ = 4.457 g

Mass of NaH_2PO_4 = 0.0686 mol * 119.98 \frac{g}{mol}

Mass of  NaH₂PO₄  = 8.23 g

5 0
2 years ago
What term defines a compound which contains a carbonyl bonded to an -OH group?
natka813 [3]

I think that its a ketone.

3 0
2 years ago
In a chemical reaction such as photosynthesis,<br>are the substances present before the reaction​
sashaice [31]

Yes

Explanation:

In a chemical reaction such as photosynthesis, substances are present before the reaction starts.

A chemical reaction is usually represented by simple chemical equations that shows the reactants and the products.

  • The reactants are the compounds combining together at the beginning of the reaction.
  • For photosynthesis, the reactants are carbon dioxide and water.
  • The products are the results obtained from the combining reactants.
  • In photosynthesis, they are glucose and oxygen gases.

Learn more:

Chemical reaction brainly.com/question/3953793

#learnwithBrainly

3 0
3 years ago
An 18.0 g piece of an unidentified metal was heated from 21.5 °C to 89.0 °C. If 789.75 J of heat energy was absorbed by the meta
nydimaria [60]

Answer: The metal is Calcium.

Explanation:

To calculate the specific heat of substance during the reaction.

q=m\times c\times \Delta T

where,

q = heat absorbed = 789.75 J

c = specific heat of metal = ?

m = mass of substance = 18.0 g

\Delta T_f = final temperature  - initial temperature  = (89.0-21.5)^0C=67.5^0C

Now put all the given values in the above formula, we get:

789.75J=18.0g\times c\times 67.5^0C

c=0.65J/g^0C

As specific heat is characteristic of each metal and thus the metal is calcium which has  specific heat of 0.65J/g^0C

5 0
2 years ago
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