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s344n2d4d5 [400]
3 years ago
7

Why is water a special chemical as a solid versus a liquid?

Chemistry
2 answers:
Andrew [12]3 years ago
7 0

Answer:

it is iceeeeeeeeeee

Explanation:

ice ice baby

Taya2010 [7]3 years ago
4 0

Answer: Solid water, or ice, is less dense than liquid water. Ice is less dense than water because the orientation of hydrogen bonds causes molecules to push farther apart, which lowers the density. ... Because ice is less dense than water, it is able to float at the surface of water.

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What is a positive effect of nuclear power plants?
vagabundo [1.1K]

Answer:

MRCORRECT has answered the question

Explanation:

Nuclear Energy Pros and Cons

As of today, nuclear energy is considered as one of the most environmentally friendly source of energy as it produces fewer greenhouse gas emissions during the production of electricity as compared to traditional sources like coal power plants

7 0
4 years ago
5
belka [17]

Answer:B

Explanation:

The metals are to the left of the linethe nonmetals are to the right of the line, and the elements immediately adjacent to the line are the metalloids.

6 0
3 years ago
At STP, which element is brittle and not a conductor of electricity
Sever21 [200]
Your answer is the element of S I think.
5 0
3 years ago
How do you solve this?4. How many km are equal to 1.55 x 10^4 m?
katrin [286]

Answer:

15.5 km

Explanation:

1.55 x 10^4 m means 1.55 x 10000 = 15,500 m

1 km = 1000 meters

To convert to km we divide by 1000

So 15,500/1000

= 15.5 km

4 0
3 years ago
Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring
ludmilkaskok [199]

Answer:

1.20 V

Explanation:

Pb(s) + Br_2(l)\rightarrow Pb^{2+}(aq) + 2Br^-(aq)

Here Pb undergoes oxidation by loss of electrons, thus act as anode. Bromine undergoes reduction by gain of electrons and thus act as cathode.

E^0=E^0_{cathode}- E^0_{anode}

Where both E^0 are standard reduction potentials.

Given,

Pb^{2+}(aq) + 2 e^-\rightarrow Pb(s)

E^0_{[Pb^{2+}/Pb]}= -0.13\ V

Br_2(l) + 2 e^-\rightarrow 2 Br(aq)

E^0_{[Br_2/Br^{-}]}=+1.07\ V

E^0=E^0_{[Br_2/Br^{-}]}- E^0_{[Pb^{2+}/Pb]}

E^0=+1.07- (-0.13)\ V=1.20\ V

6 0
4 years ago
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