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3 years ago

A decrease in the concentration of reactants causes the rate of the _____ reaction to slow.

1 answer:

nataly862011 [7]3 years ago

6 0

The correct answer is a Forward reaction to slow.
According to Le Chatelier's principle as the reactant decrease so the reaction should go leftward and this is called backward or reverse reaction so the reverse reaction will be fast to achieve equilibrium and the forward reaction will be slow.

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Consider the equilibrium reaction. 2 A + B − ⇀ ↽ − 4 C After multiplying the reaction by a factor of 2, what is the new equilibr

vredina [299]

Answer : The correct expression for equilibrium constant will be:

$K_c=\frac{[C]^8}{[A]^4[B]^2}$

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

$4A+2B\rightleftharpoons 8C$

The expression of $K_c$ will be,

$K_c=\frac{[C]^8}{[A]^4[B]^2}$

Therefore, the correct expression for equilibrium constant will be, $K_c=\frac{[C]^8}{[A]^4[B]^2}$

4 0

3 years ago

Vitamin D is produced in the skin when 7-dehydrocholesterol reacts with UVB rays (ultraviolet B) having wavelengths between

goblinko [34]

The energy range expected is 6.6 × 10^-19 J < E < 7.33 × 10^-19 J

The energy of the photon is given by;

E = hc/λ

E = energy of the photon

h = Plank's constant

c = speed of light

λ = wavelength of light

For the upper boundary range;

E = ?

h = 6.6 × 10^-34 Js

c = 3 × 10^8 m/s

λ = 270 × 10^-9

E = 6.6 × 10^-34 Js × 3 × 10^8 m/s / 270 × 10^-9

E = 7.33 × 10^-19 J

For the lower range;

E = ?

h = 6.6 × 10^-34 Js

c = 3 × 10^8 m/s

λ =300 × 10^-9

E = 6.6 × 10^-34 Js × 3 × 10^8 m/s / 300 × 10^-9

E = 6.6 × 10^-19 J

Hence, the energy range 6.6 × 10^-19 J < E < 7.33 × 10^-19 J

Learn more: brainly.com/question/24857760

8 0

2 years ago

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa

notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

$2NaOH+H_2SO_4-->Na_2SO_4+2H_2O$

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

$n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\$

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

$0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4$

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

$0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4$

Finally, the percent yield turns out into:

$Y=\frac{1.92g}{7.1g} *100$

$Y=27.0$ %

Best regards.

6 0

3 years ago

A three-component mixture of organic compounds in diethyl ether consists of a carboxylic acid, an amine base, and a neutral comp

Dmitrij [34]

Answer:the sodium carboxylate salt

Explanation:

The reaction between the carboxylic acid and the sodium hydroxide yields a sodium carboxylate. This sodium carboxylate is an ionic in nature; RCCOO-Na+. This can effectively interact with water and remain in the aqueous phase since it is composed of the carboxylate ion and sodium ion in solution. The aqueous phase always contains water soluble ionic substances of which the sodium carboxylate is a typical example of such.

4 0

3 years ago

How many grams are in 1.60 moles Hg(NO3)2

Ilia_Sergeevich [38]

Molar mass=MM

MM(Hg)=200.59g/mol
MM(N)=14.01g/mol
MM(O)=16.00g/mol

MM(Hg(NO3)2)=MM(Hg)+2*MM(N)+6*MM(O)=200.59g+2*14.01g+6*16g=324.61g/mol

1.60mol Hg(NO3)2*(324.61g/mol)=519.376 g=519 g Hg(NO3)2

6 0

3 years ago