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Alex73 [517]
3 years ago
13

Sheena has a lump of sodium, a lump of potassium and a lump of lithium, but they’ve got mixed up and she doesn’t know which one

happens 1. Which group of periodic table are all these elements in? 2. What id the pattern of reactivity in this group -answer for brainlest
Chemistry
1 answer:
Arada [10]3 years ago
5 0

Answer:

Explanation:

Just saw your request regarding answering this so here it is:

All of them belong of Group 1 in periodic table and thus are highly reactive! Pattern of reactivity for Group 1 (Alkali metals) increases as you move down the group as their radius keeps increasing and thus electrons can be easily lost. Thus, to ID the lumps, Sheena should look at their reactivity and she should get the following trend:

Most reactive: Potassium (K)

Intermediate: Sodium (Na)

Least reactive: Lithium (Li)

Hope it helps!

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A block of pine wood has a mass of 120 g and a volume of 300 cm3. What is the density of wood? (Show your work)
Zielflug [23.3K]

Answer:

The answer is

<h2>0.4 g/cm³</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}

From the question

mass of wood = 120 g

volume = 300 cm³

So the density of the wood is

density =  \frac{120}{300}  \\  =  \frac{12}{30}  =  \frac{2}{5}

We have the final answer as

<h3>0.4 g/cm³</h3>

Hope this helps you

5 0
3 years ago
Calculate the ph of a 0.60 m h2so3, solution that has the stepwise dissociation constants ka1 = 1.5 × 10-2 and 1.82 1.06 1.02 2.
Vsevolod [243]
Missing in your question Ka2 =6.3x10^-8
From this reaction:
 H2SO3 + H2O ↔ H3O+  + HSO3-
by using the ICE table :
                H2SO3     ↔    H3O     +    HSO3- 
intial         0.6                     0                  0
change     -X                      +X                +X
Equ         (0.6-X)                  X                   X

when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
∴[H2SO3] = 0.6 - 0.088 = 0.512
[HSO3-] = [H3O+] = 0.088

by using the ICE table 2:
                 HSO3-     ↔   H3O     +     SO3-
initial        0.088              0.088              0
change    -X                      +X                   +X
Equ         (0.088-X)          (0.088+X)          X

Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] =  0.088 as the value of Ka2 is very small
 6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
               = 0.088 m ( because X is so small)
∴PH= -㏒[H3O+]
       = -㏒ 0.088 = 1.06 
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