Your cell membrane is found in all cells, yes. Why do you ask? Well... technically why do you put a question mark down in a sentence?
1. A polymer of simple sugars ≡ polysaccharide
2. A simple sugar (C_6H_12O_6) occurring in plant and animal tissues
≡ glucose
3. A two-sugar polymer ≡ disaccharide
4. The building block of carbohydrates ≡ a simple sugar monosaccharide
5. A bonding of subunits to form a polymer by the loss of water
≡ dehydration synthesis
6. A sugar or polymer of sugar ≡ carbohydrate
Answer:
ΔH3 = -110.5 kJ.
Explanation:
Hello!
In this case, by using the Hess Law, we can manipulate the given equation to obtain the combustion of C to CO as shown below:
C(s) + 1/2O2(g) --> CO(g)
Thus, by letting the first reaction to be unchanged:
C(s) + O2(g)--> CO2 (g) ; ΔH1 = -393.5 kJ
And the second one inverted:
CO2(g) --> CO(g) + 1/2O2(g) ; ΔH2= 283.0kJ
If we add them, we obtain:
C(s) + O2(g) + CO2(g) --> CO(g) + CO2 (g) + 1/2O2(g)
Whereas CO2 can be cancelled out and O2 subtracted:
C(s) + 1/2O2(g) --> CO(g)
Therefore, the required enthalpy of reaction is:
ΔH3 = -393.5 kJ + 283.0kJ
ΔH3 = -110.5 kJ
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Answer:
38.3 miles
Explanation:
First, we <u>convert 1.95 gallons to mililiters</u>:
- 1.95 gallons * = 7380.75 mL
Then we <u>calculate how many grams of octane are available for the reaction</u>, using its density:
- 0.6916 g/mL * 7380.75 mL = 5104.53 g C₈H₁₈
Now we <u>convert octane grams into octane moles</u>, using its molar mass:
- 5104.53 g C₈H₁₈ ÷ 114 g/mol = 44.78 mol C₈H₁₈
Then we <u>calculate how many kJ are produced from the combustion of 44.78 mol C₈H₁₈</u>, <em>if 2 moles produce 10900 kJ</em>:
- 44.78 mol * 10900 kJ / 2 mol = 244032 kJ
We<u> calculate how many seconds is the car available to keep going</u>, <em>if it spends 115 kJ per second</em>:
- 244032 kJ * 1 s / 115 kJ = 2122.02 s
We convert seconds to hours:
- 2122.02 / 3600 = 0.59 hours
Finally we calculate the distance:
- 65 mi/hour * 0.59 hour = 38.3 mi