Which tool or tools would you use to measure and compare the mass of a cup of fresh water and a cup of salt water

Answer is (2) - blue.

Thymol blue is a pH indicator which 3 colors according to the pH.

Thymol blue shows red color at below pH than 1.2 and shows yellow color at the pH between 2.8 and 8.0. For the pH above 9.6, thymol blue shows blue color.

Hence, the thymol blue has 2 pH ranges as 1.2 - 2.8 and 8.0 - 9.6.

Since, pH = 11 is above the pH = 9.6, thymol blue shows blue color.

Answer:

46

Explanation:

Sodium metal has a molar mass of  

22.99

Answer:

               2.873 g of CO₂

Explanation:

This problem will be solved in two steps.

Step 1: Calculating mass of Octane:

Data Given:

                  Volume = 1 L = 1000 cm³

                  Density  =  0.703 g/cm³

                  Mass  =  ??

Formula Used:

                       Density = Mass ÷ Volume

Solving for Mass,

                       Mass  =  Density × Volume

                       Mass  =  0.703 g/cm³ × 1000 cm³

Putting Values,

                       Mass  =  703 g

Step 2: Calculating Mass of Oxygen:

Data:

                  Volume =  V  = 5.0 L

                  Temperature = T = 25 °C = 298.15 K

                  Pressure = P = 1.0 atm

                  Moles = n = ?

Assuming that the gas is acting as Ideal gas so, we will use Ideal gas equation i.e.

                 P V = n R T

Solving for n,

                  n = P V / RT

Putting values,

                  n = 1.0 atm × 5.0 L / 0.0821 atm.L.mol⁻¹.K⁻¹ × 298.15 K

                  n = 0.204 moles

As,

            Moles = Mass / M.Mass

So,

            Mass = Moles × M.Mass

            Mass = 0.204 mol × 16 g/mol       ∴ M.Mass of O₂ = 16g.mol⁻¹

            Mass = 3.26 g

Step 3: Calculating mass of CO₂:

The balance chemical equation is follow,

                            2 C₈H₁₈ + 25 O2 = 16 CO₂ + 18 H₂O

According to equation

        228.45 g (2 mol) of C₈H₁₈  reacts with = 799.97 g (16 mol) of O₂

So,

                       703 g of C₈H₁₈ will react with = X g of O₂

Solving for X,

                  X  =  703 g × 799.97 g ÷ 228.45

                  X  =  2461 g of O₂

While, we are only provided with 3.26 g of O₂. This means O₂ is the limiting reactant and will control the yield of the final product. Therefore,

According to balance equation,

          799.97 g (16 mol) of O₂ produced  =  704.152 g (16 mol) of CO₂

So,

   3.26 g (0.204 mol) of O₂ will produce  =  X g of CO₂

Solving for X,

                 X =  3.26 g × 704.152 g ÷ 799.97 g

                  X =  2.873 g of CO₂