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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 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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Chemistry
3 years
2021-08-11T11:30:17+00:00
2021-08-11T11:30:17+00:00 1 Answers
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## Answers ( )

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 g2. 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 moles3. 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 LHope this helps