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USPshnik [31]
3 years ago
8

What are some examples of gases at room temperature

Chemistry
1 answer:
Lady_Fox [76]3 years ago
6 0

Answer:

Nitrogen, Hydrogen, Oxygen, Chlorine, and Fluorine are all gases at room temperature.

Explanation:

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what is the molar mass of a gaseous flouride of sulfur containing 70.4% F and having a density of approximately 4.5g/L at 20 deg
zlopas [31]

Answer:

The molar mass is 180.2 g/mol

Explanation:

<u>Step 1:</u> Data given

% of F = 70.4 %

Density = 4.5 g/L

Temperature = 20 °C

Pressure = 1 atm

<u>Step 2:</u> Calculate the number of moles

PV = nRT

 ⇒ with P = the pressure = 1.00 atm

⇒ with V = the volume = Assume this is 1L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ R = the gas constant = 0.08206 L*atm/K*mol

⇒ T = the temperature : 20°C = 293 Kelvin

1 atm*1L= n(0.08206 L-atm/mol-K)*(293 K)

n = 0.04159 moles

<u>Step 3</u>: Calculate molar mass

Molar mass = Mass / moles

4.5 grams / 0.04159 moles = 108.2 g/mol

<u>Step 4:</u> Calculate moles of F

Moles = Mass / molar mass

Moles F = 70.4 g / 19 g/mol

Moles F =  3.70 moles

Moles S = 29.6g / 32.07 g/mol

Moles S = 0.923 moles S

<u>Step 5:</u> Divide by the smallest amount of moles

F = 3.70 / 0.923 = 4

S = 0.923 / 0.923 = 1

The empirical formula is SF4

The molar mass of SF4 = 32.07 + 4*19 = 108.07 g/mol

This means the empirical formula is the same as the molecular formula SF4

The molar mass is 180.2 g/mol

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
Why does salt dissolve in water?
tigry1 [53]
The water dissolves the salt because the water molecules are able to interact with the salt-forming particles, called ions. When the water Interacts with the trainers of the salt ions, the solid salt crystal structure Suffers breakdown, until the trainers of the salt ions are completely surrounded by water molecules. At this time the salt is fully dissolved by water.

hope this helps!
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The answer is A

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