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Rudiy27
3 years ago
12

Why would 18 be wrong for the atomic mass of an atom with 6 neutrons, 6 electrons, 6 protons?

Chemistry
1 answer:
11Alexandr11 [23.1K]3 years ago
6 0
Atomic mass is just protons plus neutrons
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What volume of 24% trichloroacetic acid (tca) is needed to prepare eight 3 ounce bottles of 10% tca solution?
Sever21 [200]

Answer:

295.7 mL of 24% trichloroacetic acid (tca) is needed .

Explanation:

Let the volume of 24% trichloroacetic acid solution be x

Volume of required 10% trichloroacetic acid solution =8 bottles of 3 ounces

= 24 ounces = 709.68 mL

(1 ounces =  29.57 mL)

Amount of trichloroacetic acid in 24% solution of x volume of solution will be equal to amount of trichloroacetic acid in 10% solution of volume 709.68 mL.

x\times \frac{24}{100}=709.68 mL\times \frac{10}{100}

x = 295.7 mL

295.7 mL of 24% trichloroacetic acid (tca) is needed .

6 0
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

E₁ ≅ 28.96 kJ/mol

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

8 0
3 years ago
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