A sample of phosphonitrilic bromide, PNBr2, contains 2.01 mol of the compound. Determine the amount (in mol) of each element present and the number of atoms of each element present in the sample.

2.01 moles of P → 1.21×10²⁴ atoms

2.01 moles of N → 1.21×10²⁴ atoms

4.02 moles of Br → 2.42×10²⁴ atoms

Explanation:

We begin from this relation:

1 mol of PNBr₂ has 1 mol of P, 1 mol of N and 2 moles of Br

Then 2.01 moles of PNBr₂ will have:

2.01 moles of P

2.01 moles of N

4.02 moles of Br

To determine the number of atoms, we use the relation:

1 mol has NA (6.02×10²³) atoms

Then: 2.01 moles of P will have (2.01  . NA) = 1.21×10²⁴ atoms

2.01 moles of N (2.01  . NA) = 1.21×10²⁴ atoms

4.02 moles of Br (4.02 . NA) = 2.42×10²⁴ atoms

For 2.01 moles PNBr2 we have 1.21 *10^24 molecules

P =  1.21 *10^24 atoms

N = 1.21 *10^24 atoms

Br = 2.42*10^24 atoms

Explanation:

Step 1: Data given

Number of moles of PNBr2 = 2.01 moles

Step 2: Calculate moles per element

For 1 mol PNBr2 we have 1 mol P, 1 mol N and 2 moles Br

For 2.01 moles PNBr2 we have:

P = 2.01 moles

N = 2.01 moles

Br = 2* 2.01 = 4.02 moles

Step 3: Calculate atoms of each element

For 2.01 moles PNBr2 we have 6.022 *10^23 * 2.01 = 1.21 *10^24 molecules

P = 2.01 moles = 6.022 *10^23 * 2.01 = 1.21 *10^24 atoms

N = 2.01 moles = 6.022 *10^23 * 2.01 = 1.21 *10^24 atoms

Br = 4.02 moles = 6.022 *10^23 * 4.02 = 2.42*10^24 atoms

answer: semi-infinite solid initially at a uniform temperature of 400 ◦c is cooled at its surface by convection. the cooling fluid at a constant temperature of 200 ◦c has a convective coefficient (ha) of 250 w/m2.k. the physical properties of the solid: k = 20 w/m.k and α = 4 * 10-5  m2/s. using δx = 0.04 m and δt = 10 s. write the finite difference equations describing the transient heat transfer of the first five δx and the temperature profile after 30 s.