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Degger [83]
3 years ago
12

Which of these processes helps extract energy from food?

Chemistry
1 answer:
Stels [109]3 years ago
7 0

Answer:  Proteins extract the energy from food via Enzymes.

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20 cubic inches of a gas with an absolute pressure of 5 psi is compressed until its pressure reaches 10 psi. What's the new volu
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10 cubic inches
We will use Boyle's law that states that for a fixed amount of an ideal gas kept at a fixed temperature, pressure and volume are inversely proportional. 
P1 V1 = P2 V2
Where
 P1 is initial pressure = 5 psi
V1 is initial volume = 20 cubic inch
P2 is final pressure = 10 psi
V2 is final volume = unknown
V2 = P1,V1 / P2
V2 = 20 × 5 / 10
V2 = 100/10
V2 = 10 cubic inches


7 0
3 years ago
a mixture containing 21.4 g of ice. (0.00 c) and 75.3 g of water (55.3 c) is placed in an insulatated container
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7 0
3 years ago
A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s
Alla [95]

<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For NaBr:</u>

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol

The chemical equation for the reaction of 1-butanol and NaBr is:

\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = \frac{1}{1}\times 0.108=0.108 moles of 1-bromobutane

  • Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g

  • To calculate the percentage yield of 1-bromobutane, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0
3 years ago
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