What volume is occupied by 3.00 moles of nitrogen under a pressure of 202 kPa and a temperature of 323K? R = 8.314 kPa L/mol K INICI

39.9 L

Explanation:

Use PV = nRT to solve this problem.

Substitute the values known.

P is pressure, which is 202 kPa, V is volume, which we do not know, n is moles, which is 3, R is 8.314, and T is temperature, which is 323K.

Set up your equation as follows

Solve for V, multiply the right side of the equation and divide by the pressure, 202 kPa.

This results in Volume being 39.9 L

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b