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adelina 88 [10]
4 years ago
12

A student identifies a compound that is composed of glycerol and three fatty acids. what property would the student expect this

compound to have if one of the fatty acids is unsaturated? the compound would
Chemistry
2 answers:
Svetlanka [38]4 years ago
7 0

Answer:

decompose at room temperature

 

be a solid at room temperature

 

be a liquid at room temperature

 

be a gas at room temperature

Explanation:

vladimir1956 [14]4 years ago
3 0
Triglycerides are formed when three fatty acids are condensed with glycerol with elimination of three moles of water.

Depending upon the chemical composition of fatty acids the triglyceride may exist either in semi solid / solid form or in liquid form.

Fats (semi solid / solid form):

If all the three fatty acids forming triglyceride are saturated then the triglyceride will be in solid state. This is because the long chains of fatty acids are closely packed and increases the intermolecular interactions.

Oils (liquid form):

If one or more of the fatty acid is unsaturated having cis configuration the triglyceride will exist in a liquid state. This happens because the long chains of fatty acid get kinked, resulting in decreasing the interactions between molecules hence exist as a liquid at room temperature.
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How much ice (in grams) would have to melt to lower the temperature of 353 mL of water from 26 ∘C to 6 ∘C? (Assume the density o
Rashid [163]

Answer:

The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg

Explanation:

Heat gain by ice = Heat lost by water

Thus,  

Heat of fusion + m_{ice}\times C_{ice}\times (T_f-T_i)=-m_{water}\times C_{water}\times (T_f-T_i)

Where, negative sign signifies heat loss

Or,  

Heat of fusion + m_{ice}\times C_{ice}\times (T_f-T_i)=m_{water}\times C_{water}\times (T_i-T_f)

Heat of fusion = 334 J/g

Heat of fusion of ice with mass x = 334x J/g

For ice:

Mass = x g

Initial temperature = 0 °C

Final temperature = 6 °C

Specific heat of ice = 1.996 J/g°C

For water:

Volume = 353 mL

Density (\rho)=\frac{Mass(m)}{Volume(V)}

Density of water = 1.0 g/mL

So, mass of water = 353 g

Initial temperature = 26 °C

Final temperature = 6 °C

Specific heat of water = 4.186 J/g°C

So,  

334x+x\times 1.996\times (6-0)=353\times 4.186\times (26-6)

334x+x\times 11.976=29553.16

345.976x = 29553.16

x = 85.4197 kg

Thus,  

<u>The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg</u>

7 0
4 years ago
What volume (in mL ) of 0.200   M NaOH do we need to titrate 40.00 mL of 0.140   M HBr to the equivalence point?
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NaOH + HBr =⇒ NaBr + H2O

35.0 ml HBr x 1 liter/1000 mL x 0.140 moles HBr/ liter = 0.0049 moles HBr

0.0049 moles HBr x 1mole NaOH/1mole HBr = 0.0049 moles HBr

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