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PolarNik [594]
3 years ago
7

An example of a disaccharide is saccharin-aspartame molecule True False

Chemistry
1 answer:
Flura [38]3 years ago
6 0
I think is False! Because it is not example of a disaccharide it is not saccharin-aspartame molecule.
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Hiii please help me to balance the chemical equation:
Fed [463]

Answer:

2Na + 2H2O = 2NaOH + H2

Explanation:

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3 years ago
Which statements are correct about silicon?
ki77a [65]

Answer:

A. Metalloid

E. Has similar properties as Ge

F. Belongs to Period 3​

Explanation:

Silicon is the 14th element on the periodic table. Its unit is SI. Its properties straddles between those of metals and non-metals and it is described as a non-metal.

It's atomic weight or mass number is 28u. It has an atomic number of 14 i.e in its neutral state, the number of protons and electrons are equal to 14.

Silicon belongs to the 4th group and the 3rd period on the periodic table. Elements in the same group share similar chemical properties. The elements in Si group are: C, Ge, Sn and Pb. The properties of Si is similar to these elements because they all have a valency of 4. Across the period, the properties varies this is why Si would have a very different property from Al and P.

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3 years ago
Helpppppppppp please <br> what did max planck discovery?
tensa zangetsu [6.8K]

Answer:

Max Planck discovered the quantum of action

Explanation:

4 0
3 years ago
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The mass of an electron is approximately equal to 1 over 1,836 of the mass of 1. a positron 2. a proton 3. an atom of helium 4.
lianna [129]
Hello!

It would be A, the proton.
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Combustion analysis is performed on 0.50 g of a hydrocarbon, and 1.47 g of CO2 and 0.902 g of H2O are produced. What is the empi
JulijaS [17]

1. The empirical formula of the hydrocarbon is CH₃

2. The molecular formula of the hydrocarbon is C₂H₆

<h3>How to determine the mass of Carbon </h3>
  • Mass of CO₂ = 1.47 g
  • Molar mass of CO₂ = 44 g/mol
  • Molar of C = 12 g/mol
  • Mass of C =?

Mass of C = (12 / 44) × 1.47

Mass of C = 0.4 g

<h3>How to determine the mass of H</h3>
  • Mass of compound = 0.5 g
  • Mass of C = 0.4 g
  • Mass of H = ?

Mass of H = (mass of compound) – (mass of C)

Mass of H = 0.5 – 0.4

Mass of H =0.1 g

<h3>1. How to determine the empirical formula </h3>
  • C = 0.4 g
  • H = 0.1 g
  • Empirical formula =?

Divide by their molar mass

C = 0.4 / 12 = 0.03

H = 0.1 / 1 = 0.1

Divide by the smallest

C = 0.03 / 0.03 = 1

H = 0.1 / 0.03 = 3

Thus, the empirical formula of the compound is CH₃

<h3>2. How to determine the molecular formula</h3>
  • Empirical formula = CH₃
  • Molar mass = 30 g/mol
  • Molecular formula =?

Molecular formula = empirical × n = mass number

[CH₃]n = 30

[12 + (3×1)]n = 30

15n = 30

Divide both side by 15

n = 30 / 15

n = 2

Molecular formula = [CH₃]n

Molecular formula = [CH₃]₂

Molecular formula = C₂H₆

Learn more about empirical formula:

brainly.com/question/24297883

#SPJ1

4 0
2 years ago
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