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2 years ago

PLA is a bioplastic that is biodegradable. Explain how is is created and why it is biodegradable.

Chemistry

1 answer:

Ratling [72]2 years ago

3 0

Explanation:

<h3>PLA is a polyester produced by fermentation under controlled conditions of a carbohydrate source like corn starch or sugarcane. ... The starch is then mixed with acid or enzymes and heated. This process “breaks” starch into dextrose (D-glucose), or corn sugar.</h3>

<h3>PLA is a polymer made from high levels of polylactic acid molecules. </h3><h3>For PLA to biodegrade, you must break up the polymer by adding </h3><h3>water to it (a process known as hydrolyzing). Heat and moisture are required for hydrolyzing to occur.</h3>

<h3>PLA consists of renewable raw materials and is biodegradable in industrial composting plants. </h3><h3>However, due to the lack of infrastructure, it is difficult to </h3><h3>compost PLA industrially or to</h3><h3> recycle it.</h3>

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A sample of glucose contains 1.250x10^21 carbon atoms, how many atoms of hydrogen does it contain?

dybincka [34]

Answer:

Hydrogen = 2.5 * 10^21

Explanation:

Chemical Formula Glucose: C₆H₁₂O₆

One of the ways you could do this is to notice that for every carbon atom there are two Hydrogen atoms. You can state this more formally by using the formula to set up a ratio: 12/6 = hydrogen to Carbon

So if there are 1.250 * 10^21 Carbon atoms in the Glucose sample, then there will be twice as many hydrogen atoms.

H = 2 * 1.25 * 10^21 = 2.5 * 10^21 atoms

You could do this more formally by setting up a proportion.

6 Carbon / 12 Hydrogen = 1.25*10^21 / x Cross Multiply

6*x = 12 * 1.25*10^21 Combine the right

6x = 1.5 * 10^22 Divide by 6

x = 2.5 * 10^21

5 0

2 years ago

Which of the following molecules is responsible for transmitting the energy needed by cells

kolbaska11 [484]

Adenosine triphosphate (ATP)

7 0

3 years ago

Read 2 more answers

During the combustion of a peanut that weighed 0.341 g, the temperature of the 100 mL of water in the calorimeter rose from 23.4

blondinia [14]

Answer:

119 kCal per serving.

Explanation:

The heat energy necessary to elevates water's temperature from 23.4°C to 37.9°C can be calculated by the equation below:

Q = mcΔT

Q: heat energy

m: mass in g

c: specific heat capacity in cal/g°C

ΔT = temperature variation in °C

m is the mass of water, considering the density of water to be 1g/mL, 100 mL of water weights 100g. Therefore:

Q = 100 g x 1.00 cal/g°C x (37.9 - 23.4)°C

Q = 1450 cal

1450 cal ____ 0.341 g peanuts

x ____ 28 g peanuts

x = 119061.58 cal

This means that the cal from fat per serving of peanuts is at least 119 kCal.

3 0

3 years ago

A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The

V125BC [204]

Taking into account the definition of calorimetry, the specific heat of metal is 0.165 $\frac{cal}{gC}$ .

<h3>Definition of calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where:

Q is the heat exchanged by a body of mass m.
C is the specific heat substance.
ΔT is the temperature variation.

<h3>Specific heat capacity of the metal</h3>

In this case, you know:

For metal:

Mass of metal = 50 g
Initial temperature of metal= 45 °C
Final temperature of metal= 11.08 ºC
Specific heat of metal= ?

For water:

Mass of water = 250 g
Initial temperature of water= 10 ºC
Final temperature of water= 11.08 ºC
Specific heat of water = 1.035 $\frac{cal}{gC}$

Replacing in the expression to calculate heat exchanges:

For metal: Qmetal= Specific heat of metal× 50 g× (11.08 C - 45 C)

For water: Qwater= 1.035 $\frac{cal}{gC}$ × 250 g× (11.08 C - 10 C)

If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.

Then, the heat that the gold gives up will be equal to the heat that the water receives. Therefore:

- Qmetal = + Qwater

- Specific heat of metal× 50 g× (11.08 C - 45 C)= 1.035 $\frac{cal}{gC}$ × 250 g× (11.08 C - 10 C)

Solving:

- Specific heat of metal× 50 g× (-33.92 C)= 1.035 $\frac{cal}{gC}$ × 250 g× 1.08 C

Specific heat of metal× 1696 g×C= 279.45 cal

Specific heat of metal= $\frac{279.45 cal}{1696 gC}$