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Liula [17]
2 years ago
15

Alai is comparing the physical property of two materials. He is hitting each with a hammer to observe what happens. What physica

l property of the materials is Alai most likely observing?
a)solublity
b)conductivity
c)hardness
d)odor
Chemistry
2 answers:
maxonik [38]2 years ago
7 0
Since Alai is hitting each with a hammer, the physical property which he must be comparing on the two materials must be hardness. Solubility can be tested when you put them in a solvent. Odor can be tested with smell. Conductivity can be tested with electricity. Hardness can be tested with its resistance to the force that hits it.
PolarNik [594]2 years ago
7 0

C) HARDNESS

- odor is out because you are not smelling anything

- solubility is when the material dissolves, so that's out

- conductivity has to do with heat and energy

- the only available option is <u><em>hardness</em></u>

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The H3O+ in a 0.050M  solution of Ba(OH)2  is calculated as below

write  the equation for the dissociation of Ba(OH)2

Ba(OH)2  =  Ba^2+  +2OH^-

calculate the OH-  concentration

by use of mole ratio between Ba(OH)2  to OH^- which is 1:2 the concentration of OH  =  0.050 x2  = 0.1 M

by  use of  the  formula ( H3O+)(OH-) =  1 x10 ^-14

by  making H3O+ the subject of the formula
H3O+ = 1 x10^-14/ OH-

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2 years ago
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A 500 mL sample of a 0.100 M formate buffer, pH 3.75, is treated with 5 mL of 1.00 M KOH. What is the pH following this addition
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<u>Answer:</u> The pH of the solution after addition of KOH is 3.84

<u>Explanation:</u>

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3.75=3.75+\log(\frac{[HCOO-]}{[HCOOH]})\\\\\log(\frac{[HCOO-]}{[HCOOH]})=0\\\\\frac{[HCOO-]}{[HCOOH]}=1

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Molarity of KOH = 1.00 M

Volume of solution = 5 mL

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1.00M=\frac{\text{Moles of KOH}\times 1000}{5mL}\\\\\text{Moles of KOH}=0.005mol

The chemical reaction for formic acid and KOH follows the equation:

                  HCOOH+KOH\rightarrow HCOO^-+H_2O

Initial:       0.05    0.005               0.05

Final:         0.045          -                0.055          

Volume of solution = 500 + 5 = 505 mL = 0.505 L    (Conversion factor:  1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[salt]}{[acid]})

pH=pK_a+\log(\frac{[HCOO^-]}{[HCOOH]})

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pK_a = negative logarithm of acid dissociation constant of formic acid = 3.75

[HCOO^-]=\frac{0.055}{0.505}

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pH = ?

Putting values in above equation, we get:

pH=3.75+\log(\frac{0.055/0.505}{0.045/0.505})\\\\pH=3.84

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