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photoshop1234 [79]
3 years ago
5

What is the melting point of a solution in which 2.5 grams of sodium chloride is added to 230 mL of water?

Chemistry
1 answer:
Lemur [1.5K]3 years ago
5 0
To get the melting point of a solution so, we will use this formula:

ΔT = - mKf

when:

m is molality of the solution

Kf is cryoscopic constant  of water  = 1.86 C/m

and ΔT is the change in melting point (T2 - 0 °C)

so, now we need to calculate the molality to substitute:

when the molality = moles NaCl / Kg of water

 and when moles NaCl  = mass / molar mass

                                       = 2.5 g / 58 g/mol

                                       = 0.043 mol
∴ Kg water = volume *density /1000

                   = 230 mL * 1 g/mL / 1000

                  = 0.23 Kg
∴ molality = 0.043 / 0.23 =0.187 M

by substitution:

T2-0°C = - 0.187 * 1.86

∴T2 = - 0.348 °C

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Which equation is balanced?
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8 0
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what volume of co2 is produced at stp when 270g of glucose are consumed in the following reaction? c6h12o6 + 6o2(g) -> 6co2 (
lana [24]

Answer:

202 L

Explanation:

Step 1: Write the balanced equation

C₆H₁₂O₆ + 6 O₂(g) ⇒ 6 CO₂(g) + 6 H₂O(l)

Step 2: Calculate the moles corresponding to 270 g of C₆H₁₂O₆

The molar mass of C₆H₁₂O₆ is 180.16 g/mol.

270 g × 1 mol/180.16 g = 1.50 mol

Step 3: Calculate the moles of CO₂ generated from 1.50 moles of glucose

The molar ratio of C₆H₁₂O₆ to CO₂ is 1:6. The moles of CO₂ formed are 6/1 × 1.50 mol = 9.00 mol

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4 0
3 years ago
1. Draw a wedge/dash structure for trans-1,2-dimethylcyclohexane.
Angelina_Jolie [31]

Answer:

The structure with the ring flipped is the most stable

Explanation:

We have the  trans 1,2 - dimethylcyclohexane. With the wedge/dash structure we could not figure is this form is stable (If we do a comparison with the cis structure). But when we do a chair structure and ring flipped structure, this is easier to look.

The picture attached shows the structures, they are labeled as 1, 2 and 3, according to this problem.

In the chair structure, according to the picture below, you can see that both methyls are heading in the axial positions of the ring (One facing upward and the other downward). This is pretty stable, however, when the methyls are in those positions, the methyl position 1, can undergoes an 1,3 diaxial interactions with the hydrogens atoms (They are not drawn, but still are there), so this interaction makes this structure a little less stable that it can be.

On the other side, the ring flipped structure, we can see that both methyls are in the equatorials positions of the ring, and in these positions, it can avoid the 1,4 diaxial interactions with the hydrogens atoms, making this structure the most stable structure.

Hope this helps

6 0
3 years ago
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