3. what is the concentration of glucose (c6h1206, mm - 180.16 g/mol) solution that was prepared by
dissolving 50.500 g of glucose in 500.0 ml of solution? assume the solution has a density of 1.00
g/ml.

0.56 M

We are given;

Mass of glucose solute is 50.50 g

Volume of solution is 500 mL

We are required to calculate the concentration of glucose.

We need to know that, molarity is the concentration of a solution in moles per liter.

Molarity = Number of moles ÷ Molar mass

Molar mass of glucose = 180.16 g/mol

Mass of glucose = 50.50 g

Therefore, Moles of Glucose = 50.50 g ÷ 180.16 g/mol

                                                = 0.280 Moles

We can use the formula to calculate the molarity of glucose solution

Moles of glucose = 0.280 moles

Volume of solution = 500 mL or 0.5 L

Molarity = 0.280 moles ÷ 0.5 L

             = 0.56 moles/L

             = 0.56 M

Therefore, the concentration of glucose is 0.56 M

3 hours because it is a decent distance.

the ozone layer acts as a filter for the shorter wavelength and highly hazardous ultraviolet radiation (uvr) from the sun, protecting life on earth from its potentially harmful effects. when the sky is clear, there is an inverse relationship between stratospheric ozone and solar uvr measured at the earth's surface.