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nadezda [96]
3 years ago
12

BTW IT IS SCIENCE

Chemistry
2 answers:
labwork [276]3 years ago
8 0
The warm air rises and the cold air sinks.
Sliva [168]3 years ago
4 0
The answer is A warm air rises cool air sinks

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What are the chemical processes used to make chloroquine?
4vir4ik [10]
You spell Chlorine wrong but the process is called electrolysis  
6 0
3 years ago
Carbon-14 has a half-life of 5,730 years. How long will it take for 112. 5 g of a 120. 0-g sample to decay radioactively? 5,730
vichka [17]

The time taken by Carbon-14 to decay radioactively from 120g to 112.5g is 22,920 years.

<h3>How do we calculate the total time of decay?</h3>

Time required for the whole radioactive decay of any substance will be calculated by using the below link:

T = (n)(t), where

  • t = half life time = 5730 years
  • n = number of half life required for the decay

Initial mass of Carbon-14 = 120g

Final mass of Carbon-14 = 112.5g

Left mass = 120 - 112 = 7.5g

Number of required half life for this will be:

  • 1: 120 → 60
  • 2: 60 → 30
  • 3: 30 → 15
  • 4: 15 → 7.5

4 half lives are required, now on putting values we get

T = (4)(5730) = 22,920 years

Hence required time for the decay is 22,920 years.

To know more about radioactive decay, visit the below link:

brainly.com/question/24115447

#SPJ1

3 0
2 years ago
HELP PLEASE
ololo11 [35]
Third choice because the others make no sense
8 0
2 years ago
Read 2 more answers
How can I balance this equation? TELL ME HOW you did it and all the steps and you will be the brainliest.
lianna [129]

Answer:

3 CH3CH2OH + 4 H2CrO4 + 6 H2SO4 --> 3 CH3COOH + 2 Cr2(SO4)3 + 13 H2O

Explanation:

To balance, start of with the groups that are common on both sides of the reaction equation;

In this case, these are the SO4 groups treating them as single units;

There are 3 on the right side and 1 on the left side, so we put 3 in front of the H2SO4 to balance this first;

Next, deal with the Cr, there are 2 Cr on the right side and 1 on the other side, so we put a 2 in front of the H2CrO4 to balance that;

Thirdly, we notice the C are already balanced as there are 2 on each side so this is fine;

Lastly, we can deal with the O and H;

Bearing in mind the numbers that are in front of the molecules now from prior balancing, there are 9 O and 16 H on the left side and 3 O and 5 H on the right;

7 H2O on the right side would balance the O, but gives us 18 H, which is 2 too many H;

If we were to put 2 in front of the two organic molecules (the ones with C) on either side, we would balance the O by having 6 H2O, but this gives 2 fewer H than necessary;

In order for the H to balance, we need to have 13/2 (or 6.5) H2O, which means we need 3/2 (or 1.5) in front of each organic molecule;

Since, it is not sensible to have 13/2 water molecules or 3/2 organic molecules, we can just multiply everything by 2;

Thus we end up with:

3 CH3CH2OH + 4 H2CrO4 + 6 H2SO4 --> 3 CH3COOH + 2 Cr2(SO4)3 + 13 H2O

Rules of thumb:

- When there are common chemical groups (e.g. SO4) on both sides of a reaction equation, treat them as single units

- Start of with balancing these common groups

- Thereafter, balance the atoms that appear in only one reactant and one product

- Proceed to balance the atoms that appear in more than one reactant or product

- Typically, you should deal with the O and H last

4 0
3 years ago
Which describes a chemical reaction? A. Raw eggs are fried in a pan B. Raw eggs are scrambled with a fork C. Raw eggs are placed
Otrada [13]
Raw eggs are scrambled with a fork
8 0
3 years ago
Read 2 more answers
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