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Ostrovityanka [42]
4 years ago
12

What will the final volume of a gas be if it is heated from 9 K to 117 K, and the initial volume is 85.5 L?

Chemistry
1 answer:
TiliK225 [7]4 years ago
6 0

Answer:

1111.5L

Explanation:

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Within a molecule of nitrous oxide, N2O, an atom of nitrogen combines with an atom of oxygen in a(n)
Marianna [84]
Can be more specific?
4 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
Consider the evaporation of methanol at 25.0∘c: ch3oh(l)→ch3oh(g). you may want to reference (pages 826 - 829) section 18.9 whil
monitta
The reaction is 
CH3OH(l)→CH3OH(g)
To calculate ΔG°
ΔG°=ΔG°f of CH3OH(g)-ΔG°f of CH3OH(l)
=162.3kJ/mol - (-166.6kJ/mol
=4.3kJ/mol.
7 0
4 years ago
Read 2 more answers
What is the pH at the half-equivalence point in the titration of a weak base with a strong acid? The pKb of the weak base is 7.9
ValentinkaMS [17]

Hey there!

Given the reaction:

B + H⁺   => HB⁺


At half-equivalence point :  [B] = [HB⁺]

=> [B] / [HB⁺] = 1

Henderson-Hasselbalch equation :


pH = pKa + log ( [B] ) / ( HB⁺)]

pH = 14 - pKb + log ( 1 )

pH = 14 -  7.95 + 0

pH = 6.05


Answer C


Hope that helps!



5 0
3 years ago
A student carefully mixes two clear liquids in a beaker. The
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Answer: The student should have worn splash goggles, an apron, and gloves during the investigation.

Explanation:

7 0
3 years ago
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