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nikklg [1K]
3 years ago
6

Plsssssssssssssssssssssss herlppppppppppppppppppppppppp meeeee

Chemistry
1 answer:
madreJ [45]3 years ago
3 0
Vas happenin!!

I would say the second one

Hope this helps

-Zayn Malik
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Downwelling is the process that moves cold, dense water from the ocean surface to the sea floor near the polar regions. How can
Finger [1]

Answer:

downwelling also allows for deep ocean oxygenation to occur because these waters are able to bring dissolved oxygen down from the surface to help facilitate aerobic respiration in organisms throughout the water column

7 0
3 years ago
Identify the Lewis acid and Lewis base in each reaction:Na⁺ + 6H₂O ⇆ Na (H₂O)⁺₆
klemol [59]

In the reaction, Na⁺ + 6H₂O ⇆ Na (H₂O)⁺₆, Na⁺ is a Lewis acid and H₂O is a Lewis base.

<h3>What is a Lewis acid-base reaction?</h3>
  • According to the Lewis theory of acid-base reactions, acids accept pairs of electrons and bases donate pairs of electrons.
  • Any substance like H+ ion, which is capable of accepting a pair of nonbonding electrons or an electron-pair acceptor is known as a Lewis acid.
  • Any substance, like the OH- ion, that is capable of donating a pair of nonbonding electrons or an electron-pair donor is a Lewis base.
  • Here Na⁺ that is electron deficient accepts electrons from the electron donor, H₂O
  • From the Lewis theory, with no change in the oxidation numbers of any atoms, acids react with bases to share a pair of electrons.

To learn more about Lewis acid-base reactions: brainly.com/question/14861040  

#SPJ4                                                                                              

4 0
2 years ago
If a particular ore contains 58.6 % calcium phosphate, what minimum mass of the ore must be processed to obtain 1.00 kg of phosp
rjkz [21]
Answer is: mass of the ore is 8.54kg.<span>

</span>ω(Ca₃(PO₄)₂ - calcium phosphate) = 58.6% ÷ 100% = 0.586.
m(P) = 1.00 kg · 1000 g/kg.
m(P) = 1000 g.
In one molecule of calcium phosphate there are two phosphorus atoms:
M(Ca₃(PO₄)₂) = 310.18 g/mol.
M(P) = 30.97 g/mol.
For one kilogram of phosphorus, we need:
M(Ca₃(PO₄)₂) : 2M(P) = m(Ca₃(PO₄)₂) : m(P).
310.18 g/mol : 61.94 g/mol = m(Ca₃(PO₄)₂) : 1000 g.
m(Ca₃(PO₄)₂) = 5007.75 g ÷ 1000 g/kg = 5.007 kg.
Mass of ore find from proportion:
m(Ca₃(PO₄)₂) : m(ore) = 56% : 100%.
m(ore) = 100% · 5.007 kg ÷ 58.6%.
m(ore) = 8.54kg.

5 0
3 years ago
"What makes a bond polar"?
r-ruslan [8.4K]

Answer:

The answer to your question is: letter A.

Explanation:

A Covalent bond polar is between 2 non metals where one atom is bigger than the other one so the distribution of charges creates this polarity.

A. One atom attracts shared electrons more strongly than the other atom  This is the correct definition of bond polar, one element is bigger and stronger than the other element.

B. One atom has transferred its electrons completely to another atom  This definition is incorrect, it is the definition of ionic bonding.

C. A sea of electrons has been created between the elements  This definition is incorrect for the polar bond, it describes a metallic bonding.

D. Two atoms are sharing electrons with equal attraction This definition is incorrect for a polar bond, but is the correct definition for nonpolar bonding.

8 0
3 years ago
4) The initial rate of the reaction between substances P and Q was measured in a series of
ASHA 777 [7]

Answer:

The initial rate of the reaction between substances P and Q was measured in a series of

experiments and the following rate equation was deduced.

rate = k[P]^{2} [Q]

Complete the table of data below for the reaction between P and Q

Explanation:

Given rate of the reaction is:

rate= k[P]^{2} [Q]\\=>[Q]=\frac{rate}{k.[P]^{2} } \\and \\\\\\\ [P]=\sqrt{\frac{rate}{k.[Q]} }

Substitute the given values in this formulae to get the [P], [Q] and rate values.

From the first row,

the value of k can be calulated:

k=\frac{rate}{[P]^{2}[Q] } \\  =\frac{4.8*10^-3}{(0.2)^{2} 2. (0.30)} \\ =0.4

Second row:

2. Rate value:

rate =0.4* (0.10)^{2} * (0.10)\\\\        =4.0*10^-3mol.dm^-3.s^-1

3.Third row:

[Q]=\frac{rate}{k.[P]^{2} } \\     =9.6*10^-3 / (0.4 *(0.40)^{2} \\    =0.15mol.dm^{-3}

4. Fourth row:

[P]=\sqrt{\frac{rate}{k.[Q]} }\\=>[P]=\sqrt{\frac{19.2*10^-3}{0.60*0.4} } \\=>[P]=0.283mol.dm^{-3}

6 0
3 years ago
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