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aniked [119]
3 years ago
14

What could we do to help restore some balance to the carbon cycle?

Chemistry
1 answer:
katen-ka-za [31]3 years ago
3 0

Answer:

The two most obvious ways to restore balance to carbon cycles are:

1. To move away from our dependence on fossil fuels.

  2.  To farm in ways that retain more carbon for longer periods than conventional farming methods currently do.

Explanation:

look up.

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Which statement best describes a typical difference that could be found between the "Analysis" and "Conclusion" sections of a
IgorC [24]

The statement “Only the “Conclusion” section discusses whether the original hypothesis was supported, and both sections suggest further  research”, best describes the difference between analysis and conclusion.

Answer: Option 4

<u>Explanation: </u>

In research, we do experiments and derive the results. Then, those results were analyzed by us. In this analysis part, we compare our results with the related results published elsewhere. Also, we correlate the similarities and point out the differences between our analysis and other reported results.

In conclusion part, we have to check hypothesis or it supported. And, we summarise our analysis and figure out the further research need to be done on that to improvise our research. So, the final statement is the correct option which best describes the difference between analysis and conclusion.

3 0
3 years ago
For the purposes of determining electron configuration of ions, when electrons are added to a neutral atom, they will inhabit or
irga5000 [103]

Answer: Aufbau principle

Explanation: Aufbau Principle suggests that in filling electrons into orbitals, lower energy orbitals must be filled before higher energy orbitals.

8 0
3 years ago
Read 2 more answers
How many atoms are in 165 g of calcium?
Whitepunk [10]
Atomic mass Calcium = 40.078 a.m.u

40.078 g ---------------- 6.02x10²³ atoms
165 g -------------------- ??

165 x ( 6.02x10²³) / 40.078 => 2.47x10²⁴ atoms

hope this helps!
4 0
3 years ago
(3) A 10.00-mL sample of 0.1000 M KH2PO4 was titrated with 0.1000 M HCl Ka for phosphoric acid (H3PO4): Ka1= 7.50x10-3; Ka2=6.20
shtirl [24]

Answer:

The pH of this solution is 1,350

Explanation:

The phosphoric acid (H₃PO₄) has three acid dissociation constants:

HPO₄²⁻ ⇄ PO4³⁻ + H⁺        Kₐ₃ = 4,20x10⁻¹⁰  (1)

H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺    Kₐ₂ = 6,20x10⁻⁸   (2)

H₃PO₄ ⇄ H₂PO4⁻ + H⁺       Kₐ₁ = 7,50x10⁻³   (3)

The problem says that you have 10,00 mL of KH₂PO₄ (It means H₂PO₄⁻) 0,1000 M and you add 10,00 mL of HCl (Source of H⁺) 0,1000 M. So you can see that we have the reactives of the equation (3).

We need to know what is the concentration of H⁺ for calculate the pH.

The moles of H₂PO₄⁻ are:

10,00 mL × ( 1x10⁻⁴ mol / mL) = 1x10⁻³ mol

The moles of H⁺ are, in the same way:

10,00 mL × ( 1x10⁻⁴ mol / mL) = 1x10⁻³ mol

So:

H₃PO₄   ⇄      H₂PO4⁻         +        H⁺           Kₐ₁ = 7,50x10⁻³   (3)

X mol     ⇄  (1x10⁻³-X) mol  + (1x10⁻³-X) mol                            (4)

The chemical equilibrium equation is:

Kₐ₁ = ([H₂PO4⁻] × [H⁺] / [H₃PO₄]

So:

7,50x10⁻³ = (1x10⁻³-X)² / X

Solving the equation you will obtain:

X² - 9,5x10⁻³ X + 1x10⁻⁶ = 0

Solving the quadratic formula you obtain two roots:

X = 9,393x10⁻³ ⇒ This one has no chemical logic because solving (4) you will obtain negative H₂PO4⁻ and H⁺ moles

X = 1,065x10⁻⁴

So the moles of H⁺ are : 1x10⁻³- 1,065x10⁻⁴ : 8,935x10⁻⁴ mol

The reaction volume are 20,00 mL (10,00 from both KH₂PO₄ and HCL)

Thus, the molarity of H⁺ ([H⁺]) is: 8,935x10⁻⁴ mol / 0,02000 L = 4,468x10⁻² M

pH is -log [H⁺]. So the obtained pH is 1,350

I hope it helps!

5 0
3 years ago
A vial containing radioactive selenium-75 has an activity of 3.0 mCi/mL. If 2.6 mCi are required for a leukemia test, how many m
oksian1 [2.3K]

Answer : The 866.66\mu L must be administered.

Solution :

As we are given that a vial containing radioactive selenium-75 has an activity of 3.0mCi/mL.

As, 3.0 mCi radioactive selenium-75 present in 1 ml

So, 2.6 mCi radioactive selenium-75 present in \frac{2.6mCi}{3.0mCi}\times 1ml=0.86666ml\times 1000=866.66\mu L

Conversion :

(1ml=1000\mu L)

Therefore, the 866.66\mu L must be administered.

4 0
3 years ago
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