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2 years ago

Plz help with this........

Chemistry

1 answer:

Elodia [21]2 years ago

3 0

<h2>GREETINGS!</h2><h2>_____________________________________</h2><h3>Answer:</h3>

ITS $C_{4}H_{10}$ AND $C_{5}H_{12}$

<h2>_____________________________________</h2><h3><u>GENERAL FORMULA OF ALKANE</u></h3><h3> $C_{n}H_{2n+2}$ </h3>

Where n is the number of CARBONS

By using the General Formula of Alkanes having 4 carbon atoms is

$C_{4}H_{2x4+2}$

$C_{4}H_{10}$

By using the General Formula of Alkanes having 5 carbon atoms is

$C_{5}H_{2x5+2}$

$C_{5}H_{12}$

IF YOU USE THE SAME FORMULA FOR THE OTHER TWO OPTIONS THE OUTCOME WILL NOT BE AS SAME AS GIVEN IN THE QUESTION SO THOSE TWO ARE CAN NOT BE OUR ANSWER

<h2>_____________________________________</h2><h2>HOPE THIS HELPS!</h2>

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2 years ago

What is the energy of a wave with wavelength of 4.2 x 10-7 m. (Hint: Calculate for frequency first.)

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Answer:

Option B. 4.74×10¯¹⁹ J.

Explanation:

The following data were obtained from the question:

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Wavelength (λ) = 4.2×10¯⁷ m

Velocity (v) = constant = 3×10⁸ m/s

Frequency (f) =.?

v = λf

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f = 3×10⁸ / 4.2×10¯⁷

f = 7.143×10¹⁴ s¯¹

Therefore, the frequency of the wave is 7.143×10¹⁴ s¯¹.

Finally, we shall determine the energy of the wave using the following formula

E = hf

Where

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Thus, the enery of the wave can be obtained as follow:

Frequency (f) = 7.143×10¹⁴ s¯¹.

Planck's constant = 6.63×10¯³⁴ Js

Energy (E) =..?

E = hf

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E = 4.74×10¯¹⁹ J

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2 years ago

15.1 L N2 at 25 °C and 125 kPa and 44.3 L O2 at 25 °C and 125 kPa were transferred to a tank with a volume of 6.25 L. What is th

Y_Kistochka [10]

Answer:

The total pressure of the mixture in the tank of volume 6.25 litres at 51°C is 1291.85 kPa.

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Temperature (T₁)=25°C=25+273 K=298 K

Similarly, for Oxygen,

Pressure(P₂)= 125 kPa

Volume(V₂)= 44.3 L

Temperature(T₂)=25°C= 298 K

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Then, By Combined gas laws,

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2 years ago

How does the density of a gas depend on the molar mass of the gas?

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Answer:

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