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aivan3 [116]
3 years ago
15

Mina is attending cooking school, where she prepares different mixtures of ingredients. She works with both homogenous and heter

ogeneous mixtures. Today she is using honey and salsa. What kinds of mixtures are these ingredients?
Chemistry
1 answer:
rodikova [14]3 years ago
3 0

Answer:

Honey is an homogeneous mixture

Salsa is an heterogeneous mixture

Explanation:

Honey is a sweet uniformly colored liquid that can be of a dark variety or of clear golden color. Honey is made in nature by bees from flower nectar and is used as a food additive or sweetener

Honey is a homogeneous mixture because the concentration of the components of honey are uniformly distributed throughout the mixture. Every portion has the same concentration of components

Salsa is a sauce made by mixing chopped tomatoes, onions, chilies, lime juice and seasoning and therefore consists of both solid and liquid components mixed in varying proportion such that part will have continuous  that comes in between different types of solid and no two parts have exactly the same composition

Therefore, salsa is an heterogenous liquid.

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Answer:

Meeee:(

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3 years ago
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Give an example where handpicking is used for separation. (1)​
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Answer:

If you want to separate black grapes from the mixture of black and green grapes, then you will simply pick black grapes using your hands from the mixture. In this way you are actually using handpicking separation method.

Explanation:

5 0
2 years ago
_______ was responsible for first grouping elements into triads based on similar properties.
frez [133]
German scientist Döbereiner was one responsible for grouping elements into triads based on most notably atomic mass, many of which can be found in the periodic table to be in a pattern (for example <span><span>Iron </span><span>Cobalt </span><span>Nickel, elements 26, 27, 28)</span></span>
7 0
3 years ago
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A volume of 90.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What
Gemiola [76]

Answer:

The answer to your question:  0.7 M

Explanation:

Data

V of KOH = 90 ml

[KOH] = ?

V H2SO4 = 21.2 ml

[H2SO4] = 1.5 M

                       2KOH(aq)  +  H₂SO₄(aq)   →   K₂SO₄(aq)  +  2H₂O(l)

Molarity = moles / volume

moles of H₂SO₄ = (1.5) (21.2)

                           = 31.8

                    2 moles of KOH --------------  1 mol of H₂SO₄

                   x                           --------------  31.8 mol of H₂SO₄

                    x = (31.8)(2) / 1

                    x = 63.8 moles of KOH

Molarity = 63.8 / 90

             = 0.7 M

7 0
3 years ago
how to determine the net charge of the tripeptide Asp-Gly-Leu at pH 7. Can someone show in details and tricks on how to solve it
Ugo [173]

Answer:

0!

Explanation:

  • You need to search your pKa values for Asn (2.14, 8.75), Gly (2.35, 9.78) and Leu(2.33, 9.74), the first value corresponding to -COOH, the second to -NH3 (a third value would correspond to an R group, but in this case that does not apply), and we'll build a table to find the charges for your possible dissociated groups at indicated pH (7), we need to remember that having a pKa lower than the pH will give us a negative charge, having a pKa bigger than pH will give us a positive charge:            

           

                   -COOH         -NH3              

pH 7------------------------------------------------------              

Asn               -                      +

Gly                -                      +

Leu               -                      +

  • Now that we have our table we'll sketch our peptide's structure:

<em>HN-Asn-Gly-Leu-COOH</em>

This will allow us to see what groups will be free to react to the pH's value, and which groups are not reacting to pH because are forming the bond between amino acids. In this particular example only -NH group in Ans and -COOH in Leu are exposed to pH, we'll look for these charges in the table and add them to find the net charge:

+1 (HN-Asn)

-1 (Leu-COOH)

=0

The net charge is 0!

I hope you find this information useful and interesting! Good luck!

5 0
3 years ago
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