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aleksley [76]
3 years ago
14

Use you observation to determine which polymorph is in the customer’s returned chocolate sample.

Chemistry
2 answers:
Butoxors [25]3 years ago
6 0

Answer:

Cocoa butter, the fat in chocolate, can crystallize in any one of 6 different forms (polymorphs, as they are called). Unfortunately, only one of these, the beta crystal (or Form V), hardens into the firm, shiny chocolate that cooks want. Form VI is also a stable hard crystal, but only small amounts of it form from the good beta (Form V) crystals upon lengthy standing. When you buy commercial chocolate it is in the form of beta crystals.

When you melt chocolate and get it above 94° F, you melt these much desired beta crystals and other types of crystals can set up. If you simply let melted chocolate cool, it will set up in a dull, soft, splotchy, disgusting-looking form. Even the taste is different. Fine chocolate has a snap when you break it and a totally different mouthfeel from the other cocoa butter forms.

Alinara [238K]3 years ago
3 0

Answer:

When you melt chocolate and get it above 94° F, you melt these much desired beta crystals and other types of crystals can set up. If you simply let melted chocolate cool, it will set up in a dull, soft, splotchy, disgusting-looking form. Even the taste is different. Fine chocolate has a snap when you break it and a totally different mouthfeel from the other cocoa butter forms.

Explanation:

<em><u>hope </u></em><em><u>it's </u></em><em><u>help </u></em><em><u>you</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>!</u></em><em><u>!</u></em><em><u>!</u></em><em><u>!</u></em><em><u>!</u></em><em><u>!</u></em><em><u>!</u></em>

<em><u>#</u></em><em><u>rishu</u></em>

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When a 3.00 grams sample of a compound containing only c, h, and o was completely burned, 1.17 grams of h2o and 2.87 grams of co
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Answer:- CH_2O_2

Solution:- From given masses of carbon dioxide and water we could calculate the moles that helps to calculate the moles of C and H.

Molar mass of carbon dioxide = 44 gram per mol

molar mass of water = 18.02 gram per mol

From given info, combustion of compound gives 1.17 grams of water and 2.87 grams of carbon dioxide. Let's calculate the moles of these:

1.17gH_2O(\frac{1mol}{18.02g})

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Similarly, 2.87gCO_2(\frac{1mol}{44g})

= 0.0652molCO_2

One mol of water has two moles of H. So, the moles of H would be two times the moles of water as calculated above.

So, moles of H = 2* 0.0649 = 0.1298 mol

One mol of carbon dioxide contains one mol of C. So, the moles of C would be equal to the moles of carbon dioxide calculated above.

moles of C = 0.0652 mol

Let's convert the moles of H and C to grams so that we could calculate the amount of oxygen present in the sample as:

grams of H in sample = 1.008 x 0.1298 = 0.1308 g

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If we subtract the sum of the masses of C and H from sample mass then it would give as the mass of oxygen since the sample has only C, H and O.

mass of O in sample = 3.00g - (0.1308 g + 0.7824 g)

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Let's convert these grams of oxygen to moles on dividing by it's atomic mass as:

2.0868gO(\frac{1mol}{15.999g})

= 0.130 mol O

Now, we have the moles of all the three atoms and we know that an empirical formula is the simplest whole number ratio of the moles of atoms. So, let's calculate the ratio. For this, we divide the moles of each by the least one of them.Looking at the moles, the least value is for carbon. So, let's divide the moles of each by the moles of C as:

C = \frac{0.0652}{0.0652}  = 1

H = \frac{0.1298}{0.0652}  = 2

O = \frac{0.130}{0.0652}  = 2

The ratio of C, H and O is 1:2:2. So, the simplest formula of the compound is CH_2O_2 .



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