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)

1. m(NH₄)₂HPO₄ = 1.62 g

2. molesC₃H₇NO₂ = 1.79 moles

3. V = 4.87x10⁹ 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

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

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 = 1.612x10¹⁰ pounds / 12.53 pounds/gallons

V = 1.287x10⁹ gallons

Converting this to liters:

V = 1.287x10⁹ * 3.7854

Hope this helps

answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.

a:

by the combination of an unlimited number of elements