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Anettt [7]
2 years ago
13

Otis watches a cooking show on making mayonnaise. The chef dissolves salt and sugar in vinegar. She mixes in an egg yolk, and th

en mixes in oil until the ingredients are completely blended together. The chef says that it’s great on sandwiches and that it can be stored without separating. Which best describes the purpose of the egg yolk in the recipe?
Chemistry
2 answers:
RUDIKE [14]2 years ago
3 0

We are using vinegar ( a polar compound) and oil (non polar compound) in the making of mayonnaise. The two will form emulsion when mixed together and will be separated as two distinct layers

Here chef uses egg to hold or homogenize the two immiscible substances together.

Thus egg is stabilizing the colloid formed between vinegar and oil.

ki77a [65]2 years ago
3 0

Answer:

B

Explanation:

It acts to hold the vinegar and oil together in a colloid.

You might be interested in
How many moles are there in 140.2 g of Ca
olga nikolaevna [1]

Answer:

3.5

Explanation:

7 0
3 years ago
Read 2 more answers
2.3 Zinc has five naturally occurring isotopes: 48.63% of 64 Zn with an atomic weight of 63.929 amu; 27.90% of 66Zn with an atom
lakkis [162]

<u>Answer:</u> The average atomic mass of element Zinc is 65.40 amu.  

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i .....(1)

  • <u>For _{30}^{64}\textrm{Zn} isotope:</u>

Mass of _{30}^{64}\textrm{Zn} isotope = 63.929 amu

Percentage abundance of _{30}^{64}\textrm{Zn} isotope = 48.63 %

Fractional abundance of _{30}^{64}\textrm{Zn} isotope = 0.4863

  • <u>For _{30}^{66}\textrm{Zn} isotope:</u>

Mass of _{30}^{66}\textrm{Zn} isotope = 65.926 amu

Percentage abundance of _{30}^{66}\textrm{Zn} isotope = 27.90 %

Fractional abundance of _{30}^{66}\textrm{Zn} isotope = 0.2790

  • <u>For _{30}^{67}\textrm{Zn} isotope:</u>

Mass of _{30}^{67}\textrm{Zn} isotope = 66.927 amu

Percentage abundance of _{30}^{67}\textrm{Zn} isotope = 4.10 %

Fractional abundance of _{30}^{67}\textrm{Zn} isotope = 0.0410

  • <u>For _{30}^{68}\textrm{Zn} isotope:</u>

Mass of _{30}^{68}\textrm{Zn} isotope = 67.925 amu

Percentage abundance of _{30}^{68}\textrm{Zn} isotope = 18.75 %

Fractional abundance of _{30}^{68}\textrm{Zn} isotope = 0.1875

  • <u>For _{30}^{70}\textrm{Zn} isotope:</u>

Mass of _{30}^{70}\textrm{Zn} isotope = 69.925 amu

Percentage abundance of _{30}^{70}\textrm{Zn} isotope = 0.62 %

Fractional abundance of _{30}^{70}\textrm{Zn} isotope = 0.0062

Putting values in equation 1, we get:

\text{Average atomic mass of Zinc}=[(63.929\times 0.4863)+(65.926\times 0.2790)+(66.927\times 0.0410)+(67.925\times 0.1875)+(69.925\times 0.0062)]

\text{Average atomic mass of Zinc}=65.40amu

Hence, the average atomic mass of element Zinc is 65.40 amu.

7 0
3 years ago
) Experimental evidence indicates that the nucleus of an atom
geniusboy [140]

Answer:

A (contains most of the mass of the atom)

Evidence has it that a proton is about 2000 times as massive as an electron.

And there is usually multiple protons and neutrons in the nucleus

From what I just said, you can say that B is wrong

C however is also wrong because protons have a +charge and neutrons are neutrle which means you always have a charge > (greater than) 0

And D is wrong because electrons (which are not in the nucleus) have a neg charge. and protons have a + charge and are in the nucleus

So your answer is A

Hope it helped

Spiky Bob

5 0
2 years ago
What's meant by the term enantiotropy?​
Mila [183]

Answer:

the relation of two different forms of the same substance (such as two allotropic forms of tin) that have a definite transition point and can therefore change reversibly each into the other — compare monotropy.

7 0
2 years ago
Read 2 more answers
A sample of pure calcium fluoride with a mass of 15.0 g contains 7.70 g of calcium. how much calcium is contained in 45.0 g of c
kolbaska11 [484]

In a sample of pure calcium fluoride of mass 15.0 g, 7.70 g of calcium is present. First convert the mass into number of moles as follows:

n=\frac{m}{M}

Here, m is mass and M is molar mass.

Molar mass of Ca is 40 g/mol, putting the values,

n=\frac{7.70 g}{40 g/mol}=0.1925 mol

Similarly, molar mass of CaF_{2} is 78.07 g/mol thus, number of moles will be:

n=\frac{15.0 g}{78.07 g/mol}=0.1921 mol.

Thus, 0.1921 mol of CaF_{2} have 0.1925 mol of Ca, or 1 mole of CaF_{2} will have approximately 1 mole of Ca.

Now, mass of Ca needs to be calculated in 45.0 g of CaF_{2}. Converting mass into number of moles first,

n=\frac{45.0 g}{78.07 g/mol}=0.5764 mol

Thus, number of moles of Ca will also be 0.5764 mol, converting number of moles into mass,

m=n\times M=0.5764 mol\times 40 g/mol=23.06 g

Therefore, mass of Ca will be 23.06 g.

6 0
3 years ago
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