Answer:
42.5 g
Explanation:
Calculate the mass of the soft drink given the density and volume:
355 mL × 1.04 g/mL = 369.2 g
Now calculate the mass of sucrose given the percentage:
0.115 × 369.2 g = 42.46 g
Rounded to 3 significant figures, the mass is 42.5 g.
Answer:
H2O<en<phen
Explanation:
The degree of d- splitting is observed from the intensity of colour. The order of d splitting from least to greatest is H2O<en<phen. Phen shows the greatest d-splitting. The degree of splitting of d- orbitals by ligands depends on their relative positions in the spectrochemical series. The spectrochemical series is an experimentally determined series. The series separates the ligands into strong field and weak field ligands. Strong field ligands are found towards the end of the series. Strong field ligands such as en and phen can participate in metal to ligand or ligand to metal pi-bonding. Hence they cause more d-splitting. Ethylendiamine and phenanthroline occur towards the end of the spectrochemical series hence the higher order of d-splitting.
The mean is simply the arithmetic average of all your raw data. This can be solved methodically by summing up all of the raw data points that you have. Take note how many raw data points you used, because this will be used to divide the sum. You will obtain the mean.
The percent A by mass for substance AB =<u> 75%</u>
<h3>Further explanation</h3>
Proust states the Comparative Law that compounds are formed from elements with the same Mass Comparison, so that compounds have a fixed composition of elements
Empirical formula is the mole ratio of compounds forming elements.
From Substance AB₂ is 60.0% A by mass.
Let's say that AB₂ mass = 100 gram, then
mass A = 60 gram
mass B = 40 gram : 2 (coefficient in compound AB₂ = 2) = 20 gram
In compound AB:
Total mass = mass A + mass B
Total mass = 60 + 20 grams = 80 grams
Then the percentage of compound A = (60: 80) = 75%
<h3>Learn more</h3>
Grams of KO₂ needed to form O₂
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Keywords : percent mass, substance
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Answer:
The possible structures are ketone and aldehyde.
Explanation:
Number of double bonds of the given compound is calculated using the below formula.
=Number of double bonds
= Number of carbon atoms
= Number of hydrogen atoms
= Number of nitrogen atoms
The number of double bonds in the given formula - 
The number of double bonds in the compound is one.
Therefore, probable structures is as follows.
(In attachment)
The structures I and III are ruled out from the probable structures because the signal in 13C-NMR appears at greater than 160 ppm.
alkene compounds I and II shows signal less than 140 ppm.
Hence, the probable structures III and IV are given as follows.
The carbonyl of structure I appear at 202 and ketone group of IV appears at 208 in 13C, which are greater than 160.
Hence, the molecular formula of the compound having possible structure in which the signal appears at greater than 160 ppm are shown aw follows.
