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ozzi
3 years ago
7

What is the approximate volume of 19g of fluorine gas at a pressure of 4.0 atmospheres and a temperature of 127o C?

Chemistry
1 answer:
mr Goodwill [35]3 years ago
5 0

Answer:

V = 8.21 L

Explanation:

Given data:

Volume of fluorine = ?

Mass of fluorine = 19 g

Pressure = 4.0 atm

Temperature = 127 °C(127+ 273 =400 K)

Solution:

Number of moles of fluorine:

Number of moles = mass/molar mass

Number of moles = 19 g/ 19 g/mol

Number of moles = 1 mol

PV = nRT

V = nRT/P

V = 1 mol ×0.0821 atm. L. mol⁻¹. k⁻¹ × 400 K/ 4 atm

V = 32.84 L/ 4

V = 8.21 L

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The answer is volume.
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2 years ago
Once an enzymatic reaction is completed, the enzyme releases what?
sleet_krkn [62]

Answer:

Once an enzymatic reaction is completed, the enzyme releases substrates.

Explanation:

The enzyme will always return to its original state at the completion of the reaction. One of the important properties of enzymes is that they remain ultimately unchanged by the reactions they catalyze. After an enzyme is done catalyzing a reaction, it releases its products (substrates).

5 0
2 years ago
Volume of 8.29ml and a mass of 16.31g
strojnjashka [21]
Density = mass/volume = 16.31/8.29 = 1.96 g/mL.

Hope this helps!
3 0
3 years ago
Describe what is happening in this chemical equation: Al2(SiO3)3 + NaOH → Na(SiO3)3 + Al2OH Please Help ASAP
WITCHER [35]

Answer:

Double replacement reaction

Explanation:

This is what is known as a double replacement reaction. The two parts of each molecule separate and recombine to form two new compounds. For instance, Al_2(SiO_3)_3 separates into Al_2 and (SiO_3)_3, while NaOH separates into Na and OH, and then they recombine with the other compound.

Hope this helps!

7 0
3 years ago
Read 2 more answers
The activation energy of an uncatalyzed reaction is 70 kJ/mol. When a catalyst is added, the activation energy (at 20 °C) is 42
denis23 [38]

Answer:

T = 215.33 °C

Explanation:

The activation energy is given by the Arrhenius equation:

k = Ae^{\frac{-Ea}{RT}}

<u>Where:</u>

k: is the rate constant

A: is the frequency factor    

Ea: is the activation energy

R: is the gas constant = 8.314 J/(K*mol)

T: is the temperature

We have for the uncatalyzed reaction:

Ea₁ = 70 kJ/mol

And for the catalyzed reaction:

Ea₂ = 42 kJ/mol

T₂ = 20 °C = 293 K

The frequency factor A is constant and the initial concentrations are the same.

Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

k_{1} = k_{2}

Ae^{\frac{-Ea_{1}}{RT_{1}}} = Ae^{\frac{-Ea_{2}}{RT_{2}}}   (1)

By solving equation (1) for T₁ we have:

T_{1} = \frac{T_{2}*Ea_{1}}{Ea_{2}} = \frac{293 K*70 kJ/mol}{42 kJ/mol} = 488. 33 K = 215.33 ^\circ C  

Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.

I hope it helps you!      

4 0
3 years ago
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