What is the scaling factor if the molar mass of this compound, having an empirical formula of ch2o, is 150 g/mol

The scaling factor is 5

Considering the question given above, the following data were obtained:

Empirical formula = CH₂O

Molar mass of compound = 150 g/mol

The scaling factor can be obtained as follow:

[CH₂O]n = 150

[12 + (2×1) + 16]n = 150

[12 + 2 + 16]n = 150

30n = 150

Therefore, the scaling factor is 5

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The answer to your question is Scaling factor = 5

Explanation:

Data

Empirical formula CH₂O

molar mass = 150 g/mol

Empirical formula is a chemical formula that shows the simplest proportions of the elements of that formula.

Process

1.- Calculate the molar mass of the Empirical formula

    CH₂O = (12 x 1) + (2 x 1) + (16 x 1)

               = 12 + 2 + 16

               = 30 g

2.- Divide the molar mass of the molecular formula by the molar mass of the empirical formula.

                   150g / 30g = 5      Scaling factor = 5

3.- Molecular formula

                    5(CH₂O) = C₅H₁₀O₅

fe 55.85 g/mol and h2o 18.02 g/mol.

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.