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Diano4ka-milaya [45]
3 years ago
7

If the freezing point of an aqueous 0.10 m glucose solution is −x°c, what is the approximate freezing point of a 0.10 m nacl sol

ution?
Chemistry
1 answer:
maks197457 [2]3 years ago
8 0
Answer is: the approximate freezing point of a 0.10 m NaCl solution is -2x°C.
V<span>an't Hoff factor (i) for NaCl solution is approximately 2.
</span>Van't Hoff factor (i) for glucose solution is 1.<span>
Change in freezing point from pure solvent to solution: ΔT = i · Kf · m.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
m -  molality, moles of solute per kilogram of solvent.
</span>Kf and molality for this two solutions are the same, but Van't Hoff factor for sodium chloride is twice bigger, so freezing point is twice bigger.
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If I have 12 protons, how many neutrons and electrons would I need to make a stable atom?
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The two rate statements below are not the same. Use the concept of unit rates to explain
olganol [36]
It’s definitely B but I’m not sure
8 0
3 years ago
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balu736 [363]

Explanation:

Given -

  • An organic compound gives H₂ gas with Na
  • On treatment with alkaline iodine it gives yellow ppt.
  • On oxidation with CrO₃/H⁺ forms an aldehyde (C₂H₄O)

To Find -

  • Name the compound and write the reaction involved

Now,

Let A be the organic compound.

Then,

  1. A + Na → + H₂↑
  2. A + I₂ → CHI₃ (yellow ppt.)
  3. A + CrO₃ + H⁺ → C₂H₄O

Now,

Here we see that compound A reacts with chromic oxide (CrO₃) in the presence of acidic medium gives aldehyde.

  • Functional group of aldehyde = —CHO

And It forms only 2 Carbon aldehyde it means, It is Ethanal (CH₃CHO).

Compound A reacts with chromic oxide (CrO₃) in the presence of acidic medium gives ethanal.

It means,

We know that 1° alcohol on oxidation gives aldehyde.

Here it gives 2 Carbon aldehyde.

It means,

Here 2 Carbon and 1° alcohol is used.

Now,

Its cleared that Compound A is Ethanol.

Reaction Involved -

  1. CH₃CH₂OH + Na → CH₃CH₂O⁻Na⁺ + H₂↑
  2. CH₃CH₂OH + I₂ + OH⁻ → CHI₃↓ + HCOO⁻ + HI + H₂O
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6 0
3 years ago
The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that
kakasveta [241]

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is 1.1\times 10^6 M.

HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)

This value indicates that

A. CN^- is a stronger base than ONO^-

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is ONO^-

D. The conjugate acid of CN- is HCN

Answer: A. CN^- is a stronger base than ONO^-

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When K_{p}>1; the reaction is product favoured.

When K_{p}; ; the reaction is reactant favored.

When K_{p}=1; the reaction is in equilibrium.

As, K_p>>1, the reaction will be product favoured and as it is a acid base reaction where HONO acts as acid by donating H^+ ions and CN^- acts as base by accepting H^+

Thus HONO is a strong acid thus ONO^- will be a weak conjugate base and CN^- is a strong base which has weak HCN conjugate acid.

Thus the high value of K indicates that CN^- is a stronger base than ONO^-

7 0
3 years ago
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