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

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"