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Bess [88]
3 years ago
14

Suppose now that you wanted to determine the density of a small crystal to confirm that it is sulfur. From the literature, you k

now that sulfur has a density of 2.07 . How would you prepare 20.0 mL of the liquid mixture having that density from pure samples of ( = 1.492 g/mL) and ( = 2.890 g/mL)? (Note: 1 mL = 1 .)
Chemistry
1 answer:
ivolga24 [154]3 years ago
7 0

Answer:

Mix 11.73 mL of CHCl₃ and 8.27 mL of CHBr₃

Explanation:

To begin, it seems your question lacks the name of the pure samples that will be mixed to prepare the liquid mixture. The names are not necessary, as the numerical values are there. However, an internet search tells me they are CHCl₃ (d=1.492g/mL) and CHBr₃ (d=2.890 g/mL).

The mixture has a density of 2.07 g/mL, so <u>20 mL of the mixture would weigh</u>:

20 mL * 2.07 g/mL = 41.4 g

Let X be the volume of CHCl₃ and Y the volume of CHBr₃:

X + Y = 20 mL

The mass of CHCl₃ and CHBr₃ combined have to be equal to the mass of the mixture. We can write that equation using the volume of the samples and their density:

X * 1.492 + Y * 2.890 = 41.4 g

So now we have a system of two equations and two unknowns, we use algebra to solve it:

   1. Express Y in terms of X:

X + Y = 20

  • Y = 20 - X

   2. Replace Y in the second equation:

X * 1.492 + Y * 2.890 = 41.4

  • 1.492*X + 2.890*(20-X) = 41.4

   3. Solve for X:

  • 1.492*X + 57.8 - 2.890*X = 41.4
  • 1.398*X = 16.4
  • X = 11.73 mL

   4. Using the now known value of X, solve for Y:

X + Y = 20

  • 11.73 + Y = 20
  • Y = 8.27 mL

So, to prepare the liquid mixture we would mix 8.27 mL of CHBr₃ and 11.73 mL of  CHCl₃.

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The specific heat of aluminum is 0.897 j/g•°c. which equation would you use to calculate the amount of heat needed to raise the
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<span>c. q = 0.75 g x 0.897 j/g•°c x 22°c</span>
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A pan containing 40 grams of water was allowed to cool from a temperature of 91.0C. If the amount of heat repressed is 1,300 jou
Vinvika [58]

Answer:

83°C

Explanation:

The following were obtained from the question:

M = 40g

C = 4.2J/g°C

T1 = 91°C

T2 =?

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Q = MCΔT

ΔT = Q/CM

ΔT = 1300/(4.2x40)

ΔT = 8°C

But ΔT = T1 — T2 (since the reaction involves cooling)

ΔT = T1 — T2

8 = 91 — T2

Collect like terms

8 — 91 = —T2

— 83 = —T2

Multiply through by —1

T2 = 83°C

The final temperature is 83°C

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3 years ago
Give the structure of the principal organic product formed by free-radical bromination of 2,2,4−trimethylpentane.
Fudgin [204]

the principal organic product formed by free-radical bromination of 2,2,4−trimethylpentane is 2-bromo-2,4,4-trimethylpentane.

what is free radical halogenation?

A substitution reaction in which a hydrogen atom is replaced with a halogen atom, via a free radical reaction mechanism. when this reaction is carrid out by bromine radical, it is called free radicle bromination. When bromine (Br{2}) treated with light (hν) it comes to homolytic cleavage of the Br-Br bond and give rise to bromine radicles.

free-radical bromination of 2,2,4−trimethylpentane

Bromination of an alkane includes the substitution of a bromine atom for a hydrogen atom. The following stages will be taken by 2,2,4-trimethylpentane during this reaction:

Initiation reaction:  The production of a bromine free radical requires the initiation of heat or light.

Br - Br ⇒ 2Br·

Propagation: This reaction relies heavily on hydrogen. This reaction is impossible if hydrogen is not present. Because tertiary free radicals are more stable than secondary and primary free radicals, they are favoured in this reaction.

Termination: The remaining free radical of bromide reacts with the tertiary free radical of 2,2,4-trimethylpentane to form 2-bromo-2,4,4-trimethylpentane.

the principal organic product formed by free-radical bromination of 2,2,4−trimethylpentane is 2-bromo-2,4,4-trimethylpentane.

To know more about free radical halogenation, check out:

brainly.com/question/13046867

#SPJ4

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The damage caused by an earthquake depends on which factors? Select all that apply.
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