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olga2289 [7]
3 years ago
11

Please select the word from the list that best fits the defintion

Chemistry
1 answer:
lutik1710 [3]3 years ago
8 0
D. Chemical formula





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Give the effect on the melting point of the presence of a cis double bond in a fatty acid.
vovikov84 [41]

Answer:

The cis double bond present in unsaturated fatty acids acids results in lower melting point when compared to saturated fatty acids of the same chain length.

Explanation:

Melting point of a fatty acids are affected by the length and degree of unsaturation of the hydrocarbon chain.

At room temperature, saturated fatty acids with hydrocarbon chain lengths between 12-24 are waxy solids whereas unsaturated atty acids of the same chain length are liquids. This is due to the nature of the packing of the fatty acid molecules in the saturated and unsaturated compounds.

In the saturated compounds, the molecules are tightly packed side by side with minimal steric hindrance and maximal van der Waals forces of attraction between molecules. However, in unsaturated fatty acids, the cis double bond introduces a bend or kink in the molecules which then interferes with the tight packing of the molecules and reducing interaction between molecules. Therefore, less energy is required to cause a disorder in the arrangement of unsaturated fatty acids, leading to a lowering of melting point.  

8 0
3 years ago
Consider two processes: sublimation of I2(s) and melting of I2(s) (Note: the latter process can occur at the same temperature bu
AVprozaik [17]

Answer:

positive

positive

Explanation:

Entropy -

In a system, the randomness is measured by the term entropy .

Randomness basically refers as a form of energy that can not be used for any work.

The change in entropy is given by amount heat per change in temperature.

  • When solid is converted to gas entropy increases,

As the molecules in solid state are tightly packed and has more force of attraction between the molecules, but as it is converted to gas, the force of attraction between the molecule decreases and hence entropy increases.

So,

The particles of the substance , if are tightly held by strong force of attraction will decrease the entropy ,

And

If the particles are loosely held , the entropy will increase , i.e. , positive entropy .

Similar with solid converting to liquid , the entropy will increases , i.e. , positive entropy .

Hence ,

The correct sign of entropy for both the process is positive .

4 0
3 years ago
Read 2 more answers
What's the easiest way to find balance in a chemical equation?
ivann1987 [24]
Well not calculus because that has nothing, well mostly nothing to do with balancing chemical equation, so B or C. Now for me personally B is way faster, though C is sometimes faster if you get lucky the way to solve it is B
6 0
3 years ago
What mass of hydrogen sulfide, H2S, will completely react with 2.00 moles of silver nitrate, AgNO3?
hodyreva [135]

Answer:

34g

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

H2S + 2AgNO3 —> 2HNO3 + Ag2S

Next, we shall determine the number of mole of H2S required to react with 2 moles of AgNO3.

This is illustrated below:

From the balanced equation above,

We can see that 1 mole of H2S is required to react completely with 2 moles of AgNO3.

Finally, we shall convert 1 mole of H2S to grams. This is shown below:

Number of mole H2S = 1 mole

Molar mass of H2S = (2x1) + 32 = 34g/mol

Mass = number of mole x molar Mass

Mass of H2S = 1 x 34

Mass of H2S = 34g

Therefore, 34g of H2S is needed to react with 2 moles of AgNO3.

6 0
3 years ago
Determine the acid dissociation constant for a 0.0250 M weak acid solution that has a pH of 2.37 . The equilibrium equation of i
Oksanka [162]

Answer:

For part (a): pHsol=2.22

Explanation:

I will show you how to solve part (a), so that you can use this example to solve part (b) on your own.

So, you're dealing with formic acid, HCOOH, a weak acid that does not dissociate completely in aqueous solution. This means that an equilibrium will be established between the unionized and ionized forms of the acid.

You can use an ICE table and the initial concentration ofthe acid to determine the concentrations of the conjugate base and of the hydronium ions tha are produced when the acid ionizes

HCOOH(aq]+H2O(l]⇌ HCOO−(aq] + H3O+(aq]

I 0.20 0 0

C (−x) (+x) (+x)

E (0.20−x) x x

You need to use the acid's pKa to determine its acid dissociation constant, Ka, which is equal to

3 0
2 years ago
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