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3 years ago

Difference between hygroscopic and deliquescent substance with of each

Chemistry

2 answers:

iVinArrow [24]3 years ago

7 0

Answer:

The water soluble substance which absorb moisture from the air and then dissolve on the absorbed moisture to change into liquid taste are called deliquescent substances whereas the substances which absorb moisture from air but do not change their state are called hygroscopic substances.

melamori03 [73]3 years ago

5 0

Answer:

Hygroscopic substances absorb moisture from the air but do not dissolve in it, whereas substance that undergo deliquescence dissolves in the water vapour that is absorbed from the air, forming a liquid solution

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Un tecnico di laboratorio deve preparare una soluzione di carbonato di sodio decaidrato, Na2CO3⋅ 10 H2O per eseguire alcune anal

fiasKO [112]

Answer:

1. 3.70 g Na₂CO₃·10H₂O

2. 50.0 mL of the first solution

Explanation:

1. Prepare the solution

(a) Calculate the molar mass of Na₂CO₃·10H₂O

$\begin{array}{rrr}\textbf{Atoms} &\textbf{M}_{\textbf{r}} & \textbf{Mass/u}\\\text{2Na} & 2\times22.99 & 45.98\\\text{1C} & 1\times 12.01 & 12.01\\\text{13O}&13 \times16.00 & 208.00\\\text{20H}&20 \times 1.008 & 20.16\\&\text{TOTAL =} & \mathbf{286.15}\\\end{array}$

The molar mass of Na₂CO₃·10H₂O is 286.15 g/mol.

(b) Calculate the moles of Na₂CO₃·10H₂O

$\text{Moles of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}\\= \text{0.250 L solution} \times \dfrac{\text{0.0500 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}{\text{1 L solution}}\\\\= \text{0.0125 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}$

(c) Calculate the mass of Na₂CO₃·10H₂O

$\text{Mass of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O }\\= \text{0.012 50 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O } \times \dfrac{\text{296.15 g Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}{\text{1 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}\\\\= \text{3.70 g Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}\\\text{You need $\large \boxed{\textbf{3.70 g}}$ of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}$

2. Dilute the solution

We can use the dilution formula to calculate the volume needed.

V₁c₁ = V₂c₂

Data:

V₁ = ?; c₁ = 0.0500 mol·L⁻¹

V₂ = 100 mL; c₂ = 0.0250 mol·L⁻¹

Calculation:

$\begin{array}{rcl}V_{1}c_{1} & = & V_{2}c_{2}\\V_{1}\times \text{0.0500 mol/L} & = & \text{100 mL} \times\text{0.0250 mol/L}\\0.0500V_{1}& = & \text{2.500 mL}\\V_{1}&=& \dfrac{\text{2.500 mL}}{0.0500}\\\\& = & \text{50.0 mL}\\\end{array}\\\text{You need $\large \boxed{\textbf{50.0 mL}}$ of the concentrated solution.}$

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2 years ago

How would you prepare the following aqueous solutions?

Xelga [282]

The molarity of the solution is 0.1625M.

<h3>What do you meant by Molarity</h3>

Molarity is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of solute per litres of a solution .

Molarity → no of moles / Volume (L)

SI unit of Molarity is M or mol/ L

We have given here the mass of solution is 3.10×10²g .

molality of the solution is 0.125m

Molality → no of moles / mass in kg

→ 0.125×3.10×10²/ 1000

→ no.of moles = 0.0162

For molarity we can assume volume as 1000 ml .

Molarity = 0.0162×1000/ 100

Molarity →0.162 M.

So, the molarity of solution will be 0.162M.

to Learn more about Molarity click here brainly.com/question/8732513

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1 year ago

A 10 gram sample of material, a, was placed in water. four grams of it, b, did not dissolve, even after being separated and plac

kicyunya [14]

Sksksk isjsjs jsjsjs jsjsjs jsjsjs jsjsjs ksjss

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2 years ago

13. An organic compound is found to contain 77.42% of C, 7.53% of H and

ivann1987 [24]

Answer

7.53% 97% if you divide it you can get the answer

Explanation:

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2 years ago

The chemical formula for artificial sweetener is C7H5NO3S. How many carbon atoms will be found in 5 molecules of the artificial

irga5000 [103]

35

Because there’s 7 carbon atoms in every molecule of artificial sweetener

And if you have 5 molecules of that

7x5 =35

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2 years ago

Difference between hygroscopic and deliquescent substance with of each​

Difference between hygroscopic and deliquescent substance with of each