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3 years ago

Zn(s) + HCl(aq) → ZnCl2(aq) + H2(g) ↑

Chemistry

1 answer:

WINSTONCH [101]3 years ago

6 0

Hey there!

Zn + HCl → ZnCl₂ + H₂

First, balance Cl.

1 on the left, 2 on the right. Add a coefficient of 2 in front of HCl.

Zn + 2HCl → ZnCl₂ + H₂

Next, balance H.

2 on the left, 2 on the right. Already balanced.

Lastly, balance Zn.

1 on the left, 1 on the right. Already balanced.

Our final balanced equation:

Zn + 2HCl → ZnCl₂ + H₂

Hope this helps!

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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

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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

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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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Answer : The concentration of $H_3O^+$ and $OH^-$ are $7.94\times 10^{-5}$ and $1.258\times 10^{-10}$ respectively.

Solution : Given,

pH = 4.10

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Formula used : $pH=-log[H_3O^+]$

First we have to calculate the hydronium ion concentration by using pH formula.

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$[H_3O^+]=antilog(-4.10)$

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Now we have to calculate the pOH.

As we know, $pH+pOH=14$

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