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3 years ago

If you have 6 moles of reactant A and excess of B and C, how much product E would be formed?

Chemistry

1 answer:

riadik2000 [5.3K]3 years ago

3 0

To be able to solve this problem, we must be given the balanced chemical equation:

3A + 5B + 1C ---> 2D + 1E

Our reactant A is called the limiting reactant because it is limited in amount which is only 6 moles. So we based our calculation on that reactant alone.

From stoichiometry, we have 1 mole of E for every 3 moles of A, therefore:

moles E = 6 moles A * (1 mole E / 3 moles A)

moles E = 2 moles

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The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the r

rodikova [14]

Answer:

E₁ ≅ 28.96 kJ/mol

Explanation:

Given that:

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,

Let the activation energy for a catalyzed biochemical reaction = E₁

E₁ = ??? (unknown)

Let the activation energy for an uncatalyzed biochemical reaction = E₂

E₂ = 50.0 kJ/mol

= 50,000 J/mol

Temperature (T) = 37°C

= (37+273.15)K

= 310.15K

Rate constant (R) = 8.314 J/mol/k

Also, let the constant rate for the catalyzed biochemical reaction = K₁

let the constant rate for the uncatalyzed biochemical reaction = K₂

If the rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:

K₁ = 3.50 × 10³

K₂ = 1

Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;

we can use the formula for Arrhenius equation;

$K=Ae^{\frac{-E}{RT}}$

If $K_1=Ae^{\frac{-E_1}{RT}} -------equation 1$ &

$K_2=Ae^{\frac{-E_2}{RT}} -------equation 2$

$\frac{K_1}{K_2} = e^{\frac{-E_1-E_2}{RT}$

$E_1= E_2-RT*In(\frac{K_1}{K_2})$

$E_1= 50,000-8.314*310.15*In(\frac{3.50*10^3}{1})$

$E_1 = 28957.39292$ $J/mol$

E₁ ≅ 28.96 kJ/mol

∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol

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3 years ago

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Answer:

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2 years ago

How to round the number 314.77 to 2 significant figures

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Answer:

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Arrange the aqueous solutions from the most acidic to the most basic, at 25 C.

Vaselesa [24]

Answer:

See explanation below

Explanation:

If [OH | = 1x 10^-7 M then [H^+] = 1 * 10^-14/1x 10^-7 = 1x 10^-7 M

pH = - log [H^+] = -log [1x 10^-7]

pH = 7

If [OH | = 1x 10^-9 M then [H^+] = 1 * 10^-14/1x 10^-9 = 1x 10^-5 M

pH = - log [H^+] = -log [1x 10^-5]

pH = 5

[H^+] = 1 x 10^-3 M

pH = - log [H^+] = -log [1x 10^-3]

pH = 3

Hence the decreasing acidity;

pH 3 > pH 5 > pH 7 > pH 11

The most acidic is pH 3 while the most basic is pH 11

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3 years ago