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2 years ago

A chemist is working on a reaction represented by this chemical equation:

Chemistry

2 answers:

erastova [34]2 years ago

6 0

Answer is: A. 1.81 mol.

Balanced chemical reaction: FeCl₂ + 2KOH → Fe(OH)₂ + 2KCl.

n(FeCl₂) = 4.15 mol; amount of iron(II) chloride.

n(KOH) = 3.62 mol; amount of potassium hydroxide, limiting reactant.

From chemical reaction: n(KOH) : n(Fe(OH)₂) = 2 : 1.

n(Fe(OH)₂) = n(KOH) ÷ 2.

n(Fe(OH)₂) = 3.62 mol ÷ 2.

n(Fe(OH)₂) = 1.81 mol; amount of iron(II) hydroxide.

diamong [38]2 years ago

4 0

Answer:

It's A. 1.81 mol. This answer is on Plato.

#PlatoLivesMatter

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Suppose a soap manufacturer starts with a triglyceride that has the fatty acid chains arachidic acid, palmitic acid and palmitic

DIA [1.3K]

Answer:

Sodium arachidate; Sodium palmitate and Sodium palmitate

Explanation:

Triglycerides are esters of fatty acids with glycerol. In triglycerides, three fatty acid molecules are linked by ester bonds to each of the three carbon atoms in a glycerol molecule. The fatty acids may be same or different fatty acid molecules. Hydrolysis of triglycerides yields the three fatty acid molecules and glycerol.

Saponification is the process by which a base is used to catalyst the hydrolysis of the ester bonds in glycerides. The products of this base-catalyzed hydrolysis of triglycerides are the metallic salts of the three fatty acids and glycerol. The salts of the fatty acids are known as soaps.

For a triglyceride that has the fatty acid chains arachidic acid, palmitic acid and palmitic acid attached to the three backbone carbons glycerol, the saponification of the triglyceride with NaOH will yield the sodium salts or soaps of the three fatty acids as well as glycerol.

Arachidic acid will react with NaOH to yield sodium arachidate.

The two palmitic acid molecules will each react with NaOH to yield sodium palmitate.

8 0

1 year ago

Nitrogen dioxide decomposes to nitric oxide and oxygen via the reaction: 2NO2 → 2NO + O2 In a particular experiment at 300 °C, [

serious [3.7K]

Answer: The rate of disappearance of $NO_2$ is $3.5\times 10^{-5}M/s$

Explanation:

The given chemical reaction is:

$2NO_2\rightarrow 2NO+O_2$

The rate of the reaction for disappearance of $NO_2$ is given as:

$\text{Rate of disappearance of }NO_2=-\frac{\Delta [NO_2]}{\Delta t}$

Or,

$\text{Rate of disappearance of }NO_2=-\frac{C_2-C_1}{t_2-t_1}$

where,

$C_2$ = final concentration of $NO_2$ = 0.00650 M

$C_1$ = initial concentration of $NO_2$ = 0.0100 M

$t_2$ = final time = 100 minutes

$t_1$ = initial time = 0 minutes

Putting values in above equation, we get:

$\text{Rate of disappearance of }NO_2=-\frac{0.00650-0.0100}{100-0}\\\\\text{Rate of disappearance of }NO_2=3.5\times 10^{-5}M/s$

Hence, the rate of disappearance of $NO_2$ is $3.5\times 10^{-5}M/s$

6 0

2 years ago

What do you think happens to Difluoroethane at –24°C? Provide evidence to support your claim.

balandron [24]

Answer:

The following subsections explain the explanation according to the particular circumstance.

Explanation:

The boiling point seems to be the temperature beyond which the working fluid as well as the boiling phase would be at a predetermined pressure or voltage at equilibrium among one another and.
The vapor or boiling temperature of 1,1 difluoroethane seems to be -25oC at 1 atm, although as a gas it can remain at a higher temperature around -24oC.

6 0

2 years ago

Complete and balance the following redox reaction in acidic solution As2O3(s) + NO3- (aq) → H3AsO4(aq) + N2O3(aq)

scoray [572]

Answer:

$As_{2}O_{3}(s)+2NO_{3}^{-}(aq)+2H_{2}O(l)+2H^{+}(aq)\rightarrow 2H_{3}AsO_{4}(aq)+N_{2}O_{3}(aq)$

Explanation:

Oxidation: $As_{2}O_{3}(s)\rightarrow H_{3}AsO_{4}(aq)$

Balance As: $As_{2}O_{3}(s)\rightarrow 2H_{3}AsO_{4}(aq)$
Balance H and O in acidic medium: $As_{2}O_{3}(s)+5H_{2}O(l)\rightarrow 2H_{3}AsO_{4}(aq)+4H^{+}(aq)$
Balnce charge: $As_{2}O_{3}(s)+5H_{2}O(l)-4e^{-}\rightarrow 2H_{3}AsO_{4}(aq)+4H^{+}(aq)$ ......(1)

Reduction: $NO_{3}^{-}(aq)\rightarrow N_{2}O_{3}(aq)$

Balance N: $2NO_{3}^{-}(aq)\rightarrow N_{2}O_{3}(aq)$
Balance H and O in acidic medium: $2NO_{3}^{-}(aq)+6H^{+}(aq)\rightarrow N_{2}O_{3}(aq)+3H_{2}O(l)$
Balance charge: $2NO_{3}^{-}(aq)+6H^{+}(aq)+4e^{-}\rightarrow N_{2}O_{3}(aq)+3H_{2}O(l)$ ......(2)