Answer: potassium iodide is the basic test for starch,and the positive test is blue-black coloration, any other test substance which is not starch will give a negative results.
Explanation:
Starch is an example of polysaccharide and since the standard test for it is potassium iodide solution, it gives a positive test.
Diasaccharides e.g maltose are reducing sugars.their standard test is BENEDICT test .
Therefore, in the hydrolysis; starch should give a positve test, while Diasaccharides should give negative rest.
Answer:
Explanation:
We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.
Carbonate is CO₃ and has an oxidation number of -2. Sodium is Na and has an oxidation number of +1. There must be 2 moles of sodium so the charge of the sodium balances the charge of the carbonate. The formula is Na₂CO₃.
We will convert grams to moles using the molar mass or the mass of 1 mole of a substance. They are found on the Periodic Table as the atomic masses, but the units are grams per mole instead of atomic mass units. Look up the molar masses of the individual elements.
- Na: 22.9897693 g/mol
- C: 12.011 g/mol
- O: 15.999 g/mol
Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.
- Na₂ = 22.9897693 * 2= 45.9795386 g/mol
- O₃ = 15.999 * 3= 47.997 g/mol
- Na₂CO₃= 45.9795386 + 12.011 + 47.997 =105.9875386 g/mol
We will convert using dimensional analysis. Set up a ratio using the molar mass.
We are converting 57.3 grams to moles, so we multiply by this value.
Flip the ratio so the units of grams of sodium carbonate cancel.
The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.
There are approximately <u>0.541 moles of sodium carbonate</u> in 57.3 grams.
Answer:
12 moles of F₂
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
N₂ + 3F₂ —> 2NF₃
From the balanced equation above,
3 moles of F₂ reacted to produce 2 moles of NF₃.
Finally, we shall determine the number of mole of F₂ needed to produce 8 moles of NF₃. This can be obtained as illustrated below:
From the balanced equation above,
3 moles of F₂ reacted to produce 2 moles of NF₃.
Therefore, Xmol of F₂ will react to produce 8 moles of NF₃ i.e
Xmol of F₂ = (3 × 8)/2
Xmol of F₂ = 12 moles
Thus, 12 moles of F₂ is needed for the reaction.
The radius of the electron's or basically the energy level for which the electron is found orbiting the nucleus of he hydrogen atom, as the principal quantum number tells us primarily the energy level that the electron will be found, is it the 1st, 2nd, and 3rd. The other quantum numbers tells us more specifically as per the subshell of the main shell the electron is in, the spin of the electron etc.
Answer:
Close to the calculated endpoint of a titration - <u>Partially open</u>
At the beginning of a titration - <u>Completely open</u>
Filling the buret with titrant - <u>Completely closed</u>
Conditioning the buret with the titrant - <u>Completely closed</u>
Explanation:
'Titration' is depicted as the process under which the concentration of some substances in a solution is determined by adding measured amounts of some other substance until a rection is displayed to be complete.
As per the question, the stopcock would remain completely open when the process of titration starts. After the buret is successfully placed, the titrant is carefully put through the buret in the stopcock which is entirely closed. Thereafter, when the titrant and the buret are conditioned, the stopcock must remain closed for correct results. Then, when the process is near the estimated end-point and the solution begins to turn its color, the stopcock would be slightly open before the reading of the endpoint for adding the drops of titrant for final observation.
