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2 years ago

Consider the potassium permanganate reaction again.

2 answers:

6 0

Hey there !

Mole ratio :

<span>2 KMnO4 + 16 HCl → 2 KCl + 2 MnCl2 + 8 H2O + 5 Cl2

2 moles KMnO4 ----------------- 8 moles H2O
3.45 moles KMnO4 ------------- (moles H2O )

Moles H2O = 3.45 * 8 / 2

Moles H2O = 27.6 / 2

= 13.8 moles of H2O

</span>The option that was given is wrong , <span>You're right.</span>

puteri [66]2 years ago

3 0

Answer:

Moles of H2O produced = 13.8

Explanation:

Given:

Moles of KMnO4 reacted = 3.45

To determine:

moles of H2O produced

Explanation:

Given reaction:

2KMnO4 + 16HCl →2KCl + 2MnCl2 + 8H2O + 5Cl2

Based on the reaction stoichiometry:

2 moles of KMnO4 produces 8 moles of H2O

Therefore, moles of H2O produced when 3.45 moles of KMnO4 react is:

$= \frac{3.45\ moles\ KMnO4 * 8\ moles\ H2O}{2\ moles\ KMnO4} = 13.8$

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which of the following amino acids are used to measure the protein concentration by UV? A. His B. Pro C. Trp D. Phe E. Tyr

Goryan [66]

Answer:

C. Trp D. Phe E. Tyr

Explanation:

The concentration of a protein has a direct relation with absorbance of the protein in a UV spectrophotometer. The formula which relates concentration with absorbance is described as under:

A = ∈ x c x l

where, A = Absorbance

∈ = Molar extinction co-efficient

c = Concentration of absorbing species i.e. protein

l = Path length of light

Tryptophan (Trp), phenylalanine (Phe ) and tyrosine (Tyr) are three aromatic amino acids which are used to measure protein concentration by UV. It is mainly because of tryptophan (Trp), protein absorbs at 280 nm which gives us an idea of protein concentration during UV spectroscopy.

The table depicting the wavelength at which these amino acids absorb and their respective molar extinction coefficient is as under:

Amino acid Wavelength Molar extinction co-efficient (∈)

Tryptophan 282 nm 5690

Tyrosine 274 nm 1280

Phenylalanine 257 nm 570

In view of table above, we can easily see that Molar extinction co-efficient (∈) of Tryptophan is highest amongst all these 3 amino acids that is why it dominates while measuring concentration.

7 0

3 years ago

Which condition must be met in order for an equation to be balanced? The elements in the reactants are the same as the elements

Nookie1986 [14]

Answer:

There must be an equal amount of each element on both sides of the equation.

Hope this helps

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8 0

3 years ago

Read 2 more answers

An experiment applies heat to a substance and produces a gas that is a new and different substance. Choose the true statement.

sukhopar [10]

I believe it is C; reasoning being that the hint for physical change is," the producing of a gas," chemical "that's new and diff. substance. "

5 0

3 years ago

A solid precipitates from a solution, absorbing heat as it does so. a. Yes b. No c. Can't decide with information given.

inn [45]

Answer:

c. Can't decide with information given.

Explanation:

The chemical and physical processes can be classified as endothermic or exothermic. The first one happens when the system absorbs heat, so the temperature of the surroundings will decrease, and the other one happens when the system releases heat, then the temperature of the surrounds will increase.

Precipitation is the formation of a solid in a solution. The process can happen with absorption or release of heat, it depends on the substance. So, with the information given it's impossible to say it.

3 0

3 years ago

A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor

AlexFokin [52]

Answer : The partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of $C_2HBrClF_3$ = 15.0 g

Mass of $O_2$ = 22.6 g

Molar mass of $C_2HBrClF_3$ = 197.4 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_2HBrClF_3$ and $O_2$ .

$\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole$

and,

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole$

Now we have to calculate the mole fraction of $C_2HBrClF_3$ and $O_2$ .

$\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971$