For many purposes we can treat methane as an ideal gas at temperatures above its boiling point of . suppose the temperature of a sample of methane gas is lowered from ⁢ to , and at the same time the pressure is changed. if the initial pressure was and the volume increased by , what is the final pressure? round your answer to significant digits.

1.3 × 10 atm

Explanation:

There is some info missing. I think this is the original question.

For many purposes we can treat methane (CH₄) as an ideal gas at temperatures above its boiling point of -161. °C Suppose the temperature of a sample of methane gas is raised from -19.0 °C to 12.0 °C, and at the same time the pressure is changed. If the initial pressure was 5.6 atm and the volume decreased by 50.0%, what is the final pressure? Round your answer to the correct number of significant digits. atm x10 ?

Given data

T₁ = -19.0°C + 273.15 = 254.2 K

T₂ = 12.0°C + 273.15 = 285.2 K

V₁ (unknown)

V₂ = 50.0% V₁ = 0.500 V₁

P₁ = 5.6 atm

P₂ (what we are looking)

We can find the the final pressure P₂ using the combined law gas.

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

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i believe it is a.) released into

explanation: