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fgiga [73]
3 years ago
13

The mass of a salt in grams that will dissolve in 100 mL of water.

Chemistry
1 answer:
lys-0071 [83]3 years ago
7 0

Answer:

Solubility  : The mass of a salt in grams that will dissolve in 100 mL of water

Saturated solution : A solution that has dissolved the maximum amount of a compound at a given temperature. Any further addition of salt will remain undissolved

Solubility product constant :The product of the molarities of the dissolved ions, raised to a power equal to the ion's coefficient in the balanced chemical equation

Common ion effect : A decrease in the solubility of an ionic compound as a result of the addition of a common ion

Molar solubility : The maximum number of moles of a salt that will dissolve in 1 L of solution

Explanation:

Solubility is expressed usually as the mass of solute per 100 grams or 100 ml of solvent.

Molar solubility of a solid is expressed as the concentration of the dissolved solid in a saturated solution.

Solubility product constant is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as  

The common-ion effect is the decrease in solubility of a sparingly soluble salt by the addition to the solution of a soluble compound with an ion in common with the precipitate.

Saturated solution is defined as the solution in which no more solute particles can be dissolved in the solvent.

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The theory of evolution was proposed by Darwin. 
4 0
4 years ago
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A 50.0 mg sample of iodine-131 was placed in a container 32.4 days ago. if its half-life is 8.1 days, how many milligrams of iod
sertanlavr [38]

3.124mg of I-131 is present after 32.4 days.

The 131 I isotope emits radiation and particles and has an 8-day half-life. Orally administered, it concentrates in the thyroid, where the thyroid gland is destroyed by the particles.

What is Half life?

The time required for half of something to undergo a process: such as. a : the time required for half of the atoms of a radioactive substance to become disintegrated.

Half of the iodine-131 will still be present after 8.1 days.

The amount of iodine-131 will again be halved after 8.1 additional days, for a total of 8.1+8.1=16.2 days, reaching (1/2)(1/2)=1/4 of the initial amount.

The quantity of iodine-131 will again be halved after 8.1 more days, for a total of 16.2+8.1+8.1=32.4 days, to (1/4)(1/2)(1/2)=1/16 of the initial quantity.

If the original dose of iodine-131 was 50mg, the residual dose will be (50mg)*(1/16)=3.124mg after 32.4 days.

Learn more about the Half life of radioactie element with the help of the given link:

brainly.com/question/27891343

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7 0
1 year ago
For the chemical reaction CaI2+2AgNO3-> 2AgI+Ca(NO3)2 how many moles of silver iodide will be produced from 205g of calcium i
Colt1911 [192]

Answer:

Moles of silver iodide produced = 1.4 mol

Explanation:

Given data:

Mass of calcium iodide = 205 g

Moles of silver iodide produced = ?

Solution:

Chemical equation:

CaI₂ + 2AgNO₃     →      2AgI + Ca(NO₃)₂

Number of moles calcium iodide:

Number of moles = mass/ molar mass

Number of moles = 205 g/ 293.887 g/mol

Number of moles = 0.7 mol

Now we will compare the moles of calcium iodide with silver iodide.

                     CaI₂         :           AgI

                         1           :             2

                       0.7         :           2×0.7 = 1.4

Thus 1.4 moles of silver iodide will be formed from 205 g of calcium iodide.

6 0
3 years ago
Does anybody know how to do q4. Please show working out thanks.
insens350 [35]

Answer:

% purity of limestone = 96.53%

Explanation:

Question (4).

Weight of impure CaCO₃ = 25.9 g

Molecular weight of CaCO₃ = 40 + 12 + 3(16)

                                              = 100 g per mole

We know at S.T.P. number of moles of CO₂ = 1 and volume = 22.4 liters

From the given reaction, 1 mole of CaCO₃ reacts with 1 mole or 22.4 liters of

CO₂.

∵ 22.4 liters of CO₂ was produced from CaCO3 = 100 g

∴ 1 liter of CO₂ will be produced by CaCO₃ = \frac{100}{22.4}

∴ 5.6 liters of CO₂ will be produced by CaCO₃ = \frac{100\times 5.6}{22.4}

                                                                              = 25 g

Therefore, % purity of CaCO₃ = \frac{\text{Weight calculated}}{{\text{Weight given}}}\times 100

                                                 = \frac{25}{25.9}\times 100

                                                 = 96.53 %

7 0
4 years ago
Heat energy that is lost cannot usually be used to produce work.<br><br> True False
Veseljchak [2.6K]

Answer:

true

Explanation:

thanks for free point

7 0
3 years ago
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