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Evgen [1.6K]
3 years ago
11

Why saturated fatty acid can’t create kink ?​

Chemistry
1 answer:
Leya [2.2K]3 years ago
7 0
Saturated fatty acids exhibit a linear structure while unsaturated fatty acids bend, or kink, due to double bonds within the chemical foundation.
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Need these 2 answered asap pls
Kitty [74]

Answer:

im not a brain wiz but i think 20

Explanation:

3 0
3 years ago
A or B or C or D, fassst plzzz
olga nikolaevna [1]

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a.option is the correct answer

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How many atoms in 5.01 mole of lead?
leva [86]

Answer:

30.17 × 10²³ atoms

Explanation:

Given data:

Number of moles of lead = 5.01 mol

Number of atoms = ?

Solution:

Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

In given question:

1 mole = 6.022 × 10²³ atoms

5.01 mol × 6.022 × 10²³ atoms / 1 mol

30.17 × 10²³ atoms

3 0
3 years ago
Which piece of glassware shown below is used to hold and dispense a solution of known concentration during a titration?
Musya8 [376]

Answer:

Answer C

Explanation:

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4 0
3 years ago
Cu(s) + 4 HNO3 (aq) --> Cu(NO3)2 (aq) + 2NO2 (g) + 2H2O(l)
GREYUIT [131]

Answer:

The percent by mass of copper in the mixture was 32%

Explanation:

The ammount of HNO₃ used is:

mol HNO₃ = volume * concentration

mol HNO₃ = 0.015 l * 15.8 mol/l

mol HNO₃ = 0.237 mol

According to the reaction, 4 mol HNO₃ will react with 1 mol Cu and produce 1 mol Cu²⁺. Since we have 0.237 mol HNO₃, the amount of Cu that could react would be (0.237 mol HNO₃ * 1 mol Cu / 4 mol HNO₃) 0.06 mol. This reaction would produce 0.060 mol Cu²⁺, however, only 0.010 mol Cu²⁺ were obtained, indicating that only 0.010 mol Cu were present in the mixture. This means that the acid was in excess, so we can assume that all copper present in the mixture has reacted.

Since 0.010 mol of Cu²⁺ were produced, the amount of Cu was 0.01 mol.

1 mol of Cu has a mass of 63.5 g, then 0.01 mol has a mass of:

0.01 mol Cu * 63.5 g / 1 mol = 0.635 g.

Since this amount was present in 2.00 g mixture, the amount of copper in 100 g of the mixture will be:

100 g(mixture) * 0.635 g Cu / 2 g(mixture) = 32 g

Then, the percent by mass of Cu (which is the mass of Cu in 100 g mixture) is 32%

6 0
3 years ago
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