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taurus [48]
3 years ago
10

When chemical, transport, or mechanical work is done by an organism, what happens to the heat generated?

Chemistry
1 answer:
dalvyx [7]3 years ago
5 0

Explanation:

The heat generated when chemical, transport, or mechanical work is done by the organism is lost to the environment.

The the matter from one organism to the other is transferred via energy. The producer's produce and consumer's consume this energy according to 10% law. Most of the energy is lost (90% ) to the environment in form of heat.

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How many moles of sodium chloride can react with 18.3 liters of fluorine gas at 1.2 atmospheres and 299 Kelvin?
My name is Ann [436]

Answer:

1.79 mol.

Explanation:

  • For the balanced reaction:

<em>2NaCl + F₂ → 2NaF + Cl₂. </em>

It is clear that 2 mol of NaCl react with 1 mol of F₂ to produce 2 mol of NaF and 1 mol of Cl₂.

  • Firstly, we can get the no. of moles of F₂ gas using the general law of ideal gas: <em>PV = nRT.</em>

where, P is the pressure of the gas in atm (P = 1.2 atm).

V is the volume of the gas in L (V = 18.3 L).

n is the no. of moles of the gas in mol (n = ??? mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (299 K).

∴ no. of moles of F₂ (n) = PV/RT = (1.2 atm)(18.3 L)/(0.0821 L.atm/mol.K)(299 K) = 0.895 mol.

  • Now, we can find the no. of moles of NaCl is needed to react with 0.895 mol of F₂:

<em><u>Using cross multiplication:</u></em>

2 mol of NaCl is needed to react with → 1 mol of F₂, from stichiometry.

??? mol of NaCl is needed to react with → 0.895 mol of F₂.

∴ The no. of moles of NaCl needed = (2 mol)(0.895 mol)/(1 mol) = 1.79 mol.

4 0
3 years ago
Sharon reads two different articles about avocados. The first article, in a weight-loss magazine, claims that avocados are unhea
ser-zykov [4K]

Answer:

Is the source an authority on the subject?

Explanation:

To figure out which source is accurate, you want to find out the source's authority on the subject. For  example, if you wanted to learn how to play the piano, you wouldn't go to a dance teacher, you would go to someone who knows how to play the piano. The person who knows how to play the piano has more authority to teach the piano than a dance teacher.

3 0
3 years ago
An element has three naturally occurring isotopes. ten percent (.10) occurs as 55^x, fifteen percent (.15) occurs as isotope 56^
Leviafan [203]
The average atomic weight is, from the name itself, the average weight of all its naturally occurring isotopes. All you have to do is multiple the abundance of each isotope with its individual mass, then add them altogether.

Mass = (0.10*55)+(0.15*56)+(.75*57)
<em>Mass = 56.65 amu</em>
5 0
3 years ago
How to I answer this question! Not looking for the answers for the question just how to solve it myself!
Mrac [35]
I have not taken Chemistry in a year but I remember that if you look at a periodic table there are certain sections for polar and non polar elements. I dont know if you have learned that yet
7 0
3 years ago
Read 2 more answers
An insulated container contains 0.3 kg of water at 20 degrees C. An alloy with a mass of 0.090 kg and an initial temperature of
Lorico [155]

Answer:

The specific heat of the alloy is 2.324 J/g°C

Explanation:

<u>Step 1:</u> Data given

Mass of water = 0.3 kg = 300 grams

Temperature of water = 20°C

Mass of alloy = 0.090 kg

Initial temperature of alloy = 55 °C

The final temperature = 25°C

The specific heat of water = 4.184 J/g°C

<u>Step 2:</u> Calculate the specific heat of alloy

Qlost = -Qwater

Qmetal = -Qwater

Q = m*c*ΔT

m(alloy) * c(alloy) * ΔT(alloy) = -m(water)*c(water)*ΔT(water)

⇒ mass of alloy = 90 grams

⇒ c(alloy) = the specific heat of alloy = TO BE DETERMINED

⇒ ΔT(alloy) = The change of temperature = T2 - T1 = 25-55 = -30°C

⇒ mass of water = 300 grams

⇒ c(water) = the specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature = T2 - T1 = 25 - 20 = 5 °C

90 * c(alloy) * -30°C = -300 * 4.184 J/g°C * 5°C

c(alloy) = 2.324 J/g°C

The specific heat of the  alloy is 2.324 J/g°C

3 0
3 years ago
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