A container of nitrogen gas contains 0.55 mol at 7.0 atm and 45°C, and it has a volume of 2.0 L. What volume will 1.20 mol of nitrogen have at 7.0 atm and 45°C?

4.4 L

Explanation:

Given data:

Initial number of mole = 0.55 mol

Initial volume = 2.0 L

Final volume = ?

Final number of moles = 1.20 mol

Temperature and pressure = constant

Solution:

Formula:

V₁/n₁   = V₂/n₂

V₂ = V₁×n₂/n₁  

V₂ = 2.0 L ×1.20 mol/ 0.55 mol

V₂ = 2.4 L.mol / 0.55 mol

V₂ = 4.4 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"

because natural gas is colorless, odorless, and tasteless, distributors add mercaptan (a chemical that smells like sulfur) to give natural gas a distinct unpleasant odor (it smells like rotten eggs). they locate the types of rock that are likely to contain natural gas deposits.