What pressure will be exerted by 26g if CO2 at a temperature of 35C and a volume of 550ml?

P = 28.5 atm

Explanation:

PV = nRT => P = nRT/V

P = Pressure = ?

n = moles of sample = mass/molar mass = 26g/44g·mol⁻¹ = 0.619 mole CO₂

R = gas constant = 0.08206 L·atm/mol·K

T = temperature in Kelvin = (35 + 273)K = 308K

V = volume in Liters = 550 ml = 0.550 L

Pressure (P) = nRT/V

= (0.619 mole)(0.08206 L·atm/mol·K)(308K)/(0.550 L) = 28.5 atm

it can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.