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Sladkaya [172]
2 years ago
12

What needs to happen for a solute to dissolve in a particular solvent

Chemistry
1 answer:
sergiy2304 [10]2 years ago
7 0
<span>the atractions between the solute and solvent molecules must be greater than the atractions keeping the solute together and the atractions keeping the solvent togetherrr.</span>
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Compared with the freezing-point depression of a 0.01 m c6h12o6 solution, the freezing-point depression of a 0.01 m nacl solutio
agasfer [191]

Answer:

Twice  as much.

Explanation:

That's because the freezing point depression depends on the total number of solute particles.

C₆H₁₂O₆(s) ⟶ C₆H₁₂O₆(aq)

0.01 mol of C₆H₁₂O₆ gives 0.01 mol of solute particles.

NaCl(s) ⟶ Na⁺(aq) + Cl⁻(aq)

1 mol of NaCl gives 0.01 mol of Na⁺(aq) and 0.01 mol of Cl⁻(aq).

That's 0.02 mol of particles, so the freezing point depression of 0.01 mol·L⁻¹ NaCl will be twice that of 0.01 mol·L⁻¹ C₆H₁₂O₆.

6 0
3 years ago
Distinguish between dunes and loess
agasfer [191]

Answer:

Dunes are loose sand hills.

Loess is a compacted wind-blown formation of sediment.

Explanation:

Sand dunes are, as their name implies, made of sand which is itself made of tiny quartz pieces. Sand dunes are present where there is a ready source of broken down sandstone or other type of rock and wind to transport the sand. The dunes are mobile and loose and migrate over time.

Loess comes in vast formations and was created in a very different way. In the last ice age, winds from the north carried sediments loosened and ground by glaciers southwards. There, the sediments accumulated in large formations and became compacted over time.

6 0
3 years ago
[24 points] A sample of soil has a total volume of 205 cm3. The soil mass when saturated is 361 g. A specific yield test was con
telo118 [61]

Answer:

Follows are the solution to the given question:

Explanation:

Dry Soil weight = solid soil weight = 284 \ grams

solid soil volume =205 \ cc

saturated mass soil = 361 \ g

The weight of the soil after drainage is =295 \ g

Water weight for soil saturation = (361-284) = 77 \ g

Water volume required for soil saturation =\frac{77}{1} = 77 \ cc

Sample volume of water: = \frac{\text{water density}}{\text{water density input}}

= 361- 295 \\\\ = 66 \ cc

Soil water retained volume = (draining field weight - dry soil weight)

                                             = 295 - 284 \\\\ = 11 \ cc.

\text{POROSITY}= \frac{\text{Vehicle volume}}{\text{total volume Soil}}

                    = \frac{77}{(205 + 77)}  \\\\= \frac{77}{(282)}  \\\\ = 27.30 \%

(Its saturated water volume is equal to the volume of voids)

\text{YIELD SPECIFIC} = \frac{\text{Soil water volume}}{\text{Soil volume total}}

                              = \frac{66}{(205+77)}\\\\= \frac{66}{(282)}\\\\=0.2340\\\\ = 0.23

\text{Specific Retention}= \frac{\text{Volume of soil water}}{\text{Total soil volume}}

                            = \frac{11}{282} \\\\= 0.0390 \\\\ = 0.04

6 0
2 years ago
Please help!!!!!! I need this question answered
salantis [7]
The answer would be A) cells!!
5 0
3 years ago
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A 30-kg girl and a 25-kg boy face each other on friction-free roller blades. the girl pushes the boy, who moves away at a speed
dsp73
0.83 m/s seems the correct answer, hope it helps
6 0
2 years ago
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