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svp [43]
3 years ago
9

Somebody please help me

Chemistry
1 answer:
Oliga [24]3 years ago
8 0

Answer:

7. solar flares- f. eruptions that occur, bleh bleh

8.core- h. the center of the sun

9.chromosphere- b.the layers of the sun's surface that has reddish, bleh bleh

10.sunspot- d. that are cooler, bleh bleh

11.corona- c.looks like halo during the eclipse bleh bleh

12. nuclear fusion -j. joining hydrogens to form heliums

13.photosphere- a. gives off visible lights bleh blh

14.solar wind- g. stream charged of corona bleh bleh

15.prominence -e. reddish loops of gas

16.radiation zone- k. electromagnetic radiation bleh bleh

17. convection zone- i. outermost layer bleh bleh

Explanation:

ain't i the brainiest

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When the procedure calls for making a more dilute solution of an acid, or mixing an acid with other solutions, what is the corre
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Answer:

The correct order it b. always add acid last.

Explanation:

Adding acid first could result on a violent reaction and heat or fumes can be generated. The best approach is to always add all the water or non-acid component first, or add a significant portion before adding the acid slowly to the mixture.

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3 years ago
Find the pHpH of a solution prepared from 1.0 LL of a 0.15 MM solution of Ba(OH)2Ba(OH)2 and excess Zn(OH)2(s)Zn(OH)2(s). The Ks
charle [14.2K]

Answer:

pH  = 13.09

Explanation:

Zn(OH)2 --> Zn+2 + 2OH-   Ksp = 3X10^-15

Zn+2 + 4OH-   --> Zn(OH)4-2   Kf = 2X10^15

K = Ksp X Kf

  = 3*2*10^-15 * 10^15

  = 6

Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M

                Zn(OH)₂ + 2OH⁻(aq)  --> Zn(OH)₄²⁻(aq)

Initial:           0             0.3                      0

Change:                      -2x                     +x

Equilibrium:               0.3 - 2x                 x

K = Zn(OH)₄²⁻/[OH⁻]²

6 = x/(0.3 - 2x)²  

6 = x/(0.3 -2x)(0.3 -2x)

6(0.09 -1.2x + 4x²) = x

0.54 - 7.2x + 24x² = x

24x² - 8.2x + 0.54 = 0

Upon solving as quadratic equation, we obtain;

x = 0.089

Therefore,

Concentration of (OH⁻) = 0.3 - 2x

                                    = 0.3 -(2*0.089)

                                  = 0.122

pOH = -log[OH⁻]

         = -log 0.122

          = 0.91

pH = 14-0.91

     = 13.09

4 0
3 years ago
After undergoing alpha decay, an atom of radium-226 becomes
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The answer is b. radon-222. The alpha decay means that it will emit an alpha particle when decays. The alpha particle has two protons and two neutrons. So Radium(88) minus two protons will become Radon(86). And the atomic mass will become 226-4=222.
8 0
3 years ago
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