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Igoryamba
2 years ago
15

Potassium sulfate has a solubility of 15g/100g water at 40 Celsius. A solution is prepared by adding 39.0g of potassium sulfate

to 225g water, carefully heating the solution, and cooling it to 40 Celsius. A homogeneous solution is obtained. Is this solution saturated, unsaturated, or supersaturated? The beaker is shaken and precipitation occurs. How many grams of potassium sulfate would you except to crystallize out?please help me to understant this
Chemistry
1 answer:
nadya68 [22]2 years ago
8 0

Answer:

5.25 grams of potassium sulfate will get crystallize out.

Explanation:

Solubility of potassium sulfate at 40 °C = 15 g/100 g

This means that at 40 °C 15 g of potassium sulfate will get completely dissolved in 100 of water.

39.0 g of potassium sulfate to 225 g water, carefully heating the solution.

Amount of potassium sulphate will get dissolve in 225 g of water at 40 °C will be:

\frac{15g}{100g} × 225 = 33.75g

Amount of potassium sulfate precipitated out by the solution:

= 39.0 g-33.75 g = 5.25 g

At 40 °C 5.25 g of potassium sulfate will get precipitate out from the solution which means that solution is saturated.

Saturated solution are solution in which solute is dissolved in maximum amount. Further addition of solute results in precipitation of solute form the solution.

5.25 grams of potassium sulfate will get crystallize out.

(Hope this Helps can I pls have brainlist (crown)☺️)

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<span>0.0165 m The balanced equation for the reaction is AgNO3 + MgCl2 ==> AgCl + Mg(NO3)2 So it's obvious that for each Mg ion, you'll get 1 AgCl molecule as a product. Now calculate the molar mass of AgCl, starting with looking up the atomic weights. Atomic weight silver = 107.8682 Atomic weight chlorine = 35.453 Molar mass AgCl = 107.8682 + 35.453 = 143.3212 g/mol Now how many moles were produced? 0.1183 g / 143.3212 g/mol = 0.000825419 mol So we had 0.000825419 moles of MgCl2 in the sample of 50.0 ml. Since concentration is defined as moles per liter, do the division. 0.000825419 / 0.0500 = 0.016508374 mol/L = 0.016508374 m Rounding to 3 significant figures gives 0.0165 m</span>
5 0
3 years ago
Read 2 more answers
Can someone please help I need help?!?!
anzhelika [568]

Answer:

a: 6 moles

b: 6.75 moles

c: 5 grams

Explanation:

a: mole ratio 2:3

b: mole ratio 2:3

c: mole ratio 2:2

7 0
3 years ago
How do I find the number of moles and molar mass of Gas A and Gas B with the information provided?
drek231 [11]

Answer:

A.

Explanation:

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3 0
2 years ago
How many liters of hydrogen gas will be produced at STP from the reaction of 7.179 x 1023 atoms of magnesium with 54.21 g of pho
wlad13 [49]
<h3>Answer:</h3>

18.58 liters of hydrogen gas

<h3>Explanation:</h3>

We are given;

  • The equation;

3Mg + 2H₃(PO₄) → Mg₃(PO₄)₂ + 3H₂

  • Atoms of Magnesium = 7.179 x 10^23 atoms
  • Mass of phosphoric acid as 54.21 g

We are required to determine the volume of hydrogen gas produced;

Step 1; moles of Magnesium

1 mole of an element contains 6.02 × 10^23 atoms

therefore;

Moles of Mg = (7.179 x 10^23 ) ÷ (6.02 × 10^23)

                   = 1.193 moles

Step 2: Moles of phosphoric acid

moles = Mass ÷ Molar mass

Molar mass of phosphoric acid = 97.994 g/mol

Therefore;

Moles of Phosphoric acid = 54.21 g ÷ 97.994 g/mol

                                           = 0.553 moles

Step 3: Determine the rate limiting reagent

From the mole ratio of Mg to Phosphoric acid (3 : 2);

1.193 moles of magnesium requires 0.795 moles of phosphoric acid while,

0.0553 moles of phosphoric acid requires 0.8295 moles of Mg

Therefore, phosphoric acid is the rate limiting reagent

step 4: Determine the moles of hydrogen produced

From the equation, w moles of phosphoric acid reacts to produce 3 moles of hydrogen;

Therefore; moles of Hydrogen = moles of phosphoric acid × 3/2

                                                   = 0.553 moles × 3/2

                                                   = 0.8295 moles

Step 5: Volume of hydrogen gas

1 mole of a gas occupies a volume of 22.4 liters at STP

Therefore;

Volume of Hydrogen = 0.8295 moles × 22.4 L/mol

                                  = 18.58 Liters

Therefore; 18.58 liters of hydrogen gas  will be produced

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lutik1710 [3]

Answer:

1.05 V

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From Nernst's equation;

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Since log 1=0

Ecell= E°cell= 1.05 V

3 0
2 years ago
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