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Answers

  • Réponse publiée par: maledabacuetes
    ✏️ANSWER✏️Hindi ako tamad, Sadyang masipag lang akong matulog.
    Hindi ako tamad =)pre puts
  • Réponse publiée par: batopusong81

    a. 8.9x10^5

    b. 3.1x10^-8

    c.1.456x10^3

    d.6.7x10^2

  • Réponse publiée par: rhaineandreirefuerzo

    Hello!

    A molecule is a chemical combination of two or more atoms. which contains the largest number of molecules?  

    A.1.0g CH4 (MM=16 g/mol)

    B.1.0g HNO3 (MM=63 g/mol)

    C.1.0g H20 (MM=18g/mol)

    D.1.0g N204 (MM=92 g/mol)

    Solving:

    A)

    m1 (solute mass - CH4) = 1.0 g

    MM (molar mass - CH4) = 16 g/mol

    n (number of mols of CH4) = ? (in mol)

    Let's find the number of mol, let's see:

    n = \dfrac{m_1}{MM}

    n = \dfrac{1.0\:\diagup\!\!\!\!\!g}{16\:\diagup\!\!\!\!\!g/mol} = \boxed{0.0625\:mol\:of\:CH4}

    Knowing that by Avogadro's Law for each mole of a substance we have 6.022 * 10²³ molecules, then:

    1 mol -------------- 6.022*10²³ molecules

    0.0625 mol ------------------ y molecules

    \dfrac{1}{0.0625} = \dfrac{6.022*10^{23}}{y}

    multiply the means by the extremes

    0.0625*6.022*10^{23} = 1*y

    3.76375*10^{22} = y

    \boxed{\boxed{y = 3.76375*10^{22}\:molecules\:of\:CH_4}}\:\:\:\:\:\:\bf\green{\checkmark}

    B)

    m1 (solute mass - HNO3) = 1.0 g

    MM (molar mass - HNO3) = 63 g/mol

    n (number of mols of HNO3) = ? (in mol)

    Let's find the number of mol, let's see:

    n = \dfrac{m_1}{MM}

    n = \dfrac{1.0\:\diagup\!\!\!\!\!g}{63\:\diagup\!\!\!\!\!g/mol} \approx \boxed{0.0158\:mol\:of\:HNO_3}

    Knowing that by Avogadro's Law for each mole of a substance we have 6.022 * 10²³ molecules, then:

    1 mol -------------- 6.022*10²³ molecules

    0.0158 mol ------------------ y molecules

    \dfrac{1}{0.0158} = \dfrac{6.022*10^{23}}{y}

    multiply the means by the extremes

    0.0158*6.022*10^{23} = 1*y

    9.51476*10^{21} = y

    \boxed{\boxed{y = 9.51476*10^{21}\:molecules\:of\:HNO_3}}

    C)

    m1 (solute mass - H2O) = 1.0 g

    MM (molar mass - H2O) = 18 g/mol

    n (number of mols of H2O) = ? (in mol)

    Let's find the number of mol, let's see:

    n = \dfrac{m_1}{MM}

    n = \dfrac{1.0\:\diagup\!\!\!\!\!g}{18\:\diagup\!\!\!\!\!g/mol} \approx \boxed{0.0555\:mol\:of\:H_2O}

    Knowing that by Avogadro's Law for each mole of a substance we have 6.022 * 10²³ molecules, then:

    1 mol -------------- 6.022*10²³ molecules

    0.0555 mol ------------------ y molecules

    \dfrac{1}{0.0555} = \dfrac{6.022*10^{23}}{y}

    multiply the means by the extremes

    0.0555*6.022*10^{23} = 1*y

    3.34221*10^{22} = y

    \boxed{\boxed{y = 3.34221*10^{22}\:molecules\:of\:H_2O}}

    D)

    m1 (solute mass - N2O4) = 1.0 g

    MM (molar mass - N2O4) = 92 g/mol

    n (number of mols of N2O4) = ? (in mol)

    Let's find the number of mol, let's see:

    n = \dfrac{m_1}{MM}

    n = \dfrac{1.0\:\diagup\!\!\!\!\!g}{92\:\diagup\!\!\!\!\!g/mol} \approx \boxed{0.0108\:mol\:of\:N_2O_4}

    Knowing that by Avogadro's Law for each mole of a substance we have 6.022 * 10²³ molecules, then:

    1 mol -------------- 6.022*10²³ molecules

    0.0108 mol ------------------ y molecules

    \dfrac{1}{0.0108} = \dfrac{6.022*10^{23}}{y}

    multiply the means by the extremes

    0.0108*6.022*10^{23} = 1*y

    6.50376*10^{21} = y

    \boxed{\boxed{y = 6.50376*10^{21}\:molecules\:of\:N_2O_4}}

    Therefore:

    (A) 3.76375*10²² > (C) 3.34221*10²² > (B) 9.51476*10²¹ > (D) 6.50376*10²¹

    A.1.0g CH4 (MM=16 g/mol)

    The "A" alternative contains the largest number of molecules (3.76375*10²² molecules).

    ________________________

    \bf\gray{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

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