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ira [324]
3 years ago
7

Answer (example of a bar graph)

Chemistry
1 answer:
aniked [119]3 years ago
4 0
Basically it's a chart that shows how much something has for instance number of candy bars: let's say for instance bob has thirty you have to make a bar that is up to the 30 line. Lol hope that helps

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If fluorine 20 undergoes beta decay , what will it become ?
e-lub [12.9K]
Fluorine 20 (F - Atomic number 9 and atomic mass 20). Firstly we need to know what is beta decay. Beta decay occurs when one neutron changes into a proton and an electron therefore the atomic mass will remain the same as even though we loose a neutron it is replaced by a proton, the atomic number is always raised by 1 when one beta decay occurs. The produced electron is shot out of the nucleus at an incredible speed. This speedy electron we call a beta particle.

Ok now the reaction.

 20       20        0
F  -> Ne    +     e
 9         10       -1

Remember the atomic number determines the nature of the element ( i.e what elemnt it is).
Hope this helps :).
7 0
3 years ago
Which is an example of a polymer?
g100num [7]

Answer:

diamond

Explanation:

5 0
2 years ago
How many grams of glucose are needed to prepare 144.3 mL of a 1.4%(m/v) glucose solution?
Montano1993 [528]

Answer:

2.0202 grams

Explanation:

1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.

so ?g glucose = 144.3 mL soln

Now apply the conversion factor, and you have:

?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).

so you have (144.3x1.4/100) g glucose= 2.0202 grams

6 0
3 years ago
Read 2 more answers
The decomposition of a compound at 400⁰C is first order with half-life of 1570 seconds. what fraction of an initial amount of th
kirill115 [55]

Answer: After 4710 seconds, 1/8 of the compound will be left

Explanation:

Using the formulae

Nt/No = (1/2)^t/t1/2

Where

N= amount of the compound  present at time t

No= amount of compound present at time t=0

t= time taken for N molecules of the compound to remain = 4710 seconds

t1/2 = half-life of compound  = 1570 seconds

Plugging in the values, we have  

Nt/No = (1/2)^(4710s/1570s)

Nt/No = (1/2)^3

Nt/No= 1/8

Therefore after 4710 seconds, 1/8 molecules of the compound will be left

5 0
2 years ago
In uncompetitive inhibition, the inhibitor can bind to the enzyme only after the substrate binds first. (T/F)
Veronika [31]

Answer:

True

Explanation:

In an uncompetitive inhibition, initially the substrate [S] binds to the active site of the enzyme [E] and forms an enzyme-substrate activated complex [ES].

The inhibitor molecule then binds to the enzyme- substrate complex [ES], resulting in the formation of [ESI] complex, thereby inhibiting the reaction.

This inhibition is called uncompetitive because the inhibitor does not compete with the substrate to bind on the active site of the enzyme.

Therefore, in an uncompetitive inhibition, the inhibitor molecule can not bind on the active site of the enzyme directly. The inhibitor can only bind to the enzyme-substrate complex formed.

       

4 0
2 years ago
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