Answer:
The change in T° is 47.1 °C
Explanation:
Calorimetry formula to solve this:
Q = m . C . ΔT
We replace the data gien:
635 J = 3.50×10⁻² kg . 0.385 J/g°C . ΔT
In units of C, we have g and the mass (m) is in kg. So let's convert it from kg to g → 3.50×10⁻² kg . 1000 g / 1kg = 35 g. Now we can determine the ΔT:
635 J = 35 g . 0.385 J/g°C . ΔT
635 J / 35 g . 0.385 J/g°C = ΔT
47.1°C = ΔT
Answer:
1. Orbital diagram
2p⁴ ║ ↑↓ ║ "↑" ║ ↑
2s² ║ ↑↓ ║
1s² ║ ↑↓ ║
2. Quantum numbers
- <em>n </em>= 2,
- <em>l</em> = 1,
= 0,
= +1/2
Explanation:
The fill in rule is:
- Follow shell number: from the inner most shell to the outer most shell, our case from shell 1 to 2
- Follow the The Aufbau principle, 1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d<7p
- Hunds' rule: Every orbital in a sublevel is singly occupied before any orbital is doubly occupied. All of the electrons in singly occupied orbitals have the same spin (to maximize total spin).
So, the orbital diagram of given element is as below and the sixth electron is marked between " "
2p⁴ ║ ↑↓ ║ "↑" ║ ↑
2s² ║ ↑↓ ║
1s² ║ ↑↓ ║
The quantum number of an electron consists of four number:
- <em>n </em>(shell number, - 1, 2, 3...)
- <em>l</em> (subshell number or orbital number, 0 - orbital <em>s</em>, 1 - orbital <em>p</em>, 2 - orbital <em>d...</em>)
(orbital energy, or "which box the electron is in"). For example, orbital <em>p </em>(<em>l</em> = 1) has 3 "boxes", it was number from -1, 0, 1. Orbital <em>d</em> (<em>l </em>= 2) has 5 "boxes", numbered -2, -1, 0, 1, 2
(spin of electron), either -1/2 or +1/2
In our case, the electron marked with " " has quantum number
- <em>n </em>= 2, shell number 2,
- <em>l</em> = 1, subshell or orbital <em>p,</em>
= 0, 2nd "box" in the range -1, 0, 1
= +1/2, single electron always has +1/2
Ammonia is formed by a reaction between hydrogen and nitrogen as shown by the equation below.
N2(g) + 3H2(g) = 2NH3(g)
1 mole of ammonia contains 17 g
Therefore 10.78 g of ammonia are equivalent to 10.78/17 = 0.6341 moles
The mole ratio of hydrogen to ammonia is 3 : 2
Therefore, moles of hydrogen used will be 0.6341 × 3/2 = 0.9512 moles
1 mole of hydrogen is equivalent to 2 g
Thus, the mas of hydrogen will be 0.9512 moles × 2 = 1.9023 g