It's hard to relate a mole to carbon or sulfur. Imagine if I walked up to you and said, "What's the relation between a dozen and donuts?"
A mole is a form of measurement for atoms, more specifically, 6.02 * 10^23 atoms. I suppose you could relate it to Carbon or Sulfur, since the number of atoms of each are usually measured in moles.
Carbon and Sulfur don't have a set number of moles (Just like donuts don't have to be a dozen), so it's hard to answer your second question.
In the atomic table, the number you see under the element is the molar mass, which is the weight of an a mole of the element. In this way, I guess there's a mole of Carbon and Sulfur present, if we're looking at the periodic table.
-T.B.
Answer:
Average density for method A = 2.4 g/cm³
Average density for method B = 2.605 g/cm³
Explanation:
In order to calculate the average density for each method, we need to add the data for each method, and then divide the result by the number of measurements (in this case is 4 for both methods):
Σ = 2.2 + 2.3 + 2.7 + 2.4 = 9.6
Average = 9.6/4 = 2.4 g/cm³
Σ = 2.603 + 2.601 + 2.605 + 2.611 = 10.420
Average = 10.420/4 = 2.605 g/cm³
Answer:
-179.06 kJ
Explanation:
Let's consider the following balanced reaction.
HCl(g) + NaOH(s) ⟶ NaCl(s) + H₂O(l)
We can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.
ΔH°r = 1 mol × ΔH°f(NaCl(s)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(NaOH(s))
ΔH°r = 1 mol × (-411.15 kJ/mol) + 1 mol × (-285.83 kJ/mol) - 1 mol × (-92.31 kJ/mol) - 1 mol × (-425.61 kJ/mol)
ΔH°r = -179.06 kJ
Answer:
616,0 ng is the right answer.
Explanation:
You should know that 1 mole = 1 .10^9 nanomoles
Get the rule of three.
1 .10^9 nanomoles ...................... 56.0 gr
11 nanomoles .....................
(11 x 56) / 1 .10^9 nanomoles = 6.16 x 10^-7 gr
Let's convert
6.16 x 10^-7 gr x 1 .10^9 = 616 ngr
Answer:
See explanation below
Explanation:
The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.
Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.
For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)
For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.