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Bond [772]
3 years ago
8

Four preparations involving table sugar (sucrose) are described below. Analyze the sugar preparation processes and the end produ

cts.
Sugar Water Sugar Floss Sugar Caramel Rock Sugar

Sugar is added to water, and the water is stirred until the sugar is no longer visible. Sugar is melted and blown with a fan to make thin threads of solid sugar. Pure sugar is melted and cooked until it changes color and becomes a thick, sticky liquid. A highly concentrated solution of sugar water is prepared, large sugar crystals are allowed to form slowly, and then dye is added.


Which of the following is evidence that a chemical reaction has taken place in one of the processes?
A.
the melting and re-solidification of sugar into threads, which are different from sugar’s natural cube-shaped crystals
B.
the slow growth of giant sugar crystals from a highly concentrated sugar-water solution and added dye
C.
the dissolution of solid sugar in water to form a thin, colorless, odorless liquid
D.
the color change that takes place after heat melts and begins to decompose the sugar
Chemistry
1 answer:
11111nata11111 [884]3 years ago
8 0

Answer:

b

Explanation:

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Which element do you expect to be most like magnesium? Why?
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Consider the following reaction: 2 Bi(s) + 3 Cl2(g) → 2 BiCl3(s)
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Taking into account the reaction stoichiometry, 1.119 grams of chlorine gas are required to produce 3.32 grams of bismuth chloride is produced.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 Bi + 3 Cl₂ → 2 BiCl₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Bi: 2 moles
  • Cl₂: 3 moles
  • BiCl₃: 2 moles

The molar mass of the compounds is:

  • Bi: 209 g/mole
  • Cl₂: 70.90 g/mole
  • BiCl₃: 315.45 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Bi: 2 moles ×209 g/mole=418  g

Cl₂: 3 moles ×70.90 g/mole= 212.7 g

BiCl₃: 2 moles ×315.45 g/mole= 630.9 g

<h3>Mass of Cl₂ required</h3>

The following rule of three can be applied: If by reaction stoichiometry 630.9 grams of BiCl₃ are formed by 212.7 grams of Cl₂, 3.32 grams of BiCl₃ are formed by how much mass of Cl₂?

mass of Cl_{2} =\frac{3.32 grams of BiCl_{3}x212.7 grams of Cl_{2} }{630.9 grams of BiCl_{3}}

<u><em>mass of Cl₂= 1.119 grams</em></u>

Finally, 1.119 grams of chlorine gas are required to produce 3.32 grams of bismuth chloride is produced.

Learn more about the reaction stoichiometry:

<u>brainly.com/question/24741074</u>

<u>brainly.com/question/24653699</u>

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3 years ago
A student has 70.5 mL of a 0.463 M aqueous solution of sodium bromide. The density of the solution is 1.22 g/mL. Find the follow
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Answer:

a.) 86.01 g.

b.) 3.36 g.

c.) 0.394 m ≅ 0.40 m.

d.) 4.77%.

e.) 3.9%.

Explanation:

<em>a.) mass of the solution:</em>

The density of the solution is the mass per unit volume.

<em>∵ Density of solution = (mass of solution)/(volume of the solution).</em>

∴ Mass of the solution = (density of solution)*(volume of the solution) = (1.22 g/mL)*(70.5 mL) = 86.01 g.

<em>b.) grams of sodium bromide  :</em>

  • Molarity (M) is defined as the no. of moles of solute dissolved per 1.0 L of the solution.

∵ M = (no. of moles of NaBr)/(Volume of the solution (L))

∴ no. of moles of NaBr = M*(Volume of the solution (L)) = (0.463 M )*(0.0705 L) = 0.0326 mol.

<em>∵ no. of moles of NaBr = (mass of NaBr)/(molar mass of NaBr)</em>

∴ mass of NaBr = (no. of moles of NaBr)*(molar mass of NaBr) = (0.0326 mol)*(102.894 g/mol) = 3.36 g.

<em>c.) molality of the solution:</em>

  • Molality (m) is defined as the no. of moles of solute dissolved per 1.0 kg of the solvent.

∵ m = (no. of moles of NaBr)/(mass of the soluvent (kg))

no. of moles of NaBr = 0.0326 mol,

mass of solvent = mass of the solution - mass of NaBr = 86.01 g - 3.36 g = 82.65 g = 0.08265 kg.

∴ m = (no. of moles of NaBr)/(mass of the soluvent (kg)) = (0.0326 mol)/(0.08265 kg) = 0.394 m ≅ 0.40 m.

<em>d.) % (m/v) of the solution:</em>

∵ (m/v)% = [(mass of solute) /(volume of the solution)]* 100

∴ (m/v)% = [(3.36 g)/(70.5 mL)]* 100 = 4.77%.

<em>e.) % (m/m) of the solution:</em>

∵ (m/m)% = [(mass of solute) /(mass of the solution)]* 100

∴ (m/m)% = [(3.36 g)/(86.01 g)] * 100 = 3.9 %.

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Answer:

c. iron I hope it helped.....

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