Answer:
34 g/100 mL
Explanation:
The solubility of a compound can be expressed in g/100mL, for this we must divide the mass of the compound that dissolves in the solute by the volume of the solvent.
The solvent, in this case, is water, and that mass of the solute X that dissolved is the mass that was recovered after the solvent was drained and evaporated. So the solubility of X (S) is:
S = 0.17 kg/5L
S = 170g/5000mL
S = 170g/(5*1000)mL
S = 34 g/100 mL
Raised temperature, decreased volume.
Temperature and Pressure are directly related, when volume increases so does the your pressure.
Volume and Pressure are indirectly related. When volume decreases, your pressure will increase.
Answer:
D
Explanation:
I believe the answer is D.
Answer:
H =2; I = 2; J = 2
Explanation:
Carbon is element 6 in the Periodic Table.
Start at element 1 (H) and count from left to right until you reach element 6 (C).
You get the electron configuration
C: 1s² 2s²2p².
Thus,
H =2; I = 2; J = 2
The boiling point of water at 1 atm is 100 degrees celsius. However, when water is added with another substance the boiling point of it rises than when it is still a pure solvent. This called boiling point elevation, a colligative property. The equation for the boiling point elevation is expressed as the product of the ebullioscopic constant (0.52 degrees celsius / m) for water), the vant hoff factor and the concentration of solute (in terms of molality).
ΔT(CaCl2) = i x K x m = 3 x 0.52 x 0.25 = 0.39 °C
<span> ΔT(Sucrose) = 1 x 0.52 x 0.75 = 0.39 </span>°C<span>
</span><span> ΔT(Ethylene glycol) = 1 x 0.52 x 1 = 0.52 </span>°C<span>
</span><span> ΔT(CaCl2) = 3 x 0.52 x 0.50 = 0.78 </span>°C<span>
</span><span> ΔT(NaCl) = 2 x 0.52 x 0.25 = 0.26 </span>°C<span>
</span>
Thus, from the calculated values, we see that 0.75 mol sucrose dissolved on 1 kg water has the same boiling point with 0.25 mol CaCl2 dissolved in 1 kg water.
