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2 years ago

10

3)

1 answer:

Vikentia [17]2 years ago

6 0

Answer:

A

Explanation:

I think it is A because a dinosaurs are extinct so it wold be possible to find a dinosaur fossil in the coal mine.

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blondinia [14]

Answer:

c elevation

Explanation:

7 0

2 years ago

Read 2 more answers

What is the wavenumber of the radiation emitted when a hydrogen

LUCKY_DIMON [66]

Answer: Wavenumber of the radiation emitted is $0.08\times 10^{8}m^{-1}$

Explanation:

The relationship between wavelength and energy of the wave follows the equation:

$E=\frac{hc}{\lambda}$

where,

E = energy of the radiation = $1.634\times 10^{-18}J$

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength of radiation = ?

Putting values in above equation, we get:

$1.634\times 10^{-18}J=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{\lambda}\\\\\lambda=12.16\times 10^{-8}m$

$\bar {\nu}=\frac{1}{\lambda}=\frac{1}{12.16\times 10^{-8}}=0.08\times 10^{8}m^{-1}$

Thus wavenumber of the radiation emitted is $0.08\times 10^{8}m^{-1}$

8 0

3 years ago

An ethylene gas torch requires 300 L of gas at 0.8 atm. What will be the pressure of the gas if ethylene is supplied by a 200.0

Vlad [161]

Answer:

1.2 atm

Explanation:

Given data

Volume of the gas in the tank (V₁): 200.0 L

Pressure of ethylene gas in the tank (P₁): ?

Volume of the gas in the torch (V₂): 300 L

Pressure of the gas in the torch (P₂): 0.8 atm

If we consider ethylene gas to be an ideal gas, we can find the pressure of ethylene gas in the tank using Boyle's law.

$P_1 \times V_1 = P_2 \times V_2\\P_1 = \frac{P_2 \times V_2}{V_1} = \frac{0.8atm \times 300L}{200.0L} = 1.2 atm$

3 0

3 years ago

Coal and petroleum are useful because the both have ___ stored in them.

gizmo_the_mogwai [7]

I think the answer is 4 carbon dioxide

5 0

3 years ago

Read 2 more answers

The value of Kp for the reaction NO(g) 1 1 2 O2(g) 4 NO2(g) is 1.5 3 106 at 25°C. At equilibrium, what is the ratio of PNO2 to P

expeople1 [14]

Answer : The ratio of $p_{NO_2}$ to $p_{NO}$ is, $6.87\times 10^5$

Solution : Given,

$K_p=1.5\times 10^6$

$p_{O_2}$ = 0.21 atm

The given equilibrium reaction is,

$NO(g)+\frac{1}{2}O_2\rightleftharpoons NO_2(g)$

The expression of $K_p$ will be,

$K_p=\frac{(p_{NO_2})}{(p_{NO})\times (p_{O_2})^{\frac{1}{2}}}$

Now put all the given values in this expression, we get:

$1.5\times 10^6=\frac{(p_{NO_2})}{(p_{NO})\times (0.21)^{\frac{1}{2}}}$

$\frac{(p_{NO_2})}{(p_{NO})}=(1.5\times 10^6)\times (0.21)^{\frac{1}{2}}$

$\frac{(p_{NO_2})}{(p_{NO})}=6.87\times 10^5$

Therefore, the ratio of $p_{NO_2}$ to $p_{NO}$ is, $6.87\times 10^5$

5 0

3 years ago