Answer: i could be incorrect but im pretty sure the answer is, Gas surface with no rings.
Explanation:
s-s-s-science!
Answer:

Explanation:
We must do the conversions
mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
Mᵣ: 180.16
C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O
m/g: 24.5
(a) Moles of C₆H₁₂O₆

(b) Moles of CO₂

(c) Volume of CO₂
We can use the Ideal Gas Law.
pV = nRT
Data:
p = 0.960 atm
n = 0.8159 mol
T = 37 °C
(i) Convert the temperature to kelvins
T = (37 + 273.15) K= 310.15 K
(ii) Calculate the volume

Answer:
2
Explanation:
to separate objects or ideas into group based on ways they are alike
Answer:
<u><em>Pentane </em></u>
Explanation:
since we have in here CH3-CH2-CH2-CH2-CH3 5 Carbon atoms and 12 Hydrogen making it 
Answer:
The Bayer process is the principal industrial means of refining bauxite to produce alumina (aluminium oxide) and was developed by Carl Josef Bayer. Bauxite, the most important ore of aluminium, contains only 30–60% aluminium oxide (Al2O3), the rest being a mixture of silica, various iron oxides, and titanium dioxide.