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lesya [120]
3 years ago
11

Which example is a body fossil?

Chemistry
1 answer:
marin [14]3 years ago
3 0
Answer:
B
Explanation:
An body fossil is something that is apart of the animals b o d y that was left behind to be fossilized.
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6 Which element requires the least amount of
DaniilM [7]

Which element requires the least amount of

energy to remove the most loosely held electron

from a gaseous atom in the ground state?

<h3>Answer-</h3><h3>Na</h3>
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3 years ago
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The Difference Between an Alloy and an Amalgam
Delicious77 [7]

Mixtures or combinations of various different metals or metallic substances form things called alloys. An alloy composed of mercury and other metal (or metals) forms "amalgam". When a true alloy is created, the component metals are combined together at a temperature which is greater than the melting point of all of them.

Also, it helps to remember the word "amalgamate", which means "to alloy (a metal) with mercury" according to Dictionary.com.


Hope this helped :)


(btw I'm like 3 brainliest answers away from my next rank so could you...you know... :)

6 0
4 years ago
Why do reactions need to be balanced?
nexus9112 [7]

Answer:

A part.

Explanation:

Because the reactants must be the exact same as the the products.

5 0
3 years ago
The Periodic Table of Elements is organized by the number of:
bija089 [108]

Answer:

c) protons

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7 0
3 years ago
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Two different bromide solutions are mixed with each other: Solution 1 is an aqueous solution of 4.85 g aluminum bromidein 150. m
erma4kov [3.2K]

Answer:

M=0.380 M.

Explanation:

Hello there!

In this case, given those two solutions of aluminum bromide and zinc bromide, it is firstly necessary to compute the moles of bromide ions in each solution as shown below:

n_{Br^-}^{in\ AlBr_3}=4.85 gAlBr_3*\frac{1molAlBr_3}{266.69gAlBr_3}*\frac{3molBr^-}{1molAlBr_3}  =0.05456molBr^-\\\\n_{Br^-}^{in\ ZnBr_2}=7.75gZnBr_2*\frac{1molZnBr_2}{225.22gZnBr_2}*\frac{2molBr^-}{1molZnBr_2}  =0.06882molBr^-

Now, we compute the total moles of bromide:

n_{Br^-}=0.05456mol+0.06882mol\\\\n_{Br^-}=0.12338mol

Then, the total volume in liters:

150mL+175mL=325mL*\frac{1L}{1000mL} \\\\=0.325L

Therefore, the concentration of total bromide is:

M=\frac{0.12338mol}{0.325L}\\\\M=0.380M

Best regards!

8 0
3 years ago
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