We read your photo as:

Which of the following allotropes of carbon is used in the rubber industry?

Chemistry

ScanSolve AI Answer

Answer

D

57900 C

Step-by-step solution

6 steps

Here are the answers to the questions:

18. How many coulombs of electricity would be needed to deposit 5.4 g of aluminium by the electrolysis of aluminium oxide dissolved in molten cryolite? [Al = 27.0, F = 96500 C mol⁻¹]

  1. Write the half-reaction for the deposition of aluminium.

    Al3++3e−⟶Al\text{Al}^{3+} + 3\text{e}^- \longrightarrow \text{Al}

    From the reaction, 3 moles of electrons are required to deposit 1 mole of aluminium.

  2. Calculate the number of moles of aluminium to be deposited.

    Moles of Al=Mass of AlMolar mass of Al\text{Moles of Al} = \frac{\text{Mass of Al}}{\text{Molar mass of Al}} Moles of Al=5.4 g27.0 g/mol\text{Moles of Al} = \frac{5.4 \text{ g}}{27.0 \text{ g/mol}} Moles of Al=0.20 mol\text{Moles of Al} = 0.20 \text{ mol}
  3. Calculate the total charge (coulombs) required.

    Since 1 mole of Al requires 3 moles of electrons, and 1 mole of electrons carries 96500 C (Faraday's constant):

    Charge=Moles of Al×Moles of e− per mole of Al×Faraday’s constant\text{Charge} = \text{Moles of Al} \times \text{Moles of e}^- \text{ per mole of Al} \times \text{Faraday's constant} Charge=0.20 mol Al×3 mol e−/mol Al×96500 C/mol e−\text{Charge} = 0.20 \text{ mol Al} \times 3 \text{ mol e}^-/\text{mol Al} \times 96500 \text{ C/mol e}^- Charge=0.60×96500 C\text{Charge} = 0.60 \times 96500 \text{ C} Charge=57900 C\text{Charge} = 57900 \text{ C}

    The correct answer is D. 57900 C.

    19. Which of the following allotropes of carbon is used in the rubber industry?
    Carbon black is a form of amorphous carbon that is widely used as a reinforcing filler in rubber products, such as tires, to improve their strength, durability, and wear resistance.
    The correct answer is D. Carbon black.

    20. The two rows of elements placed below the main body of the periodic table show the
    I. lanthanides
    II. actinides
    III. f-block elements
    The two rows of elements placed below the main body of the periodic table are the lanthanides and the actinides. These elements are also collectively known as the f-block elements because their valence electrons occupy the f-orbitals.
    The correct answer is D. I, II and III.

    21. The process of converting glucose to ethanol by the use of yeast is known as
    Fermentation is a metabolic process that converts sugar (like glucose) into acids, gases, or alcohol (like ethanol). Yeast is commonly used to carry out this process.
    The correct answer is D. fermentation.

    22. The loss of some or all of water of crystallization when a hydrated compound is exposed to air is called
    Efflorescence is the process by which a hydrated salt loses its water of crystallization when exposed to dry air, often resulting in a powdery coating.
    The correct answer is D. efflorescence.

    23. The measure of the tendency of an atom or molecule to attract shared electrons to itself is
    Electronegativity is a chemical property that describes the tendency of an atom or a functional group to attract electrons (or electron density) towards itself from a shared covalent bond.
    The correct answer is A. electronegativity.

    24. What is the molecular mass for CuSO4⋅5H2O\text{CuSO}_4 \cdot 5\text{H}_2\text{O}? [Cu = 63.5, S = 32.0, O = 16.0, H = 1.0]

  4. Calculate the molar mass of CuSO4\text{CuSO}_4.

    Molar mass of Cu=1×63.5=63.5\text{Molar mass of Cu} = 1 \times 63.5 = 63.5 Molar mass of S=1×32.0=32.0\text{Molar mass of S} = 1 \times 32.0 = 32.0 Molar mass of O=4×16.0=64.0\text{Molar mass of O} = 4 \times 16.0 = 64.0 Molar mass of CuSO4=63.5+32.0+64.0=159.5 g/mol\text{Molar mass of CuSO}_4 = 63.5 + 32.0 + 64.0 = 159.5 \text{ g/mol}
  5. Calculate the molar mass of 5H2O5\text{H}_2\text{O}.

    Molar mass of H2O=(2×1.0)+(1×16.0)=2.0+16.0=18.0 g/mol\text{Molar mass of H}_2\text{O} = (2 \times 1.0) + (1 \times 16.0) = 2.0 + 16.0 = 18.0 \text{ g/mol} Molar mass of 5H2O=5×18.0=90.0 g/mol\text{Molar mass of } 5\text{H}_2\text{O} = 5 \times 18.0 = 90.0 \text{ g/mol}
  6. Add the molar masses to find the total molecular mass of CuSO4⋅5H2O\text{CuSO}_4 \cdot 5\text{H}_2\text{O}.

    Total molecular mass=Molar mass of CuSO4+Molar mass of 5H2O\text{Total molecular mass} = \text{Molar mass of CuSO}_4 + \text{Molar mass of } 5\text{H}_2\text{O} Total molecular mass=159.5 g/mol+90.0 g/mol\text{Total molecular mass} = 159.5 \text{ g/mol} + 90.0 \text{ g/mol} Total molecular mass=249.5 g/mol\text{Total molecular mass} = 249.5 \text{ g/mol}

    The correct answer is A. 249.5 g.

    25. All pure substances which are not elements are
    Pure substances are classified into two categories: elements (which cannot be broken down into simpler substances by chemical means) and compounds (which are formed when two or more elements are chemically combined in fixed proportions). Therefore, any pure substance that is not an element must be a compound.
    The correct answer is A. compounds.

    26. A neutron striking an unstable nucleus may split it, causing the process known as
    Nuclear fission is a nuclear reaction in which the nucleus of an atom splits into two or more smaller, lighter nuclei, often producing gamma photons, free neutrons, and a large amount of energy. This process is typically initiated by bombarding a heavy, unstable nucleus with a neutron.
    The correct answer is A. nuclear fission.

    27. The valence electrons in magnesium are in the
    Magnesium (Mg) has an atomic number of 12. Its electron configuration is 1s22s22p63s21s^2 2s^2 2p^6 3s^2. The valence electrons are the electrons in the outermost shell, which is the third shell in this case. These electrons are in the 3s orbital.
    The correct answer is D. 3s orbital.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp