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  1. Explanation:

    An atom:

    is the smallest whole unit of matter

    can form one or more elements

    is the smallest whole unit of matter with the properties of a specific element

    it's true

  2. There are three main types of chemical formulas: empirical, molecular and structural. Empirical formulas show the simplest whole-number ratio of atoms in a compound, molecular formulas show the number of each type of atom in a molecule, and structural formulas show how the atoms in a molecule are bonded to each other.

  3. Largest whole number that divides evenly into both numbers.

    Step-by-step explanation:

    GCF is greatest common factor. It is also known as LCM of lowest common multiple.

    It is the set of whole numbers that divides evenly into all numbers with zero remainder.

    For example, three numbers are 8,12 and 20.

    The factors of 8 are: 1, 2, 4, 8

    The factors of 12 are: 1, 2, 3, 4, 6, 12

    The factors of 20 are: 1, 2, 4, 5, 10, 20

    The GCF of these three numbers is 4. 4 divides evenly into all numbers.

    Hence, the correct option is (B).

  4. D. The empirical formula expresses the smallest whole number ratio of atoms in a compound.

    Explanation:

    empirical formula is the simplest form of molecular formula. generally empirical formula is divided by some whole number on molecular formula.

    example : C6H1206 is molecular formula while CH2O is empirical formula here moleculer formula is divided by 6.

  5. NO2- + 5H2O + 2Al + OH- → NH3 + 2Al(OH)4-

    There is 1 hydroxide ion, on the reactant side

    Explanation:

    NO2−(aq) + Al(s) → NH3(aq) + Al(OH)4−(aq)

    Step 1: The half reactions

    NO2- (aq) → NH3(g)

    Al(s) → Al(OH)4-

    Step 2: Balancing electrons

    NO2- → NH3

    On the left side N has an oxidation number of +3 and on the right side -3.

    NO2- +6e-→ NH3

    Al(s) → Al(OH)4-

    On the left side, Al has an oxidation number of 0 and on the right side +3.

    Al(s) → Al(OH)4- +3e-

    To have the same amount of electrons transfered, we have to multiply the second reaction by 2

    NO2- +6e-→ NH3

    2(Al(s) → Al(OH)4- +6e-)

    Step 3: Balance with OH/H2O

    NO2- +6e +5H2O → NH3 +7OH-

    2Al +8OH- → 2Al(OH)4- + 6e-

    Step 4: The netto reaction

    NO2- + 5H2O + 2Al + 8OH-  → NH3 +7OH- + 2Al(OH)4-

    NO2- + 5H2O + 2Al + OH- → NH3 + 2Al(OH)4-

    There is 1 hydroxide ion, on the reactant side

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