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Karo-lina-s [1.5K]
3 years ago
5

why do you need to think about polarity of the different ingredients in your salad dressing in order for it to turn out well?

Chemistry
2 answers:
Elena L [17]3 years ago
7 0

Answer:

No.

Explanation:

Hello,

In this case, a salad dressing makes up of an oil-and-vinegar heterogeneous mixture as long as if it is let to stand it is possible to decant its constituents by density difference, therefore, we do not mind about the polarity of the different ingredients. The only necessity is a well shaking before its use.

Best regards.

vazorg [7]3 years ago
3 0
So that they combine smoothly and you aren't left with separated looking salad dressing. With polarity only, like-polarities dissolve like-polarities. So you must make sure that all your ingredients are polar or all are nonpolar.
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According to the image, identify the number of neutrons in the most common isotope of aluminum.
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Explanation:

Isotopes : These are the atoms which have same atomic number but have different mass number.

<u>This image shows the average atomic mass of Al element because it is in decimals</u>.

Atomic mass = 26.98154

(Note : mass number of single isotope can never be in decimals)

It is the average of mass of different isotopes of Al

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mass of Al given in image(26.98) is nearly equal to mass of 2nd isotope(27)

mass of _{13}^{26.98154}\textrm{Al}\ \approx 27

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how many kilograms of a 35% m/m sodium chlorate solution is needed to react completely with 0.29 l of a 22% m/v aluminum nitrate
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3NaClO_3+Al(NO_3)_3\rightarrow 3NaNO_3+Al(ClO_3)_3

We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.

0.29L(\frac{1000mL}{1L})(\frac{22g}{100mL})

= 63.8 g aluminum nitrate

From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.

We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})

= 95.7gNaClO_3

sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate  and then the grams are converted to kg.

95.7gNaClO_3(\frac{100gSolution}{35gNaClO_3})(\frac{1kg}{1000g})

= 0.273 kg

So, 0.273 kg of 35% m/m sodium chlorate solution are required.

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