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Delicious77 [7]
2 years ago
10

A gas occupies a volume at 34.2 mL at a temperature of 15.0 C and a pressure of 800.0 torr. What will be the volume of this gas

at STP?
Chemistry
2 answers:
Nimfa-mama [501]2 years ago
5 0
The ideal gas law equation can be used to find the number of moles of the gas at given conditions
PV = nRT
where 
P - pressure - 800.0 torr x 133.3 Pa/torr = 106 640 Pa
V - volume - 34.2 x 10⁻³ m³
n - number of moles 
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 15 °C + 273 = 288 K
substituting these values in the equation
106 640 Pa x 34.2 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 288 K 
n = 1.52 mol

molar volume is where 1 mol of any gas occupies a volume of 22.4 L at STP
therefore if 1 mol occupies - 22.4 L
then 1.52 mol occupies - 22.4 L/mol x 1.52 mol = 34.05 L
volume of the gas at STP is 34.05 L
Grace [21]2 years ago
3 0
The answer is 34.1 mL.
Solution:
Assuming ideal behavior of gases, we can use the universal gas law equation
     P1V1/T1 = P2V2/T2
The terms with subscripts of one represent the given initial values while for terms with subscripts of two represent the standard states which is the final condition.
At STP, P2 is 760.0torr and T2 is 0°C or 273.15K. Substituting the values to the ideal gas expression, we can now calculate for the volume V2 of the gas at STP:
     (800.0torr * 34.2mL) / 288.15K = (760.0torr * V2) / 273.15K
     V2 = (800.0torr * 34.2mL * 273.15K) / (288.15K * 760.0torr)
     V2 = 34.1 mL
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A gem has a mass of 4.50 g. When the gem is placed in a graduated cylinder 12.00 mL of water, the water level rises to 13.45 mL.
Mandarinka [93]
<span>Displaced volume :

</span>Final volume - <span>Initial volume

</span>13.45 mL - 12.00 mL => 1.45 mL

Mass =  4.50 g

Therefore:

density = mass / volume

D = 4.50 / 1.45

<span>D = 3.103 g/mL </span>
6 0
1 year ago
A chemist adds 180.0 mL of a 1.42M sodium carbonate (Na CO,) solution to a reaction flask. Calculate the millimoles of sodium ca
svet-max [94.6K]

Answer: The millimoles of sodium carbonate the chemist has added to the flask are 256

Explanation:

Molarity is defined as the number of moles dissolved per liter of the solution.

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{milli moles of solute}}{\text{Volume of solution in ml}}     .....(1)

Molarity of BaCl_2 solution = 1.42 M

Volume of solution = 180.0 mL

Putting values in equation 1, we get:

1.42M=\frac{\text{milli moles of }BaCl_2}{180.0ml}\\\\\text{milli moles of }BaCl_2}={1.42M\times 180.0ml}=256milli mol

Thus the millimoles of sodium carbonate the chemist has added to the flask are 256.

6 0
2 years ago
Calculate the number of molecules in 46.0 grams of water
Natalka [10]

Answer:

Explanation:

Since water has a chemical formula of H2O , there will be 2 moles of hydrogen in every mole of water. In one mole of water, there will exist approximately 6.02⋅1023 water molecules.

4 0
2 years ago
Which of the following statements concerning hydrocarbons is/are correct?
Alona [7]

Answer:

1.  Saturated hydrocarbons may be cyclic or acyclic molecules.

2.  An unsaturated hydrocarbon molecule contains at least one double bond.

Explanation:

Hello,

In this case, hydrocarbons are defined as the simplest organic compounds containing both carbon and hydrogen only, for that reason we can immediately discard the third statement as ethylenediamine is classified as an amine (organic chain containing NH groups).

Next, as saturated hydrocarbons only show single carbon-to-carbon bonds and carbon-to-hydrogen bonds, they may be cyclic (ring-like-shaped) or acyclic (not forming rings), so first statement is true

Finally, since we can find saturated hydrocarbons which have single carbon-to-carbon and carbon-to-hydrogen bonds only and unsaturated hydrocarbons which could have double or triple bonds between carbons and carbon-to-hydrogen bonds, the presence of at least one double bond makes the hydrocarbon unsaturated.

Therefore, first and second statements are correct.

Best regards.

6 0
1 year ago
The graph shows the distribution of energy in the particles of two gas samples at different temperatures, T1 and T2. A, B, and C
coldgirl [10]

Answer:

  • <u><em>The average speed of gas particles at T</em></u><em><u>₂</u></em><u><em> is lower than the average speed of gas particles at T</em></u><em><u>₁</u></em><u><em>.</em></u>

Explanation:

<em>Particles A and C</em> are shown as if they are on the same vertical line, which means with the same kinetic energy. Both particle A and C are to the lett of <em>particle B</em>, which means that the formers have a lower kinetic energy than the latter.

Since the likelyhood of a particle to participate in the reaction increases with the kinetic energy, particle B is more likely to participate in the reaction than particles A and C. Hence, the first choice is incorrect.

The graph, although not perfectly symmetrical, does show a bell shape, hence there are many particles will low kinetic energy and many particles with high kinetic energy. You cannot assert that most of the particles of the two gases have high high speeds. Hence, second statement is incorrect, too.

At high values of kinetic energy (toward the right of the curve), the line labeled T₁ is higher than the line labeled T₂, meaning that at T₁ more particles have an elevated kinetic energy than the number of particles that have an elevated kinetic energy at T₂.

On the other hand, at low values of kinetic energy (toward the left of the curve) the line T₂ is higher than the line T₁, meaning that at T₂ more particles have a low kinetic energy than the number of particles that have low kinetic energy at T₁.

Hence, the last two paragraphs are telling that the average kinetic energy of gas particles at T₂ is is lower than the average kinetic energy of gas particles at T₁.

Since the average speed is proportional the the square root of the temperature, the same trend for the average kinetic energy is true for the average speed, and you conclude that the last statement is true: "The average speed of gas particles at T₂ is lower than the average speed of gas particles at T₁".

Since more particles at T₁ have high kinetic energy than the number of particles at T₂ that have a high kinetic energy, more particles of gas at T₁ are likely to participate in the reaction  than the gas at T₂, and the third statement is incorrect.

7 0
1 year ago
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