D
<span>Zinc, Copper,and Iron are elements
</span>
Answer:
0.11 atm
Explanation:
Given data
- Temperature (T): -16 °C + 273.15 = 257 K
- Ideal gas constant (R): 0.082 atm.L/mol.K
First, we will calculate the moles (n) of sulfur hexafluoride, considering its molar mass is 146.06 g/mol.
34 g × (1 mol/146.06 g) = 0.23 mol
Then, we can calculate the pressure (P) of sulfur hexafluoride using the ideal gas equation.
P × V = n × R × T
P = n × R × T / V
P = 0.23 mol × (0.082 atm.L/mol.K) × 257 K / 45 L
P = 0.11 atm
I think that bubbles are very sensitive when it hits the water and that the bubbles can change the flow of the water. I think that bubbles have little affect on land animals but it will possibly change the temp in the air.
Answer: A photon of energy was released.
Explanation: The energy of the orbitals keep on increasing as the size of the orbitals keep on increasing or the value of n keeps on increasing.
The electrons absorb energy when they are irradiated or put at high temperatures and move to higher energy states. The electrons then get back to their ground state by releasing energy in the form of radiations. Thus when an electron move from n = 2 to shell n = 1, the absorbed energy is given off in form of photons of energy.
Energy is released or absorbed in the form of discrete packets of energy called as photons.
The heat of reaction : 50.6 kJ
<h3>Further explanation</h3>
Based on the principle of Hess's Law, the change in enthalpy of a reaction will be the same even though it is through several stages or ways
Reaction
N₂(g) + 2H₂(g) ⇒N₂H₄(l)
thermochemical data:
1. N₂H₄(l)+O₂(g)⇒N₂(g)+2H₂O(l) ΔH=-622.2 kJ
2. H₂(g)+1/2O₂(g)⇒H₂O(l) ΔH=-285.8 kJ
We arrange the position of the elements / compounds so that they correspond to the main reaction, and the enthalpy sign will also change
1. N₂(g)+H₂O(l) ⇒ N₂H₄(l)+O₂(g) ΔH=+622.2 kJ
2. H₂(g)+1/2O₂(g)⇒H₂O(l) ΔH=-285.8 kJ x 2 ⇒
2H₂(g)+O₂(g)⇒2H₂O(l) ΔH=-571.6 kJ
Add reaction 1 and reaction 2, and remove the same compound from different sides
1. N₂(g)+2H₂O(l) ⇒ N₂H₄(l)+O₂(g) ΔH=+622.2 kJ
2.2H₂(g)+O₂(g)⇒2H₂O(l) ΔH=-571.6 kJ
-------------------------------------------------------------------- +
N₂(g) + 2H₂(g) ⇒N₂H₄(l) ΔH=50.6 kJ