Answer: The answer is NOT A. [CO2] / [CO] [O2]
just took the test and I chose A. The test did not tell me the correct answer. Sorry.
Explanation: NOT, I REPEAT NOT: A. [CO2] / [CO] [O2]
The amount of water that will be produced is 50.36 grams
<h3>Stoichiometric problems</h3>
The metabolism of glucose is represented by the following equation:
The mole ratio of glucose metabolized to the water produced is 1:6.
Mole of 84.0 g glucose = 84/180.156 = 0.4662 moles
Equivalent mole of water = 0.4662 x 6 = 2.7975 moles
Mass of 2.7975 moles water = 2.7975 x 18 = 50.36 grams
More on stoichiometric problems can be found here: brainly.com/question/14465605
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Answer:
V₂ = 648.53 mL
Explanation:
Given data:
Initial volume of gas = 490. mL
Initial temperature = -35°C (-35 + 273 = 238 k)
Final temperature = 42°C = (42+273 = 315 k)
Final volume = ?
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 490 mL × 315 K / 238 k
V₂ = 154350 mL.K / 238 K
V₂ = 648.53 mL
Answer: genus
Explanation:
Amoebas do not form a single taxonomic group; instead, they are found in every major lineage of eukaryotic organisms. Amoeboid cells occur not only among the protozoa, but also in fungi, algae, and animals.
Answer:
Δ S = 26.2 J/K
Explanation:
The change in entropy can be calculated from the formula -
Δ S = m Cp ln ( T₂ / T₁ )
Where ,
Δ S = change in entropy
m = mass = 2.00 kg
Cp =specific heat of lead is 130 J / (kg ∙ K) .
T₂ = final temperature 10.0°C + 273 = 283 K
T₁ = initial temperature , 40.0°C + 273 = 313 K
Applying the above formula ,
The change in entropy is calculated as ,
ΔS = m Cp ln ( T₂ / T₁ ) = (2.00 )( 130 ) ln( 283 K / 313 K )
ΔS = 26.2 J/K
