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tia_tia [17]
1 year ago
10

Suppose you are performing a titration. at the beginning of the titration, you read the titrant volume as 2.42 ml. after running

the titration and reaching the endpoint, you read the titrant volume as 22.23 ml. what volume, in ml, of titrant was required for the titration?
Chemistry
1 answer:
Luda [366]1 year ago
5 0

The volume of titrant required for the titration would be 19.81 mL.

Since the burette was not filled to the zero mark during the titration and the level of base titrant was not filled to the 2.42 mL mark. As a result, the difference between the two values represents the total amount of titrant used in the titration.

therefore,

Volume of titrant after running titration - Volume of titrant before running titration  = total titrant required for the titration

22.23 - 2.42 = 19.81 mL

What Exactly Is Titration?

Titration is a laboratory technique that uses a solution with known volume and concentration to determine the concentration of an unknown solution. Between the two solutions, an oxidation-reduction reaction or acid-base neutralization occurs, and the known quantities are used to calculate the unknown. The known concentration standard solution is referred to as the titrant or titrator, while the unknown concentration solution is referred to as the titrand or analyte.

Find more on titration at : brainly.com/question/21881827

#SPJ4

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The answer is C a combustion’s reaction
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Question 23 (3 points)
Mice21 [21]

Answer:

<h2>The answer is 3.0 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density}  \\

From the question

mass of aluminum = 8.1 g

density = 2.7 g/mL

It's volume is

volume =  \frac{8.1}{2.7}  \\

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<h3>3.0 mL</h3>

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3 0
2 years ago
1.How many mL of 0.401 M HI are needed to dissolve 5.97 g of BaCO3?
garri49 [273]

Answer:

The answer to your question is:

1.- volume = 0.151 l or 151 ml

2.- 0.241 l  or 241 ml of NaOH

Explanation:

1.-

Data

V = ? HI = 0.401 M

BaCO3 = 5.97 g

                     2HI(aq)    +    BaCO3(s)   ⇒   BaI2(aq) + H2O(l) + CO2(g)

MW BaCO3 = 137 + 12 + 48 = 197 g

                     197 g of BaCO3 ----------------- 1 mol

                     5.97 g                -----------------   x

                     x = (5.97 x 1) /197

                    x = 0.03 mol of BaCO3

                    2 moles of HI ----------------  1 mol of BaCO3

                    x                     ----------------  0.03 mol of BaCO3

                    x = (0.03 x 2) / 1

                   x = 0.060 mol of HI

Molarity = moles / volume

volume = moles / molarity

volume = 0.060 / 0.401

volume = 0.151 l or 151 ml

2.-

V = ?    NaoH 0.757 M

Co⁺² Volume = 167 ml   0.548 M

             CoSO4(aq) + 2NaOH(aq)   ⇒   Co(OH)2(s) + Na2SO4(aq)

Moles of Co = Molarity x  volume

Moles of Co = 0.548 x 0.167

Moles of Co = 0.092

                                 1 mol of CoSO4 -------------- 2 moles of NaOH

                                0.092 moles      ---------------   x

                                x = (0.092 x 2) /1

                               x = 0.183 moles of NaOH

Volume of NaOH = moles / molarity

                             = 0.183 / 0.757

                            = 0.241 l  or 241 ml of NaOH

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Answer:

a. liquid

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if you need an explanation to each lmk

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sp2606 [1]
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