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podryga [215]
3 years ago
14

Which of the following types of reactions would decrease the entropy within a cell? A. Dehydration reaction, B. Hydrolysis, C. R

espiration, D. Digestion, E. Catabolism
Chemistry
1 answer:
kicyunya [14]3 years ago
6 0
<h3><u>Answer;</u></h3>

A. Dehydration reactions.

<h3><u>Explanation</u>;</h3>
  • Dehydration reactions refer to chemical events where a water molecule is lost because of the presence of another reacting molecule.
  • Dehydration reaction is said to cause a decrease in entropy in various cells.
  • <em><u>Entropy within a cell refers to a state of inner instability secondary to different energies provided by the different molecular components.</u></em>
  • <em><u>When the water component is removed,  there will be less molecular energy and therefore decreased entropy.</u></em>
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Answer:

You are moving because the Earth and everything in our solar system is constantly moving. ... As the Earth rotates, it also moves, or revolves, around the Sun. The Earth's path around the Sun is called its orbit. It takes the Earth one year, or 365 1/4 days, to completely orbit the Sun.

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Consider the chemical reaction represented below. CH4 + 2O2 CO2 + 2H2O + Heat Which of the following statements about this react
dsp73
A. the reaction releases heat
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Phosphorus-32 (32P) is an isotope that is commonly used for medical and biological research. Phosphorus-32 has a half-life of 14
Pachacha [2.7K]

The activity of the sample when it was shipped from the manufacturer is 4.54 mCi

<h3>How to determine the number of half-lives that has elapsed </h3>

From the question given above, the following data were obtained:

  • Time (t) = 48 hours
  • Half-life (t½) = 14.28 days = 14.28 × 24 = 342.72 hours
  • Number of half-lives (n) =?

n = t / t½

n = 48 / 342.72

n = 0.14

<h3>How to determine the activity of the sample during shipping </h3>
  • Number of half-lives (n) = 0.14
  • Original activity (N₀) = 5.0 mCi
  • Activity remaining (N) =?

N = N₀ / 2ⁿ

N = 5 / 2^0.14

N = 4.54 mCi

Thus, the activity of the sample during shipping is 4.54 mCi

Learn more about half life:

brainly.com/question/2674699

5 0
2 years ago
Which is the metric standard for measuring energy? Which unit is used for specific heat capacity? If you wanted to compare the a
DENIUS [597]

1. Which is the metric standard for measuring energy?

The <u>metric standard for measuring of energy </u>defined by the International System of Units is the joule (J), which is defined as the work done by a force of a newton in a displacement of one meter in the direction of force. So,

1 J = 1 N m = 1 kg·m²/s²

Calorie is also frequently used in scientific and technological applications. Calorie is a <u>unit of thermal energy that is equivalent to the amount of heat needed to raise the temperature of 1 gram of water by 1 degree Celsius</u>.

1 cal = 4,184 J

2. Which unit is used for specific heat capacity?

The specific heat capacity (c) is a physical quantity that is defined as the <u>amount of heat (</u><u>q</u><u>) that must be supplied to the mass unit of a thermodynamic substance or system to raise its temperature by one unit.</u> So,

c = q / m ΔT

where m is the mass of the substance and ΔT is the temperature increase.

In this way, as heat is a form of energy, the International System of Units expresses the specific heat in <u>joules per kilogram and per kelvin</u> (J kg⁻¹ K⁻¹). Another common unit, not belonging to the SI, is the <u>calorie per gram and per degree centigrade</u> (cal g⁻¹ ° C⁻¹).

3. If you wanted to compare the abilities of olive oil and peanut oil to gain or lose thermal energy, which unit would you use?

You should use units of specific heat capacity (J kg⁻¹ K⁻¹) since, as mentioned above, this is a physical quantity that measures the amount of heat that must be supplied to an specific mass of a substance or system to raise its temperature.

Heat is a thermal energy, so <u>by using heat capacity units you can compare the ability of</u><u> </u><u>olive oil and peanut oil to gain or lose thermal energy by varying its temperature.</u>

3 0
3 years ago
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During a laboratory experiment, 9.68 g of iron reacted with excess sulfur to form 13.8 grams of iron II sulfide. Calculate the p
Daniel [21]

Answer:

Percent yield = 90.8%

Explanation:

The reaction of Fe with S to produce FeS is:

Fe + S → FeS

<em>Where the moles of Fe added in excess of S are the moles of FeS</em>

<em />

Now, percent yield is defined as 100 times the ratio between actual yield (13.8g of FeS) and theoretical yield (15.2g FeS):

Percent yield = 13.8g / 15.2g * 100

<h3>Percent yield = 90.8%</h3>
5 0
3 years ago
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