One molecule of sucrose is burned with oxygen to make carbon dioxide and water.
Disaccharide sugar sucrose is composed of glucose and fructose. It is produced naturally by plants and is the main component of white sugar. C₁₂H₂₂O₁₁ is the chemical formula for it.
Extraction and refining sucrose for human use can be done from either sugarcane or sugar beet. Raw sugar is created from crushing the cane, which is consistently delivered to other sectors to be refined into pure sucrose. Sugar mills generally are located in the tropical regions near the sugarcane plantations.
<em> C₁₂H₂₂O₁₁ + 12O₂ → 12CO₂ + 11H₂O</em>
When one molecule of sucrose is burnt, we get 12 carbon dioxide molecules.
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The OH peaks in the IR spectra of benzyl alcohol and benzoic acid should be compared and contrasted.
<h3>What is the IR spectra of Benzoic acid?</h3>
- The right-hand portion of the infrared spectrum of benzoic acid, between wavenumbers 1500 and 400 cm-1, is referred to as the fingerprint region.
- It results from a special combination of intricately overlapping vibrations of the atoms within the benzoic acid molecule.
<h3>What is the IR spectra of Benzyl alcohol?</h3>
- A C-Cl bond is frequently shown by a peak at 700.
- There are a few more peaks at 1500 that are directed at a C=C bond.
<h3>What is IR spectra?</h3>
The percent transmittance (or absorbance) of the radiation through the molecule against the radiation's wave number forms the IR spectrum.
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The answer is 2H2 + O2----> 2H2O
State the periodic law and explain how it relates to the periodic table.
The new pH is 7.69.
According to Hendersen Hasselbach equation;
The Henderson Hasselbalch equation is an approximate equation that shows the relationship between the pH or pOH of a solution and the pKa or pKb and the ratio of the concentrations of the dissociated chemical species. To calculate the pH of the buffer solution made by mixing salt and weak acid/base. It is used to calculate the pKa value. Prepare buffer solution of needed pH.
pH = pKa + log10 ([A–]/[HA])
Here, 100 mL of 0.10 m TRIS buffer pH 8.3
pka = 8.3
0.005 mol of TRIS.
∴ ![8.3 = 8.3 + log \frac{[0.005]}{[0.005]}](https://tex.z-dn.net/?f=8.3%20%3D%208.3%20%2B%20log%20%5Cfrac%7B%5B0.005%5D%7D%7B%5B0.005%5D%7D)
<em> </em>inverse log 0 = ![\frac{[B]}{[A]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BB%5D%7D%7B%5BA%5D%7D)
![\frac{[B]}{[A]} = 1](https://tex.z-dn.net/?f=%5Cfrac%7B%5BB%5D%7D%7B%5BA%5D%7D%20%3D%201)
Given; 3.0 ml of 1.0 m hcl.
pka = 8.3
0.003 mol of HCL.
![pH = 8.3 + log \frac{[0.005-0.003]}{[0.005+0.003]}\\pH = 8.3 + log \frac{[0.002]}{[0.008]}\\\\pH = 8.3 + log {0.25}\\\\pH = 8.3 + (-0.62)\\pH = 7.69](https://tex.z-dn.net/?f=pH%20%3D%208.3%20%2B%20log%20%5Cfrac%7B%5B0.005-0.003%5D%7D%7B%5B0.005%2B0.003%5D%7D%5C%5CpH%20%3D%208.3%20%2B%20log%20%5Cfrac%7B%5B0.002%5D%7D%7B%5B0.008%5D%7D%5C%5C%5C%5CpH%20%3D%208.3%20%2B%20log%20%7B0.25%7D%5C%5C%5C%5CpH%20%3D%208.3%20%2B%20%28-0.62%29%5C%5CpH%20%3D%207.69)
Therefore, the new pH is 7.69.
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