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3 years ago

Which term describes this reaction? addition condensation elimination substitution

Chemistry

2 answers:

sattari [20]3 years ago

6 0

Answer;

-Condensation

Explanation;

-Condensation reaction is a reaction that takes place when two molecules join together forming a larger molecule and also produces another smaller molecule in the process. The smaller molecule produced in the reaction is normally water, however, it can also be methanol, hydrogen chloride, among other small molecules.

-Condensation is common in many biological and chemical processes such as formation of disaccharides, and polmerization reactions, where small molecules join to form a larger molecule called a polmer.

vlada-n [284]3 years ago

5 0

The others are wrong it's B. Condensation

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How many milliliters of 0.200 M FeCl3 are needed to react with an excess of Na2S to produce 0.345 g of Fe2S3 if the percent yiel

GuDViN [60]

Answer:

25.53mL of 0.200 M FeCl₃ are needed to produce 0.345g of Fe₂S₃

Explanation:

Based on the reaction of the problem, 1 mole of Fe₂S₃ is produced from 2 moles of FeCl₃.

0.345g of Fe₂S₃ are (Molar mass: 207.9g/mol):

0.345g of Fe₂S₃ ₓ (1 mol / 207.9g) = <em>1.6595x10⁻³ moles Fe₂S₃</em>

Moles of Fe needed to produce these moles of Fe₂S₃ are:

1.6595x10⁻³ moles Fe₂S₃ ₓ ( 2 moles FeCl₃ / 1 mole Fe₂S₃) =

<em>3.3189x10⁻³ moles of FeCl₃</em>

As the percent yield of the reaction is 65.0%, the moles of FeCl₃ you need to add are:

3.3189x10⁻³ moles of FeCl₃ ₓ (100.0% / 65.0%) = <em>5.106x10⁻³ moles of FeCl₃</em>

A solution 0.200M contains 0.200 moles per L. Volume to obtain 5.106x10⁻³ moles is:

<em>5.106x10⁻³ moles of FeCl₃ ₓ ( 1L / 0.200mol) = 0.02553L = </em>

<h3>25.53mL of 0.200 M FeCl₃ are needed to produce 0.345g of Fe₂S₃</h3>

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3 years ago

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Explanation:

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3 years ago

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satela [25.4K]

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4 years ago

When a chemical reaction is run in aqueous solution inside a calorimeter, the temperature change of the water (and Ccal) can be

Yakvenalex [24]

Answer:

The total change in enthalpy for the reaction is - 81533.6 J/mol

Explanation:

Given the data in the question;

Reaction;

HCl + NaOH → NaCl + H₂O

Where initial temperature is 21.2 °C and final temperature is 28.0 °C. Ccal is 1234.28 J

Moles of NaOH = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol

Moles of HCl = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol

so, 0.0500 moles of H₂O produced

Volume of solution = 50.mL + 50.mL = 100.0 mL

Mass of solution m = volume × density = 100.0mL × 1.0 g/mL = 100 g

now ,

Heat energy of Solution q= (mass × specific heat capacity × temp Δ) + Cal

we know that; The specific heat of water(H₂O) is 4.18 J/g°C.

so we substitute

q_soln = (100g × 4.18 × ( 28.0 °C - 21.2 °C) ) + 1234.28

q_soln = 2842.4 + 1234.28

q_soln = 4076.68 J

Enthalpy change for the neutralization is ΔH $_{neutralization}$

ΔH $_{neutralization}$ = -q_soln / mole of water produced

so we substitute

ΔH $_{neutralization}$ = -( 4076.68 J ) / 0.0500 mol

ΔH $_{neutralization}$ = - 81533.6 J/mol

Therefore, the total change in enthalpy for the reaction is - 81533.6 J/mol

6 0

3 years ago