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Rama09 [41]
3 years ago
8

PLEASE HELP ME PLEASE I'M BEGGING YOU

Chemistry
2 answers:
gulaghasi [49]3 years ago
6 0
Easy B particles is the anwser
Anit [1.1K]3 years ago
4 0
The answer is B charged particles
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Which characteristic makes legumes a good food source for food-insecure<br> populations?
vova2212 [387]

Legumes are much easier to grow than other plants, and are more adaptable.

6 0
3 years ago
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What is the pH of a substance with a hydrogen concentration of 1.0 x 10-13?
Nikolay [14]

Answer:

Favorite Answer

1.0 x10^-14 = (1.0 x 10^-13) (x)

x = 1.0 x 10^-1 = 0.1 M (this is the [OH-])

pOH = -log 0.1 = 1.0

Explanation:

I hope this helps :) sorry if not :(

4 0
3 years ago
Calculate the solubility of zn(oh)2(s) in 2.0 m naoh solution. (hint: you must take into account the formation of zn(oh)2−4, whi
Brilliant_brown [7]
When we have:

Zn(OH)2 → Zn2+ 2OH-  with Ksp = 3 x 10 ^-16

and:

Zn2+ + 4OH- → Zn(OH)4 2-  with Kf = 2 x 10^15
 
by mixing those equations together:

Zn(OH)2 + 2OH- → Zn(OH)4 2- with K = Kf *Ksp = 3 x 10^-16 * 2x10^15 =0.6

by using ICE table:

         Zn(OH)2 + 2OH- → Zn(OH)4 2-

initial                     2m              0

change                  -2X                +X     

Equ                       2-2X                 X

when we assume that the solubility is X

and when K = [Zn(OH)4 2-] / [OH-]^2

        0.6 = X / (2-2X)^2    by solving this equation for X

∴ X = 0.53 m

∴ the solubility of Zn(OH)2 = 0.53 M
4 0
3 years ago
The rocks circle suggests that
Rama09 [41]

B, Rocks are always being recycled into different forms.

7 0
3 years ago
Addition of an excess of lead (II) nitrate to a 50.0mL solution of magnesium chloride caused a formation of 7.35g of lead (II) c
KiRa [710]

Answer:

[Cl⁻] = 0.016M

Explanation:

First of all, we determine the reaction:

Pb(NO₃)₂ (aq) + MgCl₂ (aq) → PbCl₂ (s) ↓  +  Mg(NO₃)₂(aq)

This is a solubility equilibrium, where you have a precipitate formed, lead(II) chloride. This salt can be dissociated as:

           PbCl₂(s)  ⇄  Pb²⁺ (aq)  +  2Cl⁻ (aq)     Kps

Initial        x

React       s

Eq          x - s              s                  2s

As this is an equilibrium, the Kps works as the constant (Solubility product):

Kps = s . (2s)²

Kps = 4s³ = 1.7ₓ10⁻⁵

4s³ = 1.7ₓ10⁻⁵

s =  ∛(1.7ₓ10⁻⁵ . 1/4)

s = 0.016 M

3 0
3 years ago
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