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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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A is correct hope this helps
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madreJ [45]

Answer:

Non-polar covalent bond

Explanation:

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3 years ago
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How much ice at a temperature of -17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.
Tanya [424]

Answer:

An insulated beaker with negligible mass contains liquid water with a mass of 0.205kg and a temperature of 79.9 °C How much ice at a temperature of −17.5 °C must be dropped into the water so that the final temperature of the system will be 31.0 °C? Take the specific heat for liquid water to be 4190J/Kg.K, the specific heat for ice to be 2100J/Kg.K, and the heat of fusion for water to be 334000J/kg.

The answer to the above question is

Therefore 0.1133 kg  ice at a temperature of -17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C

Explanation:

To solve this we proceed by finding the heat reaquired to raise the temperature of the water to 31.0 C from 79.9 C then we use tht to calculate for the mass of ice as follows

ΔH = m×c×ΔT

= 0.205×4190×(79.9 -31.0) = 42002.655 J

Therefore fore the ice, we have

Total heat = mi×L + mi×ci×ΔTi = mi×334000 + mi × 2100 × (0 -−17.5) = 42002.655 J

370750×mi = 42002.655 J

or mi = 0.1133 kg

Therefore 0.1133 kg  ice at a temperature of -17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C

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a nebula is a gaseous cloud formed from dead stars. this works as a nursery for baby stars.

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2 years ago
When multiplying 2.3mm by 8.00mm to find area, how would you report your answer
scZoUnD [109]

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8.00mm: there are three significant numbers

so we will report the answer to minimum number of significant number , which is two

the answer is 18.4

We will report it to be 18 mm^2

6 0
3 years ago
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