Asample of gas contains 6.25 × 10-3 mol in a 500.0 ml flask at 265°c. what is the pressure of the gas in kilopascals? which variables are given? pressure volume temperature moles

55.9 kPa; Variables given = volume (V), moles (n), temperature (T)

We must calculate p from V, n, and T, so we use the Ideal Gas Law:

pV = nRT

Solve for p: p = nRT/V

R = 8.314 kPa.L.K^(-1).mol^(-1)

T = (265 + 273.15) K = 538.15 K

V = 500.0 mL = 0.5000 L

∴ p = [6.25 x 10^(-3) mol x 8.314 kPa·L·K^(-1)·mol^(-1) x 538.15 K]/(0.5000 L) = 55.9 kPa

answer by yourhope:

how do models compare to real-world objects?

d. most models represent a real-world object in a simpler way.

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it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2