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azamat
3 years ago
15

Which of the following are producers?

Chemistry
2 answers:
nikklg [1K]3 years ago
7 0
All of the above is the answer
Artist 52 [7]3 years ago
7 0

Answer:

all the above

Explanation:

just got it right

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What trend in atomic radius occurs down a group on the periodic table? What causes the trend?
Svetlanka [38]

Explanation:

the further down the group you go, the larger the radius becomes. its caused by the amount of electrons needed to equal out the charge of the nucleus

3 0
3 years ago
5. Fill in the boxes below with solid, liquid, or gas.
g100num [7]

Answer:

i would say liquid

Explanation:

8 0
3 years ago
Can someone help me with this plz??
cestrela7 [59]

Answer:

I dont know at all and that is confusing.

5 0
3 years ago
Please help I’m torn between A and B. I’ve asked many and people say A, but I asked my science teacher and she said B. And also
astraxan [27]
If your science teacher says B, it’s probably because water has a negative and positive end, heat is just a form of energy, as other atoms can’t leave (they’re attracted to the ends) they are being insulated; but notice that ice will melt into gas (where atoms have tons of space) for other atoms to escape. Hence ice and gas aren’t ideal. (Air is a gas here.)
It’s not a 100% but hopefully it helps with some kind of analogy.
4 0
4 years ago
Compute the atomic density (the number of atoms per cm3 ... rather than the mass density g/cm3) for a perfect crystal of silicon
serious [3.7K]

Answer:

        \large\boxed{\large\boxed{5.00\times 10^{22}atoms/cm^3}}

Explanation:

You can convert the <em>density</em> into <em>atomic density</em> using the <em>atomic weight </em>and Avogadro's number

A dimensional analysis is very helpful:

           \dfrac{g}{cm^3}\times \dfrac{mol}{g}\times \dfrac{atoms}{mol}=\dfrac{atoms}{cm^3}

Follow the chain: g cancels with g, mol cancels with mol; at the end, what remains is atoms/cm³, which is what you want.

Use that with your data:

         \dfrac{2.33g}{cm^3}\times \dfrac{1mol}{28.09g}\times \dfrac{6.022\times 10^{23}atoms}{mol}=\approx 5.00\times10^{22}atoms/cm^3

3 0
3 years ago
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