A hypothesis
-Can be tested
-Is a prediction about the expected outcome of an experiment
-Must be stated in a form that can be either proven or disproven
Your answer is D: All of the above are true
Answer:
2–methylpropene.
Explanation:
To successfully name the compound given in the question, we must observe the following:
1. Determine the functional group of the compound.
2. Locate the longest continuous carbon chain. This gives the parent name of the compound.
3. Identify the substituent group attached and locate it's position by giving it the lowest possible count.
4. Combine the above to obtain the name of the compound.
Now, let us determine the name of the compound. This is illustrated below:
1. The functional group of the compound is the double bond i.e the compound is an alkene.
2. The longest continuous carbon chain is 3 i.e propene since it is an alkene.
3. The substituent group attached is methyl i.e CH3. In this case, we'll start counting from the side of the double bond being the functional group. Therefore, the methyl group i.e CH3 is at carbon 2.
4. Therefore, the name of the compound is:
2–methylpropene
4.20 mol Al would react completely with 4.20 x (1/2) = 2.10 mol Fe2O3, but there is not that much Fe2O3 present, so Fe2O3 is the limiting reactant. (1.75 mol Fe2O3) x (2/1) x ( 55.8452 g Fe/mol) = 195 g Fe 3 MgO + 2 H3PO4 → Mg3(PO4)2 + 3 H2O (15.0 g MgO) / (40.3045 g MgO/mol) = 0.37217 mol MgO (18.5 g H3PO4) / (97.9953 g H3PO4/mol) = 0.18878 mol H3PO4 0.18878 mol H3PO4 would react completely with 0.18878 x (3/2) = 0.28317 mole of MgO, but there is more MgO present than that, so MgO is in excess and H3PO4 is the limiting reactant. Now we must consider why the problem tells us "17.6g of Mg3(PO4)2 is obtained". The first possibility is that it's just there for the sake of confusion -- in which case ignore it and proceed this way: ((0.37217 mol MgO initially) - (0.28317 mole MgO reacted)) x (40.3045 g MgO/mol) = 3.59 g MgO left over However, if the amount of magnesium phosphate obtained is given because the reaction was stopped before it was complete, the amount obtained governs the amount reacted and the amount left over, so proceed this way: (17.6g Mg3(PO4)2) / (262.8581 g Mg3(PO4)2/mol) x (3/1) = 0.20087 mol MgO reacted ((0.37217 mol MgO initially) - (0.20087 mole MgO reacted)) x (40.3045 g MgO/mol) = 6.90 MgO left over
Answer:
Pneumonoultramicroscopicsilicovolcanoconiosis
Explanation: