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At standard temperature and pressure, 1 mole of any gas occupies 22.4 liters of space. So, 65.8 liters would be 65.8/22.4=2.9375, so approximately 3 moles.

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What is the final concentration if 75.0 mL of a 3.50 M glucose solution is diluted to a volume of 400 mL?

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To find out how many milliliters of the 0.266 M LiNO3 solution are needed, you can use the formula C1V1 = C2V2, where C1 is the concentration of the first solution, V1 is the volume of the first solution, C2 is the concentration of the second solution, and V2 is the volume of the second solution. Plugging in the values, you can solve for V1, which will give you the volume of the 0.266 M LiNO3 solution needed to make 150.0 ml of 0.075 M LiNO3 solution.


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