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3 years ago

A gas has v= 4.0L, T =27 degree C and P=2.0 atm. What is the V if T=327 degree C and P=8.0 atm?

1 answer:

3 0

As I am reading the problem,i can tell the question gives you two temperatures, two pressures, one volume and asking for the other. this should be an indication that you need to use the following gas formula

P1V1/T1= P2V2/T2

P1= 2.0 atm
V2= 4.0 L
T1= 27= 300 K
P2= 8.0 atm
V2= ?
T2= 327= 600 k

let's plug in the values into the formula

(2.0 x 4.0)/ 300= (8.0 x V2)/ 600 K

V2= 2.0 Liters

answer is D

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Answer:

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Explanation:

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3 years ago

The potential energy on a spring is proportional to the square of which of these quantities

erica [24]

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4 years ago

Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electr

Len [333]

Answer:

a. 1875 nm

b. 4051 nm

c. 1282 nm

These all are infrared electromagnetic radiation.

Explanation:

Our strategy here is to utilize the Rydberg equation for hydrogen atom electronic transition.

1/λ = Rh x (1/n₁² - 1/n₂²) where λ is the wavelength

Rh is Rydberg constant

n₁ and n ₂ are the energy levels ( n₁ < n₂ )

Now lets star the calculations.

a. n₁ = 3, n₂ = 4

1/λ = 1.097 x 10⁷ /m x (1/3² - 1/4²) = 5.333 x 10⁵/m

λ = 1/(5.333 x 10⁵ /m) = 1.875 x 10⁻⁶ m

Converting λ to nanometers:

1.875 x 10⁻⁶ m x (1 x10⁹ nm/m) = 1875 nm

b. n₁ = 4, n₂ = 5

1/λ = 1.097 x 10⁷ /m x (1/4² - 1/5²) = 2.468 x 10⁵/m

λ = 1/(2.468 x 10⁵/m) = 4.051 x 10⁻⁶ m

4.051 x 10⁻⁶ m x (1 x10⁹ nm/m) = 4051 nm

c. n₁ = 3, n₂ = 5

1/λ = 1.097 x 10⁷ /m x (1/3² - 1/5²) = 7801 x 10⁵/m

λ = 1/(7801 x 10⁵/m) = 1282 x 10⁻⁶ m

1282 x 10⁻⁶ m x (1 x10⁹ nm/m) = 1282 nm

All of these transitions fall in the infrared region of the spectrum.

8 0

3 years ago

PLS HELP

vivado [14]

Answer:

D. All of these.

Explanation:

5 0

3 years ago

Calculate the equilibrium constant Kp for this reaction, given the following information (at 299 K ): 2NO(g)+Br2(g)⇌2NOBr(g)Kc=2

Travka [436]

Answer:

$2NO_{(g)}+Br_2_{(g)}\rightleftharpoons2NOBr_{(g)}$ , Kp = 0.08967

$2NO_{(g)}\rightleftharpoons N_2_{(g)}+O_2_{(g)}$ , Kp = 2.3×10³⁰

Explanation:

The relation between Kp and Kc is given below:

$K_p= K_c\times (RT)^{\Delta n}$

Where,

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

$2NO_{(g)}+Br_2_{(g)}\rightleftharpoons2NOBr_{(g)}$

Given: Kc = 2.2

Temperature = 299 K

R = 0.082057 L atm.mol⁻¹K⁻¹

Δn = (2)-(2+1) = -1

Thus, Kp is:

$K_p= 2.2\times (0.082057\times 299)^{-1}$

For the second equilibrium reaction:

$2NO_{(g)}\rightleftharpoons N_2_{(g)}+O_2_{(g)}$

Given: Kc = 2.3×10³⁰

Temperature = 299 K

R = 0.082057 L atm.mol⁻¹K⁻¹

Δn = (2)-(2) = 0

Thus, Kp is:

$K_p= 2.2\times (0.082057\times 299)^{0}$

3 0

3 years ago