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stira [4]
3 years ago
7

What is the density of helium gas at 17 degrees celcius and 773 torr?

Chemistry
1 answer:
valina [46]3 years ago
7 0

Answer:

0.172g/L

Explanation:

Step 1:

Data obtained from the question:

Temperature (T) = 17°C

Pressure (P) = 773 torr

Step 2:

Conversion to appropriate unit:

For pressure :

760 torr = 1 atm

Therefore, 773 torr = 773/760 = 1.02 atm

For temperature:

Temperature (Kelvin) = temperature (celsius) + 273

temperature (celsius) = 17°C

Temperature (Kelvin) = 17°C + 273 = 290K.

Step 3:

Obtaining an expression for the density.

From the ideal gas equation PV = nRT, we can obtain an equation for the density as follow:

PV = nRT. (1)

But: number of mole(n) = mass (m)/Molar Mass(M) i.e

n = m/M

Substitute the value of n into equation 1

PV = nRT

PV = mRT/M

Divide both side by m

PV /m = RT/M

Divide both side by P

V/m = RT/MP

Invert the above equation

m/V = MP/RT (2)

Recall:

density (d) = mass(m) / volume(V) i.e

d = m/V

Replace m/V in equation 2 with d

m/V = MP/RT

d = MP/RT.

Step 4:

Determination of the density.

Temperature (T) = 290K

Pressure (P) = 1.02 atm

Molar Mass of helium (M) = 4g/mol

Gas constant (R) = 0.082atm.L/Kmol

Density (d) =?

d = MP/RT

d = 4 x 1.02 / 0.082 x 290

d = 0.172g/L

Therefore, the density of the helium gas is 0.172g/L

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A mixture contains 25 g of cyclohexane (C6H12) and 44 g of 2-methylpentane (C6H14). The mixture of liquids is at 35 oC . At this
Greeley [361]

Answer:

The mol fraction of cyclohexane in the liquid phase is 0.368

Explanation:

Step 1: Data given

Mass of cyclohexane = 25.0 grams

Mass of 2-methylpentane = 44.0 grams

Temperature = 35.0 °C

The pressure of cyclohexane = 150 torr

The pressure of 2-methylpentane = 313 torr

The pressure we only need for the mole fraction in gas phase.

Step 2: Calculate moles of cyclohexane

Moles cyclohexane = mass cyclohexane / molar mass

Moles cyclohexane = 25.0 g / 84 g/mol = 0.298 mol of cyclohexane

Step 3: Calculate moles of 2-methylpentane

Moles = 44.0 grams / 86 g/mol = 0.512 mol of 2-methylpentane

Step 4: Calculate mole fraction of cyclohexane in the liquid phase

Mole fraction of C6H12:

0.298 / (0.298 + 0.512) = 0.368

The mol fraction of cyclohexane in the liquid phase is 0.368

5 0
4 years ago
Using the picture below, compare and contrast the plant cell and the animal cell (What do they have different? What do they have
STALIN [3.7K]

Answer:

Animal cell lacks chloroplasts and a cell wall

Explanation:

Unlike animal cells, plant cells have chloroplasts and a cell wall. In a plant cell, the chloroplasts work to use photosynthesis in order to turn light into energy for the plant cell. Animal cells do not utilize light energy, and therefore do not carry out photosynthesis and do not have chloroplasts.

Additionally, the plant cell has a protective cell wall and a cell membrane, while animal cells only have a cell membrane. This is because the plant cells' cell wall provides additional strength and protection against outside factors, such as osmosis.

6 0
3 years ago
What is the total number of hydrogen atoms in a molecule of heptyne?
Step2247 [10]

Answer:

12

Explanation:

alkyne formula is CnH2n-2

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6 0
3 years ago
PLEASE HELP ME ASAP! CHEMISTRY TUTOR<br><br> SEE ATTACHED
Masteriza [31]

Answer:

\large \boxed{\text{-827.4 kJ}}

Explanation:

We have three equations:

1. 2H₂S(g)            + O₂(g)   ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

2. S(s, rhombic)  + O₂(g)   ⟶ SO₂(g);                                 ∆H = -296.8 kJ

3. PbO(s)             + H₂S(g) ⟶ PbS(s)               + SO₂(g);    ∆H =  -104.3 kJ

From these, we must devise the target equation:

4. 2PbS(s)            + 3O₂(g) ⟶2PbO(s)             + 2SO₂(g); ΔH = ?

The target equation has PbS(s) on the left, so you reverse Equation 3 and double it.

When you reverse an equation, you reverse the sign of its ΔH.

When you double an equation, you double its ΔH.

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

Equation 5 has 2H₂O on the left. That is not in the target equation.

You need an equation with 2H₂O on the right, so you copy Equation 1.  

6. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ  

Equation 6 has 2S(s, rhombic) on the right. That is not in the target equation.

You need an equation with 2S(s, rhombic) on the left, so you double Equation 2.  

7. 2S(s, rhombic)  + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

Now, you add equations 5, 6, and 7, cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

You get the target equation 4:

5. 2PbS(s)  + <u>2H₂O(g</u>)  ⟶ 2PbO(s) + <u>2H₂S(g</u>);  ∆H =  208.6 kJ

6. <u>2H₂S(g)</u> + O₂(g)        ⟶ <u>2S(s</u>)     + <u>2H₂O(g)</u> ; ∆H = -442.4 kJ

<u>7</u><u>. </u><u>2S(s)</u><u>      + 2O₂(g)      ⟶ 2SO₂(g);                   ∆H = -593.6 kJ </u>

4 . 2PbS(s) + 3O₂(g)      ⟶ 2PbO(s) + 2SO₂(g); ΔH = -827.4 kJ

\Delta H \text{ for the reaction is $ \large \boxed{\textbf{-827.4 kJ}}$}

8 0
3 years ago
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