The final volume of the gas that was heated from -25.0 °C to standard temperature is 2.2L.
<h3>How to calculate volume?</h3>
The volume of a given gas can be calculated using the Charles law equation as follows:
V1/T1 = V2/T2
Where;
- V1 = initial volume
- V2 = final volume
- T1 = initial temperature
- T2 = final temperature
- V1 = 2L
- V2 = ?
- T1 = -25°C + 273 = 248K
- T2 = 273K
2/248 = V2/273
273 × 2 = 248V2
546 = 248V2
V2 = 546/248
V2 = 2.2L
Therefore, the final volume of the gas that was heated from -25.0 °C to standard temperature is 2.2L
Learn more about volume at: brainly.com/question/11464844
Since
21.2 g H2O was produced, the amount of oxygen that reacted can be obtained
using stoichiometry. The balanced equation was given: 2H₂ + O₂ → 2H₂O and
the molar masses of the relevant species are also listed below. Thus, the
following equation is used to determine the amount of oxygen consumed.
Molar mass of H2O = 18
g/mol
Molar mass of O2 = 32
g/mol
21.2 g H20 x 1 mol
H2O/ 18 g H2O x 1 mol O2/ 2 mol H2O x 32 g O2/ 1 mol O2 = 18.8444 g O2
<span>We then determine that
18.84 g of O2 reacted to form 21.2 g H2O based on stoichiometry. It is
important to note that we do not need to consider the amount of H2 since we can
derive the amount of O2 from the product. Additionally, the amount of H2 is in
excess in the reaction.</span>
Answer:
The answer to your question is the letter C. three times as much
Explanation:
Data
First step = 6 m
Second step = 18 m
Potential energy is the energy stored that depends on its position.
Formula
Pe = mgh
m = mass; g = gravity; h = height
Potential energy of the first step
Pe1 = 6mg
Potential energy of the second step
Pe2 = 18mg
-Divide the Pe2 by the Pe1
Pe2/Pe1 = 18mg/6mg
= 3
Answer:
it is actually b because i did this i picked b and got it right
Explanation:
Answer:
B. They are dimensionless ratios of the actual concentration or pressure divided by standard state concentration, which is 1 M for solutions and 1 bar for gases.
Explanation:
Activity of a substance is defined as the ratio of an effective concentration or an effective pressure to a standard state pressure or a standard state pressure. It is usually a unit less ratio.
Concentrations in an equilibrium constant are really dimensionless ratios of actual concentrations divided by standard state concentrations. Since standard states are 1 M for solutes, 1 bar for gases, and pure substances for solids and liquids, these are the units to be used.
Hence, activity is a fudge factor to ideal solutions that correct the true concentration. Activity of a gas and solute concentration is a ratio with no unit.
