Answer:
1-ethyl-2-methyl cyclopropane.
Explanation:
- The structure of the molecule will be as shown in the attached image.
- The molecular formula of the compound is C₆H₁₂.
- It has 3 membered ring with 3 C atoms and two substituents one of them with one C atom (methyl) and the other with 2 C atoms (ethyl).
- The ring consist of 3 C atoms, so its name is cyclo propane.
- We numbering the atoms of the ring that give the ethyl substituent the low no. (1) and then methyl group take no. (2).
- <em>Thus, the name of the compound is 1-ethyl-2-methyl cyclopropane.</em>
Answer:
Every object that travel at the speed of light gains infinite mass. Is a law in physics that as it gains speed, it is gaining mass. Then, not is necesary that a meteorite hit the Earth at the speed of light, because just an atom traveling a this velocities could destroy our planet.
Explanation:
The speed of light is really fast so that will mostly happen
I would say C is the most correct.
In D it depends on what water source you're using. Let's say it is a waterfall, then the source of the water (melting ice or a lake) may disappear in the future.
If you're using underwater "windmills" placed in the ocean, then you would expect it to last a while as the ocean will not disappear in the near future.
Ans: 15.1 grams
Given reaction:
Na2CO3 + Ca(OH)2 → 2NaOH + CaCO3
Mass of Na2CO3 = 20.0 g
Molar mass of Na2CO3 = 105.985 g/mol
# moles of Na2CO3 = 20/105.985 = 0.1887 moles
Based on the reaction stoichiometry: 1 mole of Na2CO3 produces 2 moles of NaOH
# moles of NaOH produced = 0.1887*2 = 0.3774 moles
Molar mass of NaOH = 22.989 + 15.999 + 1.008 = 39.996 g/mol
Mass of NaOH produced = 0.3774*39.996 = 15.09 grams
Answer:
8.37 grams
Explanation:
The balanced chemical equation is:
C₆H₁₂O₆ ⇒ 2 C₂H₅OH (l) + 2 CO₂ (g)
Now we are asked to calculate the mass of glucose required to produce 2.25 L CO₂ at 1atm and 295 K.
From the ideal gas law we can determine the number of moles that the 2.25 L represent.
From there we will use the stoichiometry of the reaction to determine the moles of glucose which knowing the molar mass can be converted to mass.
PV = nRT ⇒ n = PV/RT
n= 1 atm x 2.25 L / ( 0.08205 Latm/kmol x 295 K ) =0.093 mol CO₂
Moles glucose required:
0.093 mol CO₂ x ( 1 mol C₆H₁₂O₆ / 2 mol CO₂ ) = 0.046 mol C₆H₁₂O₆
The molar mass of glucose is 180.16 g/mol, then the mass required is
0.046 mol x 180.16 g/mol = 8.37 g
