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nikdorinn [45]
4 years ago
6

A sample of N2 gas occupied 9.20 L at 21 degrees C and 0.959 atm. If the pressure is tuned to 1.15 atm at constant temperature w

hat is the newly occupied volume
Chemistry
1 answer:
Gwar [14]4 years ago
5 0

Newly occupied volume is 7.672L by N2 gas.  

<u>Explanation:- </u>

Given  

V= 9.20 L, T(i) = T (f) = 294 K, P (i) = 0.959 atm, P (f) = 1.15 atm

To Find V (f)  

Solution

According to the gas equation  

PV= nRT

Where P=Pressure

V=Volume

n=No of moles

R=Universal gas constant

T=Temperature

Also we know the relation between pressure, volume and temperature as

PV/T = constant

\frac{p(i) V(i) \times p(f) V(f)}{T(i) \times T(f)}

V(f)=[P(i) V(i) T(f)] \div[T(i) P(f)]

Where P(i)V(i)/T(i) are the initial values of Pressure, Volume and Temperature.

P(f)V(f)/T(f) are the final values of Pressure, Temperature and Volume.

Substitute the known values in the equation we get

= \frac{0.959 \times 9,2 \times 294}{294 \times 1.15}

=2574.94 \div 338.1

= 7.672 L

Thus the final volume of the system is 7.672 L

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A certain kind of light has a wavelength of 4.6 micrometers what is the frequency of this light and gigahertz? Use c= 2.998 × 10
Leno4ka [110]

Answer:

6.52×10⁴ GHz

Explanation:

From the question given above, the following data were obtained:

Wavelength (λ) = 4.6 μm

Velocity of light (v) = 2.998×10⁸ m/s

Frequency (f) =?

Next we shall convert 4.6 μm to metre (m). This can be obtained as follow:

1 μm = 1×10¯⁶ m

Therefore,

4.6 μm = 4.6 μm × 1×10¯⁶ m / 1 μm

4.6 μm = 4.6×10¯⁶ m

Next, we shall determine frequency of the light. This can be obtained as follow:

Wavelength (λ) = 4.6×10¯⁶ m

Velocity of light (v) = 2.998×10⁸ m/s

Frequency (f) =?

v = λf

2.998×10⁸ = 4.6×10¯⁶ × f

Divide both side by 4.6×10¯⁶

f = 2.998×10⁸ / 4.6×10¯⁶

f = 6.52×10¹³ Hz

Finally, we shall convert 6.52×10¹³ Hz to gigahertz. This can be obtained as follow:

1 Hz = 1×10¯⁹ GHz

Therefore,

6.52×10¹³ Hz = 6.52×10¹³ Hz × 1×10¯⁹ GHz / 1Hz

6.52×10¹³ Hz = 6.52×10⁴ GHz

Thus, the frequency of the light is 6.52×10⁴ GHz

5 0
3 years ago
Question 1 (3 points)
den301095 [7]

Answer:

Explanation:

ΔTemp => 35⁰C(108K) increases to 57.9⁰C(330.9L) => increases volume (Charles Law)

Use the Kelvin Temperature values in a ratio that will increase the original volume.

ΔVol = 6.33L(330.9/108.0) => gives a larger volume. Using 108.0/330.9 would give a smaller volume and would be contrary to what the problem is asking.

ΔPress => 342 mmHg increases to 821 mmHg => decreases volume (Boyles Law)

Use the pressure values in a ratio that will decrease the original volume.

ΔPress = 6.33L(342/821) => gives a smaller volume. Using 821/342 would give a larger volume and would be contrary to what the problem is asking.

Now, putting both ΔTemp together with ΔPress => net change in volume. (Combined Gas Law)

ΔVol = 6.33L(330.9/108.0)(342/821) = 8.08L (final volume of gas).

___________________

This problem can also be worked using the combined gas law equation:

P₁V₁/T₁ = P₂V₂/T₂ => V₂ = P₁V₁T₂/T₁P₂

V₂ = [(342mm)(6.33L)(330.9K)]/[(108K)(821mm)] = 8.08L (final volume of gas)

7 0
3 years ago
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