I have 3 questions that I will need help from you. 1. The weight of a millimeter of (NH4)2 HPO4 is? 2. How many moles of alanine, C3H7NO2 ar

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I have 3 questions that I will need help from you. 1. The weight of a millimeter of (NH4)2 HPO4 is? 2. How many moles of alanine, C3H7NO2 are there in 159g of alanine? And 3. Nitric acid is a very important industrial chemical. 1.612 * 10^10 pounds of it were produced in 1992. If the density of nitric acid is 12.53 pounds/gallon. What volume would be occupied by this quantity. (1 gallon =3.7854liters)

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Ngọc Hoa 5 months 2021-08-11T11:30:17+00:00 1 Answers 4 views 0

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    2021-08-11T11:31:47+00:00

    Answer:

    1. m(NH₄)₂HPO₄ = 1.62 g

    2. molesC₃H₇NO₂ = 1.79 moles

    3. V = 4.87×10⁹ L

    Explanation:

    These questions are pretty easy to solve, using the correct expressions and formula.

    1. To do this, we just need the density of the (NH₄)₂HPO₄. In this case, the reported density is 1.62 g/mL, therefore:

    d = m/V  (1)

    Where:

    d: density (g(mL)

    m: mass of the compound (g)

    V: volume of the compound (mL)

    As we already have the volume, we can solve for m:

    m = d * V  (2)

    m = 1.62 * 1

    m(NH₄)₂HPO₄ = 1.62 g

    2. To do this, we need the atomic weights of each element that composes the alanine.

    C: 12 g/mol;  H: 1 g/mol;   N: 14 g/mol;     O: 16 g/mol

    With these atomic weight, let’s determine the molar mass of alanine:

    MM C₃H₇NO₂ = 3(12) + 7(1) + 14 + 2(16) = 89 g/mol

    Now to get the moles, we use the following expression:

    moles = m/MM (3)

    moles = 159 / 89

    molesC₃H₇NO₂ = 1.79 moles

    3. We have the mass and density of the nitric acid, we just solve for Volume from expression (1) of part 1:

    V = m/d  (3)

    V = 1.612×10¹⁰ pounds / 12.53 pounds/gallons

    V = 1.287×10⁹ gallons

    Converting this to liters:

    V = 1.287×10⁹ * 3.7854

    V = 4.87×10⁹ L

    Hope this helps

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