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2 years ago

One mole of a metallic oxide reacts with one mole of hydrogen to produce two moles of the pure metal

Chemistry

1 answer:

WARRIOR [948]2 years ago

8 0

Answer:

Lithium oxide, Li₂O.

Explanation:

Hello!

In this case, according to the given amounts, it is possible to write down the chemical reaction as shown below:

$M_2O+H_2\rightarrow 2M+H_2O$

Which means that the metallic oxide has the following formula: M₂O. Next, we can set up the following proportional factors according to the chemical reaction:

$5.00gM_2O*\frac{1molM_2O}{(2*X+16)gM_2O}*\frac{2molM}{1molM_2O}*\frac{XgM}{1molM}=2.32gM$

Thus, we perform the operations in order to obtain:

$\frac{10X}{(2X+16)}=2.32$

So we solve for x as shown below:

$10X=2.32(2X+16)\\\\10X=4.64X+37.12\\\\X=\frac{37.12}{10-4.64}\\\\X= 6.93g/mol$

Whose molar mass corresponds to lithium, and therefore, the metallic oxide is lithium oxide, Li₂O.

Best regards!

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3 years ago

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

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

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

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

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

$E_1 = 28957.39292$ $J/mol$

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

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