Answer:
Jeweler B = more accurate
Jeweler A = more precise
Error:
0.008, 0
% error :
0.934% ; 0
Explanation:
Given that:
True mass of nugget = 0.856
Jeweler A: 0.863 g, 0.869 g, 0.859 g
Jeweler B: 0.875 g, 0.834 g, 0.858 g
Official measurement (A) = 0.863 + 0.869 + 0.859 = 2.591 / 3 = 0.864
Official measurement (B) = 0.875 + 0.834 + 0.858 = 2.567 / 3 = 0.8556
Accuracy = closeness of a measurement to the true value
Accuracy = true value - official measurement
Jeweler A's accuracy :
0.856 - 0.864 = - 0.008
Jeweler B's accuracy :
0.856 - 0.856 = 0.00
Therefore, Jeweler B's official measurement is more accurate as it is more close to the true value of the gold nugget.
However, Jeweler A's official measurement is more precise as each Jeweler A's measurement are closer to one another than Jeweler B's measurement which are more spread out.
Error:
Jeweler A's error :
0.864 - 0.856 = 0.008
% error =( error / true value) × 100
% error = (0.008/0.856) × 100% = 0.934%
Jeweler B's error :
0.856 - 0.856 = 0 ( since the official measurement as been rounded to match the decimal representation of the true value)
% error = 0%
We are asked to convert 25 cg to units of hg.
1 cg = 1 centigram = 10⁻² g
1 hg = 1 hectogram = 10² g
The options given are:
a) 1 hg/ 10² g
b) 10² cg/ 1 hg
c) 10² hg/ 1 cg
d) 10⁻² g/ 1 cg
To convert 25 cg to 1 hg, we could convert the 25 cg to grams first, then grams to hg.
25 cg · 10⁻² g/ 1cg = 0.25 g
Here we have converted our number from cg to grams. We can use another conversion of grams to hg to complete the conversion.
0.25 g · 1 hg/ 10² g = 0.0025 hg
Therefore, the first conversion we used was d) 10⁻² g/ 1 cg.
Answer: Matter is heated, and its particles spread out more
Explanation:
Thermal expansion occurs when there's an expansion of an object or material or when an object becomes bigger because of a rise in its temperature. This brings about the faster movement of the heated molecules and the atoms spreading out.
Therefore, the cause and effect of thermal expansion will be that when matter is heated, and its particles spread out more.
Answer:
(2R,3S)-2-ethoxy-3-methylpentane
and
(2S,3S)-2-ethoxy-3-methylpentane
Explanation:
For this case, we will have 
as nucleophile. Also, this compound is also in excess. So, we will have as solvent 
a protic solvent. Therefore the Sn1 reaction would be favored.
The first step would be the carbocation formation followed by the attack of the nucleophile. In this case both isomers would be produced: R and S (see figure).
