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Salsk061 [2.6K]
3 years ago
12

I need an answer ASAP What important role do sand dunes play?

Chemistry
1 answer:
Nezavi [6.7K]3 years ago
6 0

Answer:

Keeps tides on the shore and away from the city

Explanation

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Kareem wants to dissolve rock salt in a glass of water at
Mars2501 [29]
Add hot water into the glass
5 0
3 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
Carbon-14 has a half-life of 5,730 years. If scientists discover a mummified Cro-Magnon specimen, they will first want to determ
ryzh [129]

Answer:

the specimen has 11460 years old

Explanation:

if the live sample has as initial amount of Yo C14, the dead sample will have 0.25Yo of C14.

the rate of decay of radiactive matter over time is

  • Y(t) = Yo * e (( - t * Ln2 ) / T)

∴ Y(t) = 0.25Yo;  T = 5730

⇒ 0.25Yo = Yo * e (( - t * Ln2 ) / 5730 )

⇒ 0.25 = e (( - t * Ln2 ) / 5730 )

⇒ Ln(0.25) = ( -  t * Ln2 ) / 5730

⇒ - 7943.466 = - t * Ln2

⇒ 7943.466 / Ln2 = t

⇒ t = 11460 year

4 0
3 years ago
1. The elements in the modern Periodic Table are arranged according to their
Usimov [2.4K]

Answer:

A). Atomic number

Explanation:

THIS IS EASY

4 0
3 years ago
From the illustration of the potassium atom fill in the periodic table selection
Phoenix [80]

Answer:

2 electrons

8 electrons

8 electrons

1 electron (just count in the rings for these ones)

19 protons (it says "P=19" in the middle. The P stands for Protons)

K (it's from Latin medieval Latin "kalium")

39 (add proton and neutron amounts to get the approximate atomic mass--electrons are negligible)

Explanation:

6 0
3 years ago
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