All categories

3 years ago

Determine whether or not each chemical formula is an empirical formula. C6H12O6 CaCO3 NaMnO4 Ba3(PO4)2 K2C2O4

Chemistry

1 answer:

zavuch27 [327]3 years ago

5 0

Answer:

- C6H12O6 is not an empirical formula because it can be simplified to CH2O.

- CaCO3 is an empirical formula as well as molecular formula because it has been simplified up to the smallest subscript per atom.

- NaMnO4 is an empirical formula as well as molecular formula because it has been simplified up to the smallest subscript per atom.

- Ba3(PO4)2 is an empirical formula as well as molecular formula because it has been simplified up to the smallest subscript per atom.

- K2C2O4 is not an empirical formula because it can be simplified to KCO2.

Explanation:

Hello!

In this case, since the empirical formulas are referred to those formulas that have been simplified up to the smallest whole number for each subscript per atom, we can see that:

- C6H12O6 is not an empirical formula because it can be simplified to CH2O.

- CaCO3 is an empirical formula as well as molecular formula because it has been simplified up to the smallest subscript per atom.

- NaMnO4 is an empirical formula as well as molecular formula because it has been simplified up to the smallest subscript per atom.

- Ba3(PO4)2 is an empirical formula as well as molecular formula because it has been simplified up to the smallest subscript per atom.

- K2C2O4 is not an empirical formula because it can be simplified to KCO2.

Best regards!

You might be interested in

Storing sugar as long chains for later use is an example of a(n) ____________ chemical reaction.

DerKrebs [107]

Answer:

Endothermic

Explanation:

Storing sugar for later use is an example of an endothermic reaction because that energy is being absorbed.

8 0

3 years ago

Hydrocarbons do not dissolve in concentrated sulfuric acid, but methyl benzoate does. Explain this difference and write an equat

finlep [7]

Answer:

See explanation

Explanation:

For a substance to dissolve in another, there must be some sort of interaction between the substances.

Recall that like dissolves like. That is, polar substances dissolve polar substances and non polar substances dissolve nonpolar substances.

Hydrocarbons are nonpolar hence they do not dissolve in polar sulphuric acid. Methyl benzoate is polar hence it dissolve in polar sulphuric acid.

The equation showing the ions is depicted in the image attached to this answer.

7 0

2 years ago

Which scientist performed the cathode ray experiment leading to the discovery of electrons

SVETLANKA909090 [29]

The scientist that performed the cathode ray experiment leading to the discovery of electrons is J.J. Thomson.

<h3>What is a cathode ray?</h3>

A cathode ray is a tube that contains negatively charged electrode( that is the cathode) which emits electrons when heated at a low pressure.

The cathode ray was used by the scientist, J.J. Thomson to find the ratio of charge to mass (e/m) of the electrons.

Therefore, the scientist that performed the cathode ray experiment leading to the discovery of electrons is J.J. Thomson.

Learn more about electrons here:

brainly.com/question/11316046

#SPJ1

6 0

2 years ago

What is the normal boiling point of etanoic acid?

Nezavi [6.7K]

Answer: 244.6°F

Explanation:

3 0

3 years ago

what mass of sodium fluoride (FW=42.0 g/mol) must be added to 3.50 x 10^2 mL of water to give a solution with pH = 8.40?

MaRussiya [10]

Answer:

Explanation:

Sodium fluoride, being a salt, dissolves in water completely producing F ⁻ ions. Now F⁻ is the conjugate base of the weak acid HF, so in water we will have the following equilibrium:

F⁻ + H₂O ⇆ HF + OH⁻

Given this equilibrium, we need to calculate Kb from the Ka for HF, the [ OH ⁻] from the given pH, and finally the mass needed to produce that OH⁻ concentration.

The equilibrium constant, Kb , can be calculated from Kw = Ka x Kb, where Kw = 10⁻¹⁴ and Ka for HF is 6.6 x 10⁻⁴ from reference tables.

Kb = 10⁻¹⁴ / 6.6 x 10⁻⁴ = 1.5 x 10⁻¹¹

pH + pOH = 14 ⇒ pOH = 14 - 8.40 = 5.60

[ OH⁻ ] = 10^-5.60 = 2.51 x 10⁻⁶

Now we have all the information :

F⁻ HF OH⁻

Equilibrium X 2.51 x 10⁻⁶ 2.51 x 10⁻⁶

(2.51 x 10⁻⁶)² / X = 1.5 x 10⁻¹¹ ⇒ X = (2.51 x 10⁻⁶)² / 1.5 x 10⁻¹¹

X = [ F⁻ ] = 0.41 M

For 350 mL ( 0.35 L ) we need to add:

0.41 mol HF/ 1 L * 0.35 L = 0.144 mol

and finally the mass will be:

0.144 mol NaF * 42.0 g/mol NaF = 6.03 g NaF

7 0

3 years ago