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faust18 [17]
3 years ago
12

An anaerobic reaction takes place

Chemistry
1 answer:
Dafna1 [17]3 years ago
6 0
As the name implies, anaerobic means there is no presence of oxygen. So if it will take place, it will only take place in little or no oxygen which is in the last sentence in the following choices. First and second choice are wrong for there is a need of oxygen required when there is no oxygen is needed and the presence of nitrogen in water.
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Describe how radioactive isotopes are used to treaat cancer​
stich3 [128]

Answer:

Radioisotope therapy is a procedure in which a liquid form of radiation is administered internally through infusion or injection. RIT's ultimate purpose is to treat cancerous cells with minimal damage to the normal surrounding tissue. These therapies are not normally the first approach used to fight a patient's cancer.

Explanation:

5 0
2 years ago
fills a 500.mL flask with 3.6atm of carbon monoxide gas and 1.2atm of water vapor. When the mixture has come to equilibrium she
enot [183]

Answer:

The answer to the question is

The pressure of carbon dioxide after equilibrium is reached the second time is 0.27 atm rounded to 2 significant digits

Explanation:

To solve the question, we note that the mole ratio of the constituent is proportional to their partial pressure

At the first trial the mixture contains

3.6 atm CO

1.2 atm H₂O (g)

Total pressure = 3.6+1.2= 4.8 atm

which gives

3.36 atm CO

0.96 atm H₂O (g)

0.24 atm H₂ (g)

That is

CO+H₂O→CO(g)+H₂ (g)

therefore the mixture contained

0.24 atm CO₂ and the total pressure =

3.36+0.96+0.24+0.24 = 4.8 atm

when an extra 1.8 atm of CO is added we get Increase in the mole fraction of CO we have one mole of CO produces one mole of H₂

At equilibrium we have 0.24*0.24/(3.36*0.96) = 0.017857

adding 1.8 atm CO gives 4.46 atm hence we have

 (0.24+x)(0.24+x)/(4.46-x)(0.96-x) = 0.017857

which gives x = 0.031 atm or x = -0.6183 atm

Dealing with only the positive values we have the pressure of carbon dioxide = 0.24+0.03 = 0.27 atm

7 0
3 years ago
Need help with the 1st one
lutik1710 [3]

If I’m right this should be the answer

7 0
3 years ago
Without consulting Appendix B, arrange each group in order of increasing standard molar entropy (S°). Explain.(c) SF₆(g), SF₄(g)
Andre45 [30]

The increasing order of standard molar entropy (S°) is as follow:

SF₄(g) < SF₆(g) < S₂F₁₀(g)

<h3>What is Entropy? </h3>

Entropy is defined as the randomness of the particle. It depends on temperature and pressure or number of particle per unit volume.

It is directly proportional to the temperature and pressure of the gas.

<h3>What is Standard Molar Entropy? </h3>

The standard molar entropy is defined as the entropy content of the one mole of pure substance at the standard state of temperature and pressure of interest.

The standard molar entropy is also defined as the total amount of entropy which 1 mole of the substance acquire, as it is brought from 0K to standard conditions of temperature and pressure.

The standard molar entropy depends on the molas mass of atom, molecules or compound.

SF₄(g) has lower standard molar entropy. Due to less complexity of this molecules.

While, complexity increases from SF₆(g) to S₂F₁₀(g). Therefore, the standard molar entropy of S₂F₁₀(g) is greater than SF₆(g).

Thus, we concluded that the increasing order of standard molar entropy (S°) is as follow:

SF₄(g) < SF₆(g) < S₂F₁₀(g)

learn more about standard molar entropy:

brainly.com/question/15908262

#SPJ4

7 0
2 years ago
A 20g piece of lead absorbs 566 joules of heat and its temperature changes from 35 oC to 195 oC. Calculate the specific heat.
Alja [10]

Answer:

  • <u>Question 1: 0.2J/(gºC)</u>
  • <u>Question 2: 6,000J</u>
  • <u>Question 3: 300J</u>
  • <u>Question 4: 80g</u>
  • <u>Question 5: 74ºC</u>
  • <u>Question 6: 50g</u>

<u></u>

Explanation:

Question 1.<em> A 20g piece of lead absorbs 566 joules of heat and its temperature changes from 35º oC to 195º C. Calculate the specific heat.</em>

<em />

The thermal energy equation is:

  • Q = m × C × ΔT

<em />

Substitute and solve for C:

  • 566J = 20g × C × (195ºC - 35ºC)
  • C = 566J / (20g × 160ºC)
  • C = 0.177 J/(gºC) ≈ 0.2J/(gºC)

<em />

You must round to one significant figure because one factor has one significant figure).

<em />

<em />

Qustion 2.<em> 40g of water is heat at 40ºC and the temperature rise to 75ºC. What is the amount of heat needed for the temperature to rise? (specific heat of water is 4.184 J/gºC)</em>

<em />

Use the thermal energy equation again:

  • Q = m × C × ΔT

<em />

Substitute and compute:

  • Q = 40g × 4.184 J/gºC × (75ºC - 40ºC)
  • Q = 5,857.6J

Round to one significant figure: 6,000J

<em />

Question 3. <em>Graphite has a mass of 50g and a specific heat of 0.420 J/gºC. If graphite is cooled from 50ºC to 35ºC, how much energy was lost?</em>

  • Q = m × C × ΔT
  • Q = 50g × 0.420J/gºC × (35ºC - 50ºC)
  • Q = 315J

Round to one significant figure (because 50g has one significant figure)

  • Q = 300J

<em />

Question 4.<em> </em><em>Iron has a specific heat of 0.712 J/gºC. A piece of iron absorbs 3000J of energy and undergoes a temperature change totaling 50ºC, What is the mass of iron?</em>

<em />

  • Q = m × C × ΔT

Solve for m:

  • m = Q / (C × ΔT)

Substitute and compute:

  • m = 3,000J / (0.712J/gºC × 50ºC)
  • m = 84.26 g ≈ 80 g (rounded to one significant figure, because the factor 3,000J has one significant figure).

Question 5. <em>If 400g of an unknown solution at 70ºC loses 7500 J of heat, what is the final temperature of the unknown solution. The unknown solution has a specific heat of 4.184 J/gºC.</em>

<em />

  • Q = m × C × ΔT

<em />

Q is negative, since it is released.

Substitute and solve for T:

  • - 7,500J = 400g × 4.184J/gºC × (T - 70ºC)

  • T = - 7500J / 400g × 4.184J/gºC) + 70ºC

  • T = 74ºC

<em />

If you round to one significant figure you cannot tell the temperature difference, thus leave two significant figures.

<em />

Question 6. <em>How many grams of water would require 9500J of heat to raise the temperature from 50ºC to 100ºC</em>

  • Q = m × C × ΔT

Subsitute:

  • 9,500J = m × 4.184J/gºC × (100ºC - 50ºC)

Solve for m and compute:

  • m = 9,500J / (4.184J/gºC × 50ºC)

  • m = 45g

Since the temperatures indicate one singificant figure, the mass should be rounded to one significant figure:

  • m = 50g.
8 0
3 years ago
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