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leva [86]
2 years ago
14

In a coffee shop, the following coffee samples are prepared for customers. i. 10 g of sugar added to 100 g of coffee ii. 10 g of

sugar added to 200 g of coffee iii. 4 g of sugar added to 200 g of coffee iv. 4 g of sugar added to 100 g of coffee Based on the amount of sugar added, which sequence ranks the coffee from the sweetest coffee to the least-sweet coffee?
Chemistry
2 answers:
madreJ [45]2 years ago
8 0

The answer is:   i, ii, iv, iii

dem82 [27]2 years ago
4 0

So solve this we must know the amount of sugar per gram of coffee

<span>i.              </span>10 g sugar /100 g coffee = 0.1 g sugar/ 1 g coffee

<span>ii.            </span>10 g sugar / 200 g coffee = 0.05 g sugar / 1 g coffee

<span>iii.           </span>4 g sugar / 200 g coffee = 0.02 g sugar / 1 g coffee

<span>iv.           </span>4 g sugar / 100 g coffee = 0.04 g sugar / 1 g coffee

 

So the ranking from sweetest so the least is:

<span> i, ii, iv, iii</span>

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<span>Answer: .01 moles of D to .005 moles of L ~ so, .01+.005 = .015 total; using this total value, divide the portions of D and L. so .01/.015 to .005/.015 ~ 67% D to 33% L. And thus, the enantiomer excess will be 34%.</span>
4 0
2 years ago
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For the reaction: MgF2(s) ⇌ Mg2+(aq) + 2F- (aq), Ksp= 6.4 × 10-9, the addition of 0.10 M NaF to the solution cause what effect o
galina1969 [7]

Answer:

Shifts the equilibrium to the left. reduces solubility.

Explanation:

  • MgF2(s) ↔ Mg2+(aq) + 2F-(aq)

          S                   S              2S

∴ Ksp = 6.4 E-9 = [ Mg2+ ] * [ F- ]² = S * (2S)²

⇒ 4S² * S = 6.4 E-9

⇒ 4S³ = 6.4 E-9

⇒ S³ = 1.6 E-9

⇒ S = 1.1696 E-3 M

  • NaF(s) → Na+(aq)  +  F-(aq)

        0.10M     0.10M        0.10M

  • MgF2(s) ↔ Mg2+(aq)  + 2F-(aq)

          S'                 S'              2S' + 0.10

⇒ Ksp = 6.4 E-9 = (S')*(2S' + 0.10)²

If we compare the concentration (0.10 M) of the ion with Ksp ( 6.4 E-9 ); thne we can neglect S' as adding:

⇒ 6.4 E-9 = (S')*(0.10)² = 0.01S'

⇒ S' = 6.4 E-7 M

∴ % S' = ( 6.4 E-7 / 0.1 )*100 = 6.4 E-4% <<< 5%, we can make the assumption

We can observe that S >> S' ( 1.1696 E-3 M >> 6.4 E-7 M ), which shows that the solubility  is reduced by the efect of the common ion from the salt, which causes the equilibrium to shift to the left, precipitating part of MgF2(s).

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2 years ago
Write the equations that represent the first and second ionization steps for hydroselenic acid (H2Se) in water. (Use H3O+ instea
Otrada [13]

Answer:

The equations are

1) H_{2}Se+H_{2}O--> H_{3}O^{+}+HSe^{-}

2) HSe^{-} +H_{2}O--> H_{3}O^{+}+Se^{-2}

Explanation:

There are two ionization steps in the dissociation of hydroselenic acid.

In first dissociation the H₂Se loses one proton and forms hydrogen selenide ion as shown below:

H_{2}Se+H_{2}O--> H_{3}O^{+}+HSe^{-}

The next step is again removal of a proton from the base formed above.

HSe^{-} +H_{2}O--> H_{3}O^{+}+Se^{-2}

4 0
2 years ago
Which of the following pairs of compounds would be most easily separated by thin layer chromatography: n-octyl alcohol and 1-oct
vivado [14]

Answer:

B. n-octyl alcohol and 1-octene

Explanation:

Thin-layer chromatography (TLC) is a chromatography technique used to separate non-volatile mixtures. The principle is that different compounds in the sample mixture travel at different rates due to the differences in interactions with stationary phase and due to the differences in solubility in the solvent. The principal chemical property for separation using this technique is molecular polarity

You can intuit than hexadecane and octadecane don't have big polarity differences, also chlorobenzene and bromobenzene haven't.

An alcohol as n-octyl alcohol has different polarity than an alkene as 1-octene.

Thus, using  thin layer chromatography is most easy to separate:

<em>B. n-octyl alcohol and 1-octene </em>

<em></em>

I hope it helps!

<em></em>

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kirill115 [55]

Answer:

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-atmospheric science

-environmental science

-climate science

-biology

-astronomy

-human physiology and medicine

Explanation:

I just had this question on edgenuity, this is what came up as correct. I hope this helps

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