Answer:
-138.9 kJ/mol
Explanation:
Step 1: Convert 235.8°C to the Kelvin scale
We will use the following expression.
K = °C + 273.15 = 235.8°C + 273.15 = 509.0 K
Step 2: Calculate the standard enthalpy of reaction (ΔH°)
We will use the following expression.
ΔG° = ΔH° - T.ΔS°
ΔH° = ΔG° / T.ΔS°
ΔH° = (-936.92kJ/mol) / 509.0K × 0.51379 kJ/mol.K
ΔH° = -3.583 kJ (for 1 mole of balanced reaction)
Step 3: Convert -9.9°C to the Kelvin scale
K = °C + 273.15 = -9.9°C + 273.15 = 263.3 K
Step 4: Calculate ΔG° at 263.3 K
ΔG° = ΔH° - T.ΔS°
ΔG° = -3.583 kJ/mol - 263.3 K × 0.51379 kJ/mol.K
ΔG° = -138.9 kJ/mol
Answer:
You have read that EM waves can interact with a material medium in
the same ways that mechanical waves do. Three forms of interaction
play an especially important role in how people see light. One form is
reflection. Most things are visible because they reflect light. The two
other forms of interaction are transmission and absorption.
Transmission is the passage of an EM wave
through a medium. If the light reflected from objects did not pass
through the air, windows, or most of the eye, we could not see the
objects. Absorption is the disappearance of an
EM wave into the medium. Absorption affects how things look, because
it limits the light available to be reflected or transmitted.
Answer:
B) Heat is given off.
Explanation:
When heat is given off, a chemical change must have occurred and such a reaction is termed an exothermic reaction.
A chemical change is one in which;
- the process is not easily reversible
- leads to the production of new kinds of matter
- involves change in mass
- requires a considerable amount of energy.
Most chemical reactions are accompanied by heat changes.
You would need to utilize Mole ratios found in the adjusted condition;
for each mole of hydrogen utilized, 2 moles of HCl are delivered.
Thusly:
10 mol H2 x 2 mol HCl/1 mol H2 = 20 mol HCL.
For the second question:
you would need to change over 2.0x10^23 particles of Oxygen to moles of oxygen, utilizing Avogadro's number:
2.0x10^23 particles oxygen x 1 mol oxygen/6.022x10^23 atoms oxygen = 0.33 mol Oxygen
utilizing mole proportions once more:
0.66 mol H2O = 2 mol H2O/1 mol Oxygen x 0.33 mol Oxygen
45.0 mol H2O = 2 mol H2O/1 mol Oxygen x 22.5 mol Oxygen
fundamentally to answer stoichiometry, you should take a gander at the adjusted condition to make sense of the mole proportions between components/mixes, and utilizing mole proportions you can change over from moles of one component/compound to moles of another component/compound
