Sublimation and erosion cause ablation, a process that wears down glaciers. Sublimation of iodine may be used to reveal latent fingerprints on paper. Sublimation is used to purify compounds. ... Because dry ice sublimates so readily, the compound is used to produce fog effects.
Every science experiment should follow the basic principles of proper investigation so that the results presented at the end are seen as credible.
Observation and Hypothesis. ...
Prediction and Modeling. ...
Testing and Error Estimation. ...
Result Gathering and Presentation. ...
Conclusions. ...
Law Formation.
The answer is C. Condensation is a <span>change from a less condensed to a more condensed state of matter. It is a phase change from the gas state to liquid state. The gas is the less condensed state and the liquid is the more condensed state as compared to gas.</span>
Answer:
Explanation:
Given
485 L
Required
Determine the measurement not equal to 485L
<em>From standard unit of conversion;</em>
1 KL = 1000 L
<em>Multiply both sides by 485</em>
<em>Divide both sides by 1000</em>
---- This is equivalent
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<em>From standard unit of conversion;</em>
1000 mL = 1 L
<em>Multiply both sides by 485</em>
Convert to standard form
Hence; is not equivalent to 485L
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<em>From standard unit of conversion;</em>
100 cL = 1 L
<em>Multiply both sides by 485</em>
Convert to standard form
---- This is equivalent
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μL
<em>From standard unit of conversion;</em>
1000000 μL = 1 L
<em>Multiply both sides by 485</em>
Convert to standard form
---- This is equivalent
From the list of given options;
is not equivalent to 485L
Answer:
ΔG = -61.5 kJ/mol (<u>Spontaneous process</u>)
Explanation:
2 NO (g) + O₂ (g) ⇄ 2NO₂ (g)
Let's apply the thermodynamic formula to calculate the ΔG
ΔG = ΔG° + R .T . lnQ
We don't know if the gases are at equilibrium, that's why we apply Q (reaction quotient)
ΔG = - 69 kJ/mol + 8.31x10⁻³ kJ/K.mol . 298K . ln Q
How can we know Q? By the partial pressures (Qp)
P NO = 0.450atm
PO₂ = 0.1 atm
PNO₂ = 0.650 atm
Qp = [NO₂]² / [NO]² . [O₂]
Qp = 0.650² / 0.450² . 0.1 = 20.86
ΔG = - 69 kJ/mol + 8.31x10⁻³ kJ/K.mol . 298K . ln 20.86
ΔG = -61.5 kJ/mol (<u>Spontaneous process</u>)
