CHEMISTRY HIGH SCHOOL

How can compact fluorescent bulbs help the environment? (2 points) They will create new industry.
The will use less energy overall.
They will further science and technological development.
They will provide society with a product that is in short supply.

Answers

Answer 1
Answer:

Answer : fluorescent bulbs help the environment as they use less energy overall

Explanation :

Fluorescent bulbs are energy efficient. They consume less electricity as compared to conventional bulbs. They also generate less heat.

In modern world, we are facing energy crisis. The non renewable energy sources are getting depleted. It is highly necessary that we should start using alternative energy sources such as non renewable energy sources more often.

Another way to reduce our energy consumption is to use gadgets which use electricity more efficiently. One example is fluorescent bulb over traditional incandescent bulbs.

Therefore we can say that fluorescent bulbs help environment by using less energy and by producing less heat.

Answer 2
Answer: The answer to this question is " They will use less energy overall." 

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Tin-126 has a half life of 100,000 years. How many years does it take toreduce 100 g of Tin-126 to 12.5 g?
100,000 years
200,000 years
50,000 years
300,000 years

Answers

Answer:

300,000 years

Explanation:

100g=0 years

50g=100,00 years

25g=200,00 years

12.5g=300,000 years
300’000 yes I think she is correct

What is a covalent bond?​

Answers

A covalent bond is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs, and the stable balance of attractive and repulsive forces between atoms, when they share electrons, is known as covalent bonding. (hope this helps.)

Answer:

A covalent bond is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs, and the stable balance of attractive and repulsive forces between atoms, when they share electrons, is known as covalent bonding.

Explanation:

hope it helps, please mark as brainliest

What mass of oxygen is required to completely react with 2 moles of ch4

Answers

The balanced reaction that describes the reaction between methane and oxygen to produce water and carbon dioxide is expressed CH4  + 2 O2 = CO2 +2H2O. Hence for 2 moles of methane used, 4 moles of oxygen are needed to react completely.

What is the average kinetic energy of 1 mole of a gas that is at 100 degrees Celsius?

Answers

To simplify our solutions, we have to assume that the gas we are referring is ideal. The kinetic energy of of ideal gas is a funtion of moles and temperature only. 
KE = (3/2)(nRT)
 = (3/2)(1mol)(8.314J/molK)(100+273.15) = 4204.6J

The melting points of canola oil, corn oil, sunflower oil, and peanut oil are 10°C, –11°C, –17°C, and –2°C. If you cool a mixture of these oils to 5°C, which one can you separate easily?

Answers

You can separate out the canola oil easily, because it has cooled down enough to turn back into a solid.

Which of these statements about enzymes is NOT true? - If enough substrate is added, the normal Vmax of a reaction can be attained even in the presence of a competitive inhibitor.
- When [S] << Km, the reaction is second order and V0 depends on [S] and [Et].
- Their kcat is a second order rate constant.
- The lower their Km, the better they recognize their substrate, but the lower their reaction rate.
- When [S] << Km, V0 depends on [S] and [Et].

Answers

Answer:

1. True. 2. True. 3. Not true. 4. True. 5. True

Explanation:

1. Yes, because if the amount of substrate i much greater than of competitive inhibitor then the probability of substrate to bind to ferment is much higher than of inhibitor (if we have noncompetitive inhibitor it damages the structure of active site and the substrate concentration does not have a role in reaction rate).

2. Yeah, because then the michaelis-menten equation will transform into [tex} V0=(kcat*[E]*[S])/Km [/tex] and it is a second order equation.

3. No, because it is measured in sec-1 and that means it is 1 rate constant.

4. True, if the lower Km the better is binding and due to that rate is slower because it's harder for substrate to unbind.

5. The same as question two.
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