Answer: C
Explanation:
Sunlight
6 CO2 + 6 H2O --------------> C6H12O6 + 6 O2
Chlorophyll
Carbon(IV)oxide Water Glucose Oxygen
Boyle's law states that the volume of a fixed mass of a gas is inversely proportional to its temperature if<u> the temperature and the number of particles are constant.</u>
<h3>Further Explanation</h3><h3>Boyles’s law </h3>
- This gas law states that the volume of a fixed mass of a gas is inversely proportional to its pressure at constant absolute temperature.
- Therefore, when the volume of an ideal gas is increased at constant temperature then the pressure of the gas will also increase.
- Mathematically; Volume α 1/Pressure
Vα1/P
- Therefore, constant k, is = PV
<h3>Other gas Laws</h3><h3>Gay-Lussac’s law </h3>
- It states that at constant volume, the pressure of an ideal gas I directly proportional to its absolute temperature.
- Thus, an increase in pressure of an ideal gas at constant volume will result to an increase in the absolute temperature.
<h3>Charles’s law</h3>
- It states that the volume of a fixed mass of a gas is directly proportional to absolute temperature at constant pressure.
- Therefore, an increase in volume of an ideal gas causes a corresponding increase in its absolute temperature and vice versa while the pressure is held constant.
<h3>Dalton’s law </h3>
- It is also known as the Dalton’s law of partial pressure. It states that the total pressure of a mixture of gases is always equivalent to the total sum of the partial pressures of individual component gases.
- Partial pressure refers to the pressure of an individual gas if it occupies the same volume as the mixture of gases.
Keywords: Gas law, Boyles's law, pressure, volume, absolute temperature, ideal gas
<h3>Learn more about:</h3>
Level: High school
Subject: Chemistry
Topic: Gas laws
Sub-topic: Boyle's Law
Answer:
494.1 kPa
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (kPa)
P2 = final pressure (kPa)
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question,
P1 = 294 kPa
P2 = ?
V1 = 42.9 liters
V2 = 22.8 liters
T1 = 76.0°C = 76 + 273 = 349K
T2 = 38.7°C = 38.7 + 273 = 311.7K
294 × 42.9/349 = P2 × 22.8/311.7
12612.6/349 = 22.8 P2/311.7
36.14 = 22.8P2/311.7
Cross multiply
36.14 × 311.7 = 22.8P2
11264.605 = 22.8P2
P2 = 11264.605 ÷ 22.8
P2 = 494.1 kPa
Thank you for posting your question here. Below is the solution:
HNO3 --> H+ + NO3-
<span>HNO3 = strong acid so 100% dissociation </span>
<span>** one doesn't need to find the molarity of water since it is the solvent </span>
<span>0M HNO3 </span>
<span>1x10^-6M H3O+ </span>
<span>1x10^-6M NO3- </span>
<span>1x10^-8M OH-.....the Kw = 1x10^-14 = [H+][OH-] </span>
<span>you have 1x10^-6M H+ so, 1x10^-14 / 1x10^-6 = 1x10^-8M OH- </span>
<span>1x10^-6 Ba(OH)2 = strong base, 100% dissociation </span>
<span>1x10^-6M Ba2+ </span>
<span>2x10^-6M OH- since there are 2 OH- / 1 Ba2+ </span>
<span>0M Ba(OH)2 </span>
<span>5x10^-9M H3O+</span>
Answer:
-85 °C
Explanation:
O and S are in the same group( Group 16). Since S is below O it's atomic mass is higher than O. So molar mass of H2S is higher than H2O. The strength of Vanderwaal Interactions ( London dispersion forces) increases when the molar mass increases. However, only H2O can form H bonds with each other. This is because electronegativity of O is higher than S and therefore H in H2O has a higher partial positive charge than H of H2S.
H bond dominate among these 2 types of forces so the strength of attractions between molecules is higher in H2O than H2S. Therefore more energy should be supplied for H2O to break inter
molecular forces and convert from solid to liquid state than H2S. So mpt of H2O must be higher than that of H2S.
