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Nadusha1986 [10]
3 years ago
9

Do we generally break chemical bonds when we cool molecules?

Chemistry
1 answer:
olga_2 [115]3 years ago
7 0
The breaking of chemical bonds never releases energy to the external environment. Energy is only released when chemical bonds are formed. In general, a chemical reaction involves two steps: 1) the original chemical bonds between the atoms are broken, and 2) new bonds are formed. These two steps are sometimes lumped into one event for simplicity, but they are really two separate events. For instance, when you burn methane (natural gas) in your stove, the methane is reacting with oxygen to form carbon dioxide and water.
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Omg I’ve been stressing for the past minutes can someone please help me
Alona [7]

Answer:

I think it would be the last answer

Explanation:

5 0
3 years ago
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Match each of the following forms of sugar to its best description?
prohojiy [21]

Sucrose, a sweet, white crystalline substance, C12 H22 O11, OBTAINED CHIEFLY FROM THE JUICE OF THE SUGAR CANE AND SUGAR BEET, BUT ALSO PRESENT IN SORGHUM, THE sugar maple, some palms, and various other plants, and having extensive nutritional, pharmaceutical, and industrial uses; any of the class of carbohydrates to which this substance belongs, as glucose,  levulose, and lactose.
5 0
3 years ago
In an organic structure, you can classify each of the carbons as follows: Primary carbon (1o) = carbon bonded to just 1 other ca
Alla [95]

The question is incomplete, complete question is :

In an organic structure, you can classify each of the carbons as follows: Primary carbon (1°) = carbon bonded to just 1 other carbon group Secondary carbon (2°) = carbon bonded to 2 other carbon groups Tertiary carbon (3°) = carbon bonded to 3 other carbon groups Quaternary carbon (4°) = carbon bonded to 4 other carbon groups How many carbons of each classification are in the structure below? How many total carbons are in the structure? How many primary carbons are in the structure? How many secondary carbons are in the structure? How many tertiary carbons are in the structure? How many quaternary carbons are in the structure?

Structure is given in an image?

Answer:

There are 10 carbon atoms in the given structures out of which 6 are 1° , 1 is 2° , 2 are 3° and 1 is 4°.

Explanation:

Total numbers of carbon = 10

Number of primary carbons that is carbon joined to just single carbon atom = 6

Number of secondary carbons that is carbon joined to two carbon atoms = 1

Number of tertiary carbons that is carbon joined to three carbon atoms = 2

Number of quartenary carbons that is carbon joined to four carbon atoms = 1

So, there are 10 carbon atoms in the given structures out of which 6 are 1° , 1 is 2° , 2 are 3° and 1 is 4°.

3 0
3 years ago
The amount of energy required to heat water for a 10-minute shower (50 gallons) is 2.2125 kJ. How many calories is this?
Zigmanuir [339]
From the problem statement, this is a conversion problem. We are asked to convert from units of kilojoules to units of calories. To do this, we need a conversion factor which would relate the different units involved. We either multiply or divide this certain value to the original measurement depending on what is asked. From literature, we will find that 1 kilojoule is equal to 239 calories. We do as follows:
<span>
2.2125 kJ ( 239 calories / 1 kJ ) = 528.79 calories
</span><span>
</span>
6 0
3 years ago
If a radioactive isotope has a half-life of 26.5 days, how many days does it take for a sample of the isotope to decrease by 35.
dangina [55]

Answer:

16.5 days

Explanation:

Given that:

Half life = 26.5 days

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac {ln\ 2}{t_{1/2}}

k=\frac {ln\ 2}{26.5}\ days^{-1}

The rate constant, k = 0.02616 days⁻¹

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given:

35.0 % is decomposed which means that 0.35 of [A_0] is decomposed. So,

\frac {[A_t]}{[A_0]} = 1 - 0.35 = 0.65

t = 7.8 min

\frac {[A_t]}{[A_0]}=e^{-k\times t}

0.65=e^{-0.02616\times t}

<u>t = 16.5 days.</u>

7 0
3 years ago
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