A 1.2 L weather balloon on the

ground has a temperature of 25°C

and is at atmospheric pressure (1.0

atm). When it rises to an elevation

where the pressure is 0.74 atm,

then the new volume is 1.8 L. What

is the temperature (in °C) of the air

at this elevation?

330.95K

Explanation:

V₁ = 1.2L

T₁ = 25°C = (25 + 273.15)K = 298.15K

P₁ = 1.0 atm

P₂ = 0.74 atm

V₂ = 1.8L

T₂ =?

From combined gas equation,

(P₁ * V₁) / T₁ = (P₂ * V₂) / T₂

Solve for T₂

T₂ = (P₂ * V₂ * T₁) / (P₁ * V₁)

T₂ = (0.74 * 1.8 * 298.15) / (1.0 * 1.2)

T₂ = 397.1358 / 1.2

T₂ = 330.9465K

T₂ = 330.95K or T₂ = (330.95 - 273.15)°C = 57.8°C

stupid i wont let you get a chance

explanation:

in chemistry, a solution’s concentration is how much of a dissolvable substance, known as a solute, is mixed with another substance, called the solvent. the standard formula is c = m/v, where c is the concentration, m is the mass of the solute dissolved, and v is the total volume