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tangare [24]
3 years ago
11

If a gas occupies 22.4 L at 0.265 atm, then what will the pressure be in atmospheres if the gas occupies 35.52 L? Assume all oth

er
variables are held constant
A)0.167 atm
B)0.437 atm
C)0.886 atm
D)149 atm
Chemistry
1 answer:
pochemuha3 years ago
8 0

Answer:

The new pressure is 0.167 atm

The correct answer is option A

Explanation:

Step 1: Data given

Volume of the gas = 22.4 L

Pressure of the gas = 0.265 atm

The volume increases to 35.52 L

Step 2: Calculate the new pressure

P1*V1 = P2*V2

⇒with P1 = the initial pressure = 0.265 atm

⇒with V1 = the initial volume = 22.4 L

⇒with P2 = the new pressure = TO BE DETERMINED

⇒with V2 = the increased volume = 35.52 L

0.265 atm * 22.4 L = P2 * 35.52 L

P2 = (0.265 atm * 22.4 L) / 35.52 L

P2 = 0.167 atm

The new pressure is 0.167 atm

The correct answer is option A

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It can allow the molecule (like water) to be polar because it has a negative and positive side to it (oxygen holds the negatives tight causing the hydrogens to be positive).
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Calculate the molarity of 13.1 g of NaCl in 727 mL of solution
valentina_108 [34]

Answer:

The molarity of the solution is 0,31 M

Explanation:

We calculate the weight of 1 mol of NaCl from the atomic weights of each element of the periodic table. Then, we calculate the molarity, which is a concentration measure that indicates the moles of solute (in this case NaCl) in 1000ml of solution (1 liter)

Weight 1 mol NaCl= Weight Na + Weight Cl= 23 g + 35, 5 g= 58, 5 g

58, 5 g-----1 mol NaCl

13,1 g ---------x= (13,1 g x 1 mol NaCl)/58, 5 g= 0, 224 mol NaCl

727 ml solution------ 0, 224 mol NaCl

1000ml solution------x= (1000ml solutionx0, 224 mol NaCl)/727 ml solution

x=0,308 mol NaCl---> <em>The solution is 0,31 molar (0,31 M)</em>

5 0
3 years ago
How many liters of oxygen gas, at standard
Karo-lina-s [1.5K]

Answer:

Explanation:

  • For the balanced reaction:

<em>4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)​.</em>

It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.

  • Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:

no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.

  • Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:

<em><u>Using cross multiplication:</u></em>

4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.

0.64 mol of Fe is needed to react with → ??? mol of O₂.

∴ The no. of moles of O₂ needed = (3 mol)(0.64 mol)/(4 mol) = 0.48 mol.

  • Finally, we can get the volume of oxygen using the information:

<em>It is known that 1 mole of any gas occupies 22.4 L at standard P and T (STP).</em>

<em></em>

<em><u>Using cross multiplication:</u></em>

1 mol of O₂ occupies → 22.4 L, at STP conditions.

0.48 mol of O₂ occupies → ??? L.

∴ The no. of liters of O₂ = (0.48 mol)(22.4 L)/(1 mol) = 10.752 L.

5 0
3 years ago
How many atoms are in 13.97 liters of water vapor at STP
Arisa [49]
<span>Let's </span>assume that water vapor has ideal gas behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>

</span><span>Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹) and T is temperature in Kelvin.<span>
<span>
</span>P = 1 atm = 101325 Pa (standard pressure)
V = 13.97 L = 13.97 x 10</span>⁻³ m³<span>
n = ?
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
T = 0 °C = 273 K (standard temperature)
<span>
By substitution,
</span>101325 Pa x 13.97x 10</span>⁻³ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 273 K<span>
                                          n = 0.624 mol
<span>
Hence, the moles of water vapor at STP is 0.624 mol.

According to the </span></span>Avogadro's constant, 1 mole of substance has 6.022 × 10²³ particles.
<span>
Hence, number of atoms in water vapor = 0.624 mol x </span>6.022 × 10²³ mol⁻¹
<span>                                                                = 3.758 x 10</span>²³<span>

</span>
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