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nadezda [96]
3 years ago
11

Both fructose and glucose have an empirical formula of CH2O and a molecular mass of 180.15948 g/mol. Determine the molecular for

mula.
A)
CH2O


B)
C3H8O3


C)
C6H12O6


D)
C8H16O8
Chemistry
1 answer:
Basile [38]3 years ago
5 0

Answer:

C) C₆H₁₂O₆.

Explanation:

  • We can determine the molecular formula by calculating the molecular mass of different choices.

Molecular mass = ∑(no. of atoms * atomic mass).

A) CH₂O:

molecular mass = atomic mass of C + 2*atomic mass of H + atomic mass of O = (12.0 g/mol) + (2 * 1.0 g/mol) + (16.0 g/mol) = 30.0 g/mol.

B) C₃H₈O₃:

molecular mass = 3(atomic mass of C) + 8(atomic mass of H) + 3(atomic mass of O) = 3(12.0 g/mol) + 8(1.0 g/mol) + 3(16.0 g/mol) = 92.0 g/mol.

C) C₆H₁₂O₆:

molecular mass = 6(atomic mass of C) + 12(atomic mass of H) + 6(atomic mass of O) = 6(12.0 g/mol) + 12(1.0 g/mol) + 6(16.0 g/mol) = 180.0 g/mol.

D) C₈H₁₆O₈:

molecular mass = 8(atomic mass of C) + 16(atomic mass of H) + 8(atomic mass of O) = 8(12.0 g/mol) + 16(1.0 g/mol) + 8(16.0 g/mol) = 240.0 g/mol.

<em>So, the right choice is: C) C₆H₁₂O₆.</em>

<em></em>

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Bromine vs Chlorine | Br vs Cl

 

Halogens are group VII elements in the periodic table, and all are electronegative elements and have the capability to produce -1 anions.

Bromine

Bromine is denoted by the symbol Br. This is in the 4th period of the periodic table between chlorine and iodine halogens. Its electronic configuration is [Ar] 4s2 3d10 4p5. The atomic number of bromine is 35. Its atomic mass is 79.904. Bromine staChlorine is an element in the periodic table which is denoted by Cl.  It is a halogen (17th group) in the 3rd period of the periodic table. The atomic number of chlorine is 17; thus, it has seventeen protons and seventeen electrons. Its electron configuration is written as 1s2 2s2 2p6 3s2 3p5. Since the p sub level should have 6 electrons to obtain the Argon, noble gas electron configuration, chlorine has the ability to attract an electron. ys as a red-brown color liquid at room temperature.

8 0
3 years ago
If 12.5 grams of strontium hydroxide is reacted with 150 mL of 3.5 M carbonic acid, identify the limiting reactant.
vesna_86 [32]

Answer:

Sr(OH)2

Explanation:

We'll begin by calculating the number of mole of carbonic acid in 150mL of 3.5 M carbonic acid solution. This is illustrated below:

Molarity = 3.5M

Volume = 150mL = 150/1000 = 0.15L

Mole of carbonic acid, H2CO3 =..?

Mole = Molarity x Volume

Mole of carbonic acid, H2CO3 = 3.5 x 0.15 = 0.525 mole.

Next, we shall convert 0.525 mole of carbonic acid, H2CO3 to grams.

Mole of H2CO3 = 0.525 mole

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 =..?

Mass = mole x molar mass

Mass of H2CO3 = 0.525 x 62 = 32.55g

Next, we shall write the balanced equation for the reaction. This is given below:

Sr(OH)2 + H2CO3 → SrCO3 + 2H2O

Next, we shall determine the mass of Sr(OH)2 and H2CO3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Sr(OH)2 = 88 + 2(16 + 1) = 88 + 2(17) = 122g/mol

Mass of Sr(OH)2 from the balanced equation = 1 x 122 = 122g

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 from the balanced equation = 1 x 62 = 62g.

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Therefore, 12.5g of Sr(OH)2 will react with = (12.5 x 62)/122 = 6.35g.

We can see evidently from the calculations made above that it will take 6.35g out 32.55g of H2CO3 to react with 12.5g of Sr(OH)2. Therefore, Sr(OH)2 is the limiting reactant and H2CO3 is the excess reactant

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