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steposvetlana [31]
3 years ago
8

What reagent is used to test for starch​

Chemistry
1 answer:
Elodia [21]3 years ago
8 0

Answer:

iodine

Explanation:

In the presence of starch, iodine turns a blue/black colour. It is possible to distinguish starch from glucose (and other carbohydrates) using this iodine solution test. For example, if iodine is added to a peeled potato then it will turn black. Benedict's reagent can be used to test for glucose.

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Se tiene una solución acuosa 2M de carbonato de potasio. Expresar su concentración en %p/v y Normalidad.
zepelin [54]

Answer:

Normalidad = 4N

%p/V = 27.6%

Explanation:

La solución 2M de carbonato de potasio contiene 2moles de carbonato por litro de solución. La normalidad son los equivalente de carbonato de potasio (2eq/mol) por litro de solución:

2moles * (2eq/mol) = 4eq / 1L = 4N

El porcentaje peso volumen es el peso de carbonato en gramos dividido en el volumen en mL por 100:

%p/V:

Masa K2CO3 -Masa molar: 138.205g/mol-

2moles * (138.205g/mol) = 276g K2CO3

Volumen:

1L * (1000mL/1L) = 1000mL

%p/V:

276g K2CO3 / 1000mL * 100

<h3>%p/V = 27.6%</h3>
6 0
3 years ago
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Digiron [165]

Answer: wind

Explanation:

5 0
3 years ago
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Read the scientific question below.
svetlana [45]
If we fertilize a plant, then its height increases fast. Always use if then format
8 0
3 years ago
4) How many grams are there in 7.40 moles of AgNO3
Ludmilka [50]

Answer:

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3 0
3 years ago
How does 0.5 m sucrose (molecular mass 342) solution compare to 0.5 m glucose (molecular mass 180) solution?
mash [69]

Answer : Both solutions contain 3.011 X 10^{23} molecules.

Explanation : The number of molecules of 0.5 M of sucrose is equal to the number of molecules in 0.5 M of glucose. Both solutions contain 3.011 x 10^{23} molecules.

Avogadro's Number is  N_{A} =  6.022 X 10^{23} which represents particles per mole and particles may be typically molecules, atoms, ions, electrons, etc.

Here, only molarity values are given; where molarity is a measurement of concentration in terms of moles of the solute per liter of solvent.

Since each substance has the same concentration, 0.5 M, each will have the same number of molecules present per liter of solution.

Addition of molar mass for individual substance is not needed. As if both are considered in 1 Liter they would have same moles which is 0.5.

We can calculate the number of molecules for each;

Number of molecules  = N_{A} X M;

∴  Number of molecules =  6.022 X 10^{23} X 0.5 mol/L X 1 L which will be  = 3.011 X 10^{23}

Thus, these solutions compare to each other in that they have not only the same concentration, but they will have the same number of solvated sugar molecules. But the mass of glucose dissolved will be less than the mass of sucrose.

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3 years ago
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