Answer: Glucose is a simple sugar with six carbon atoms and one aldehyde group.
Explanation: hope this helps you
It will approximately take 2 hours and 50 minutes
Answer:
hexane
Explanation:
This is a 6- carbon hydrocarbon with no multiple bonds or any functional groups (such as -OH). Thus, the prefix 'hex' refers to the 6 carbons and 'ane' refers to the molecule being an alkane.
Naming molecules:
<u>Number of </u><u>carbons</u>
• pentane: 5 carbons
• hexane: 6 carbons
• heptane: 7 carbons
• octane: 8 carbons
• nonane: 9 carbons
<u>Functional </u><u>groups</u> (for 6- carbons molecules)
• Alkene (C=C): hexene
• Alcohol (-OH): hexanol
• Alkyne (C≡C): hexyne
Given data:
Distance travelled (d)= 200.0 m
Time taken (t) = 21.7 s
To determine:
The average speed in km/hr
Calculation:
Convert distance from m to km
1000 m = 1 km
200.0 m = 1 km * 200.0 m/1000 m = 0.2 km
Convert time from sec to hours
3600 sec = 1 hour
21.7 sec = 1 hour * 21.7 sec/3600 sec = 6.027 *10^-3 hrs
Speed is defined as the amount of distance travelled per unit time
Speed = distance/time = 0.2 km/6.027*10^-3 hr
= 33.18 km/hr (or 33.2 km/hr)
Answer:
Explanation:
Let's examine this reaction, especially the coefficients.
First of all, remember that reactants are used to make the product. They are found on the left of the arrow usually. Therefore, N₂ and H₂ are the reactants, while NH₃ is the product. We can automatically eliminate NH₃ from our answer choices, because it is simply not a reactant.
Next, look at the coefficients.
- N₂ has no coefficient, so a 1 is implied.
- H₂ has a coefficient of 3.
Therefore, for the reaction to work, there must be 1 mole of N₂ and 3 moles of H₂.
We have 3.2 moles of N₂ and 5.4 moles of H₂.
Divide each amount given by the required amount for completion.
- 3.2 mol N₂/ 1 mol N₂= 3.2 times
- 5.4 mol H₂/ 3 mol H₂=1.8 times
Therefore, there is enough nitrogen to complete the reaction 3.2 times, but only enough hydrogen for 1.8 times. If everything is completely reacted, we will run out of hydrogen and have excess <u>nitrogen</u>.
