Answer:
Pentane or 2,2-dimethylbutane
Explanation:
I've numbered the isomeric hexanes from 1 to 5 and labelled the sets of equivalent hydrogens.
The results are
Isomer 1— three sets of equivalent hydrogens
Isomer 2— five sets of equivalent hydrogens
Isomer 3— four sets of equivalent hydrogens
Isomer 4— two sets of equivalent hydrogens
Isomer 5— three sets of equivalent hydrogens
Each set will give one monochloro substitution product.
4 = A. Two monochloro isomers.
2 = B. Five monochloro isomers.
3 = C. Four monochloro isomers.
Isomers 1 and 5 each give three monochloro isomers.
Thus, we cannot assign Structure D definitively.
D is either pentane or 2,2-dimethylbutane.
Branched chain alkanes
The alkanes don't contain a functional group and so the branches are numbered from the end that gives the lowest set of position numbers for the branches.
Use the above rules to see how the names of the alkanes below are built up.
The structure of 2-methylbutane is a butane molecule (C4H10) but with a methyl group (CH3) replacing a hydrogen on the second carbon atom in the chain. The structure of 3-methylpentane could be drawn as butane with an ethyl group (C2H5) replacing a hydrogen on the second carbon. Note that this is not 2-ethylbutane. The structure of 2,2-dimethylbutane is butane with two methyl groups replacing the two hydrogens on the second carbon.
Let suppose the Gas is acting Ideally, Then According to Ideal Gas Equation,
P V = n R T
Solving for P,
P = n R T / V ----- (1)
Data Given;
Moles = n = 1.20 mol
Volume = V = 4 L
Temperature = T = 30 + 273 = 303 K
Gas Constant = R = 0.08206 atm.L.mol⁻¹.K⁻¹
Putting Values in Eq.1,
P = (1.20 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 303 K) ÷ 4 L
P = 7.45 atm
x2=x+2 at x=−1 and x=2 so we have no need to worry about the end-points
f(x)=x+2−x^2
df/dx=1–2x
and that is zero (indicating a maximum) at x=1/2
So the maximum distance is f(1/2)=2.5–0.25=2.25
learn more about maximum distance between curves here:
brainly.com/question/1603866
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Answer:
- <em>The average mass of calcium in each sample is: </em><u>0.978 g</u>
<em />
- <em>The absolute uncertainty is: </em><u>0.008 g</u>
Explanation:
The <em>absolute uncertainty </em>of the total samples indicated in the statement is ± 0.1 g.
When you multiply or divide quantities with uncertainties, you calculate the final uncertanty by adding the <em>relative uncertainties</em> together.
The relative uncertainty is the absolute uncertainty divided by the quantity:
- Relative uncertainty = 0.1g / 12.2 g = 0.008
The average mass of calcium is calculated using proportions, along with the molar masses:
- Molar mass of calcium: 40.078 g/ mol (from a periodic table)
- Molar mass of calcite: 100.085 g/mol (given)
Proportion:
- 40.078 g of calcium / 100.085 g of calcite = x / 12.2 g of calcite
- x = 12.2 × 40.078 / 100.085 g = 4.89 g calcium
So the total mass of calcium in the five samples is 4.89 g, and the average mass in each sample is:
- Average mass = total mass of five samples / number of samples
- Average mass = 4.89 g / 5 = <u>0.978 g of calcium</u>
So, the first answer is that the average mass of calcium in each sample is 0.978 g ( keep 3 signficant figures, such as the quntitiy 12.2 shows, as you have only used multiplication and division).
The absolute uncertainty of each sample is the relative uncertainty multiplied by the average mass of calcium of the five samples, rounded to one decimal:
- Absolute uncertainty = 0.978 g × 0.008 ≈ 0.008 g
The answer to the secon question is that the absolute uncertaingy of calcium in each sample is 0.008 g.
