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BlackZzzverrR [31]
3 years ago
12

Huryyyyyy plsssssssssssssssssss

Chemistry
1 answer:
AnnZ [28]3 years ago
8 0

Answer:

B.

Explanation:

bc it goes from having the potential to move to moving sorry if im wrong tho

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One gram is equal to ____ nanograms?
Katen [24]

Answer:

A

Explanation:

1g=1000mg

1g= 1000 000 micro gram

1g= 1000 000 000 nm = 10^9

5 0
3 years ago
What are the major problems associated with the production of nuclear energy?
Sindrei [870]

Answer:

• long time lag between planning and operation.

• cost.

• weapons proliferation risk.

• meltdown risk.

• mining lung cancer risk.

• carbon-equivalent emissions and air pollution.

• waste risk.

Explanation:

hope this help <33

7 0
3 years ago
The carbon bond strengths of graphite and diamonds are identical. True or False
Elena-2011 [213]

Answer:

false :)

Explanation:

3 0
4 years ago
Read 2 more answers
Citric acid is a naturally occurring compound. what orbitals are used to form each indicated bond? be sure to answer all parts.2
Rina8888 [55]

Answer : The orbitals that are used to form each indicated bond in citric acid is given below as per the attachment.

Answer 1) σ Bond a : C has SP^{2}  O has SP^{2} .

Explanation : The orbitals of oxygen and carbon which are involved in SP^{2} hybridization to form sigma bonds. This is observed at 'a' position in the citric acid molecule.

Answer 2) π Bond a: C has π orbitals and  O also has π  orbitals.

Explanation : The pi-bond at the 'a'position has carbon and oxygen atoms which undergoes pi-bond formation. And has pi orbitals of oxygen and carbon involved in the bonding process.

Answer 3) Bond b:  O SP^{3}  H has only S orbital involved in bonding.

Explanation : The bonding at 'b' position involves oxygen SP^{3} hydrogen atoms in it. It has SP^{3} hybridized orbitals and S orbital of hydrogen involved in the bonding.

Answer 4) Bond c:   C is SP^{3}  O is also SP^{3}

Explanation : The bonding involved at 'c' position has carbon and oxygen atoms involved in it. Both the atoms involves the orbitals of SP^{3} hybridized bonds.

Answer 5) Bond d:   C atom has SP^{3}  C atom has SP^{3}

Explanation : At the position of 'd' the bonding between two carbon atoms is found to be SP^{3}. Therefore, the orbitals that undergo SP^{3} hybridization are SP^{3}.

Answer 6) Bond e : C1 containing O SP^{2}    

C2 is SP^{3}

Explanation : The carbon atom which contains oxygen along with a double bond has SP^{2} hybridized orbitals involved in the bonding process; whereas the carbon at C2 has SP^{3} hybridized orbitals involved during the bonding. This is for the 'e' position.

7 0
4 years ago
At a transform fault boundary
tiny-mole [99]

Answer:

Most of these faults are hidden in the deep ocean, where they offset divergent boundaries in short zigzags resulting from seafloor spreading, the best-known (and most destructive) being those on land at the margins of continental tectonic plates. A transform fault is the only type of strike-slip fault that is classified as a plate boundary.

Explanation:

7 0
3 years ago
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