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shusha [124]
2 years ago
10

A gas with a volume of 20.0l at a pressure of 275 kpa is allowed to expand to a volume of 35.0l. what is the pressure in the con

tainer if the temperature remains constant?
1. 2.54 kPa

2. 157 kPa

3. 481 kPa

pls help ;(
Chemistry
1 answer:
sladkih [1.3K]2 years ago
8 0

Answer:

\boxed {\boxed {\sf 2. \ 157 \ kPa}}

Explanation:

Since temperature remains constant, the only variables that change are volume and pressure. Therefore, we are using <u>Boyle's Law.</u> This states that the pressure is inversely proportional to the volume. The formula is:

P_1V_1=P_2V_2

We know the gas starts with a volume of 20.0 liters at a pressure of 275 kPa. We can substitute these values into the left side of the formula.

275 \ kPa *20.0 \ L=P_2V_2

We know the gas expands to a volume of 35.0 Liters, but we do not know the pressure.

275 \ kPa *20.0 \ L=P_2* 35.0 \ L

Since we are solving for the new pressure, we must isolate the variable P₂. It is being multiplied by 35.0 Liters and the inverse of multiplication is division. Divide both sides by 35.0 L.

\frac {275 \ kPa * 20.0 \ L}{ 35.0 \ L}= \frac{P_2*35.0 \ L}{35.0 \ L}

\frac {275 \ kPa * 20.0 \ L}{ 35.0 \ L}=P_2

The units of liters cancel.

\frac {275 \ kPa * 20.0 }{ 35.0 }=P_2

\frac {5500}{35.0} \ kPa= P_2

157.142857 \  kPa=P_2

The original measurements of pressure and volume have 3 significant figures, so our answer must have the same. For the number we calculated, that is the ones place.

The 1 in the tenths place (157.142857) tells us to leave the 7 in the ones place.

157 \ kPa= P_2

If the gas expanded to a volume of 35.0 liters while the temperature remained constant, the pressure in container was approximately 157 kilopascals.

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Mashutka [201]

Answer is: (3, 2, 0, -1/2).

The principal quantum number (n) is one of four quantum numbers which are assigned to each electron in an atom to describe that electron's state.

For principal quantum number n=3:  

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l = 0, 1, 2.  

The azimuthal quantum number determines its orbital angular momentum and describes the shape of the orbital.  

2) magnetic quantum number (ml) can be ml = -l...+l.  

ml = -2, -1, 0, +1, +2.

Magnetic quantum number specify orientation of electrons in magnetic field and number of electron states (orbitals) in subshells.  

3) the spin quantum number (ms), is the spin of the electron.  

ms = +1/2, -1/2.  

(1, 1, 0, +1/2)  is not correct because orbital quantum number cannot be l = 1 for n = 1.

(2, 1, 2, +1/2)  is not correct because magnetic quantum number cannot be ml = 2 for orbital quantum number l = 1.

(3, -2, 1, -1/2) is not correct because orbital quantum number cannot be l = -2 for principal quantum number n = 3.

5 0
3 years ago
Un tanque de acetileno para una antorcha de soldadura de oxiacetileno proporciona 9340 L de gas acetileno, C2H2, a 0°C y 1 atm 2
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Answer:

3.33 tanques de O₂

Explanation:

Basados en la reacción:

2C₂H₂(g) + 5O₂(g) → 4CO₂(g) + 2H₂O(g)

<em>2 moles de acetileno reaccionan con 5 moles de oxígeno produciendo 4 moles de dióxido de carbono y 2 moles de agua</em>

<em />

La ley de Avogadro dice que el volumen de un gas bajo temperatura y presión constantes es proporcional a las moles de este gas. Así, como 2 moles de acetileno reaccionan con 5 moles de oxígeno, los litros de O₂ necesarios para quemar 9340L de acetileno son:

9340 L C₂H₂ × (5 moles O₂ / 2 moles C₂H₂) = <em>23350L de O₂</em>

Si un tanque contiene 7x10³ L de O₂ serán necesarios:

23350L O₂ ₓ (1 tanque / 7x10³L) =<em> 3.33 tanques de O₂</em>

6 0
3 years ago
In one experiment, the reaction of 1.00 mercury and an excess of sulfur yielded 1.16g of a sulfide of mercury
Nuetrik [128]

<u>Answer and Explanation:</u>

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Hg + S = HgS

Hg = 200,59

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5 0
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Determine the pH of a KOH solution made by mixing 0.251 g KOH with enough water to make 1.0 × 10 2 mL of solution
vesna_86 [32]

Answer:

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Explanation:

From the given information:

number of moles =mass in gram / molar mass

number of moles of KOH = mass of KOH / molar mass of KOH

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For solution :

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concetration  = 0.04474 M

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6 0
3 years ago
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7 0
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