The saturation level is only nominally dependent on the temperature of the water. At 20 °C one liter of water can dissolve about 357 grams of salt, a concentration of 26.3% w/w. At boiling (100 °C) the amount that can be dissolved in one liter of water increases to about 391 grams, a concentration of 28.1% w/w.
<h3>How do you calculate the solubility of salt in water?</h3>
Divide the mass of the compound by the mass of the solvent and then multiply by 100 g to calculate the solubility in g/100g .
<h3>How do you calculate the concentration of salt in water?</h3>
Salt is the solute (the dissolving substance), and water is the solvent (the substance that dissolves another to create a solution). To make a salt solution by weight percent (w/v), you apply the formula w/v = (mass of solute ÷ volume of solution) × 100.
Learn more about solubility here:
<h3>
brainly.com/question/23946616</h3><h3 /><h3>#SPJ4</h3>
Answer:
ΔH rx = -43.5 kJ / mol
Explanation:
In water, Xdissolves thus:
X(s) + H₂O(l) → X(aq) + H₂O(aq)
It is possible to find the heat in dissolution process using coffee cup calorimeter equation:
Q = -m×C×ΔT
<em>Where Q is heat, m is mass of solution (35.0g -density 1g/mL- + 2.20g = 37.2g), C is specific heat of solution (4.18J/g°C), and ΔT is change in temperature (26.0°C-15.0°C = 11.0°C)</em>
Replacing:
Q = -37.2g×4.18J/g°C×11.0°C
Q = -1710J = -<em>1.71kJ</em>
As enthalpy is the change in heat per mole of reaction, moles of X that reacted were:
2.20g X × (1mol / 56.0g) = <em>0.0393 moles</em>
As heat produced per 0.0393moles was -1.71kJ, heat per mole of X is:
-1.71kJ / 0.0393mol = -<em>43.5 kJ / mol = ΔH rx</em>
Answer:
x=170
The question is asking you to add them together to get your unknown number
Answer:
The <u>transition metals</u><u> </u>form colored ions.
Explanation:
hope it helps:)
