Mass NaCl = 4.348
Mass solution = 21.040 - 4.348 = 16.692
mass/mass % =
4.348/16.692
= 26%
mass/vol % =
21.7%
Molecular mass of NaCl = 58.5
Moles of NaCl = 0.074
Volume of solution = 0.02 L
Molarity = 3.7 mol/Liter
Answer:
189 Joules
Explanation:
Applying,
Q = cm(t₂-t₁)............. equation 1
Where Q = Heat, c = specific heat capacity of water, m = mass of water, t₁ = Initial Temperature, t₂ = Final temperature.
From the question,
Given: m = 15 grams = 0.015 kg, t₁ = 21 °C, t₂ = 24 °C
Constant: c = 4200J/kg.°C
Substitute these values into equation 1
Q = 0.015×4200×(24-21)
Q = 0.015×4200×3
Q = 189 Joules
Answer:
Please, see attached two figures:
- The first figure shows the solutility curves for several soluts in water, which is needed to answer the question.
- The second figure shows the reading of the solutiblity of NH₄Cl at a temperature of 60°C.
Explanation:
The red arrow on the second attachement shows how you must go vertically from the temperature of 60ºC on the horizontal axis, up to intersecting curve for the <em>solubility</em> of <em>NH₄Cl.</em>
From there, you must move horizontally to the left (green arrow) to reach the vertical axis and read the solubility: the reading is about in the middle of the marks for 50 and 60 grams of solute per 100 grams of water: that is 55 grams of grams of solute per 100 grams of water.
Assuming density 1.0 g/mol for water, 10 mL of water is:
Thus, the solutibily is:
