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Svetllana [295]
3 years ago
6

Which colligative property is employed when salt is put on an icy sidewalk to melt ice?

Chemistry
2 answers:
Iteru [2.4K]3 years ago
6 0
<span>For this number, we are to determine the colligative property that is affected by the addition of salt in the sidewalk in order to melt the ice. The answer is that the colligative property being displayed is the freezing point depression. Colligative properties are innate properties of substances and are only dependent on the amount of substance being added. </span>
ankoles [38]3 years ago
4 0
<span>The correct answer is 'freezing point depression'. Colligative properties depend on the concentration of molecules of a solute. Examples of other colligative properties are boiling point elevation or vapour pressure lowering. The salt causes ice on the side walk to melt because it lowers the freezing point. </span>
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Sort the phrases based on whether they describe or give an example of diffusion, facilitated diffusion, both, or neither. note:
Bingel [31]

Question:

a. Diffusion

b. Facilitated diffusion

c. Both

d. Neither

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2. movement across a membrane

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Answer:

The sorting is as follows

a. (1)

b. (5 and 6)

c. (1 and 2)

d. (4)

Explanation:

Diffusion is the movement of particles across a membrane from a high concentration region to one with a lower concentration of the diffusing substance

Here we have the correct sorting as follows

a. Diffusion

3. steroid transport into cell

b. Facilitated diffusion

5. movement assisted by proteins

6. glucose transport into cell

c. Both

1. movement to area of lower concentration

2. movement across a membrane

d. Neither

4. requires energy

8 0
3 years ago
A compound is 2.00% H by mass, 32.7% S by mass, and 65.3% O by mass. What is its empirical formula? The second step is to calcul
Free_Kalibri [48]
Lets take 100 g of this compound,
so it is going to be 2.00 g H, 32.7 g S and 65.3 g O.

2.00 g H *1 mol H/1.01 g H ≈ 1.98 mol H
32.7 g S *1 mol S/ 32.1 g S ≈ 1.02 mol S
65.3 g O * 1 mol O/16.0 g O ≈ 4.08 mol O

1.98 mol H : 1.02 mol S : 4.08 mol O = 2 mol H : 1 mol S : 4 mol O

Empirical formula
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