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Anika [276]
3 years ago
15

Can Magnesium hydroxide be used as the base for this titration? Why or why not? 5. Describe the procedure that you will follow t

o prepare 250 ml of a .100 M soliution of KHP. 6. 25.0 ml of .100M KHP is required to neutralize 10.0 ml of barium hydroxide. Find the molarity of the barium hydroxide solution.
Chemistry
1 answer:
kvasek [131]3 years ago
4 0

Explanation:

A.

Yes

B.

2KHP + Mg(OH)2 = 2KOH + Mg(HP)2

KHP is an acidic salt which reacts with Mg(OH)2 which is a base via double displacement/decomposition reaction.

C.

• One approach is to prepare the solution volumetrically using KHP crystals.

• Preparing 1 liter of 0.1 M KHP you would add 0.1 moles of KHP

Molar mass of KHP(C8H5KO4)

= (12*8) + (5*1) + 39 + (16*4)

= 204 g/mol

Number of moles = mass/molar mass

= 20.4/204

= 0.1 mol.

• This is added to a 1 liter volumetric flask, add deionized water until near the fill line, stopper and mix

D.

Number of moles = molar concentration * volume

= 0.025 * 0.100

= 0.0025 mol

Equation for the reaction:

Ba(OH)2 + 2KHP --> Ba(KP)2 + 2H2O

1 mole of Ba(OH)2 reacted with 2 moles of KHP. By stoichiometry,

Number of moles of Ba(OH)2 = 0.0025/2

= 0.00125 mol.

Molarity = number of moles/volume

= 0.00125/0.01

= 0.125 M of Ba(OH)2.

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Answer:

C. physical; 100°C

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3 years ago
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A particular reaction, A- products, has a rate that slows down as the reaction proceeds. The half-life of the reaction is found
Thepotemich [5.8K]

Explanation:

Half life of zero order and second order depends on the initial concentration. But as the given reaction slows down as the reaction proceeds, therefore, it must be second order reaction. This is because rate of reaction does not depend upon the initial concentration of the reactant.

a. As it is a second order reaction, therefore, doubling reactant concentration, will increase the rate of reaction 4 times. Therefore, the statement  a is wrong.

b. Expression for second order reaction is as follows:

\frac{1}{[A]} =\frac{1}{[A]_0} +kt

the above equation can be written in the form of Y = mx + C

so, the plot between 1/[A] and t is linear. So the statement b is true.

c.

Expression for half life is as follows:

t_{1/2}=\frac{1}{k[A]_0}

As half-life is inversely proportional to initial concentration, therefore, increase in concentration will decrease the half life. Therefore statement c is wrong.

d.

Plot between A and t is exponential, therefore there is no constant slope. Therefore, the statement d is wrong

8 0
3 years ago
What is the concentration of no3- ions in a solution prepared by dissolving 15.0 g of ca(no3)2 in enough water to produce 300. m
r-ruslan [8.4K]

Ca(NO3)2 -------> Ca²⁺ +2NO3⁻

M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol

15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2


Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
1 mol                               2 mol
0.100 mol                           0.200 mol

We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution,
so
0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M

Concentration of NO3⁻ is 0.667 M.



4 0
3 years ago
When a solution produces equal numbers of hydrogen and hydroxyl ions, it is said to be neutral. Select one: True False
ioda

Answer:

true

Explanation:

The hydroxyl group is a functional group formed by an oxygen atom and a hydrogen atom.

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3 years ago
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