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andreev551 [17]
3 years ago
6

A laboratory experiment requires 2.25 L of a 1.0 M solution of phosphoric acid (H3PO4), but the only available H3PO4 is a 9.0 M

stock solution. How could you prepare the solution needed for the lab experiment? Show all the work used to find your answer.
Chemistry
2 answers:
Ksenya-84 [330]3 years ago
6 0
The solution needed is prepared  as below

by use of the   M1V1 =M2 V2  formula  where
M1 = 2.25 L
v2 = 1.0M
M2 = 9.0 M
V2 =? l

make V2  the subject of the formula  V2 =M1V1/M2

= 2.25 L x  1.0M/9.0 M  = 0. 25 L

therefore  the solution  need  0.25 L  of   9.0M H3PO4   and  dilute it a final volume  of 2.25 l 
Semenov [28]3 years ago
3 0
We know that, M1V1   =   M2V2
                        (Initial)      (Final)

Given: M1 = 9.0 M
            M2 = 1.0 M
            V2 = 2.25 L

Therefore we have, 9.0 x V1 = 1.0 x 2.25
                                         V1 = 0.25 L

Thus, 0.25 L of stock solution (9.0 M), when diluted till 2.25 L, the resultant solution will have concentration 1.0 M.
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4 0
3 years ago
A 25.0-mL sample of 0.150 M hydrocyanic acid is titrated with a 0.150 M NaOH solution. The Ka of hydrocyanic acid is 4.9 × 10-10
lara [203]

Answer:

The pOH = 1.83

Explanation:

Step 1: Data given

volume of the sample = 25.0 mL

Molarity of hydrocyanic acid = 0.150 M

Molarity of NaOH = 0.150 M

Ka of hydrocyanic acid = 4.9 * 10^-10

Step 2: The balanced equation

HCN + NaOH → NaCN + H2O

Step 3: Calculate the number of moles hydrocyanic acid (HCN)

Moles HCN = molarity * volume

Moles HCN = 0.150 M * 0.0250 L

Moles HCN = 0.00375 moles

Step 3: Calculate moles NaOH

Moles NaOH = 0.150 M * 0.0305 L

Moles NaOH = 0.004575 moles

Step 4: Calculate the limiting reactant

0.00375 moles HCN will react with 0.004575 moles NaOH

HCN is the limiting reactant. It will completely be reacted. There will react 0.00375 moles NaOH. There will remain 0.004575 - 0.00375 = 0.000825 moles NaOH

Step 5: Calculate molarity of NaOH

Molarity NaOH = moles NaOH / volume

Molarity NaOH = 0.000825 moles / 0.0555 L

Molarity NaOH = 0.0149 M

Step 6: Calculate pOH

pOH = -log [OH-]

pOH = -log (0.0149)

pOH = 1.83

The pOH = 1.83

6 0
3 years ago
In a chemical reaction, ____ interact to form ____. select one:
QveST [7]
A. Reactants, products

In a chemical reaction, reactants interact to form products
7 0
3 years ago
18.2 mL of a 0.156 M solution of lead(II) nitrate are added to 26.2 mL of a 0.274 M solution of potassium iodide.
kow [346]

The mass of the Pbl2 : 1308.87

<h3>Further explanation</h3>

Given

18.2 mL of a 0.156 M Pb(NO3)2

26.2 mL of a 0.274 M KI

Reaction

Pb(NO3)2 (aq) + 2 KI (aq) - Pbl2 (s) + 2 KNO3

Required

the mass of the Pbl2

Solution

mol Pb(NO3)2 = 18.2 x 0.156 = 2.8392 mlmol

mol KI = 26.2 x 0.274 =7.1788 mol

Limiting reactant Pb(NO3)2(smaller ratio of mol : reaction coeffiecient)

mol Pbl2 based on limiting reactant (Pb(NO3)2)

From equation, mol ratio of Pb(NO3)2 : Pbl2 = 1 : 1, so mol Pbl2=mol Pb(NO3)2=2.8392

Mass Pb(NO3)2 :

\tt mass=mol\times MW\\\\mass=2.8392\times 461\\\\mass=1308.87~grams

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3 years ago
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tensa zangetsu [6.8K]
An hour is 60 minutes. So you multiple the hot by 60. You get 180 minutes plus the 75 remaining minutes. The answer is 255
5 0
3 years ago
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