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3 years ago

Concentration of 10.00 mL of HBr if it takes 5 mL of a 0.253 M LiOH solution to neutralize it?

1 answer:

3 0

In a titration, for an acid to neutralize a base, at the equivalence point, there should be an equal number of moles of H+ and OH-.

Moles of OH- can be found by multiplying the concentration of the base by the volume. (You will need to keep in mind the stoichimetric coefficients if the strong base is Ca(OH)₂, Ba(OH)₂, or Sr(OH)₂.

Moles of OH- = moles of H+
(0.253 M) * 0.005 L = 0.01000 L * c
c = 0.1265 M

The concentration of HBr is 0.127 M.

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Lesechka [4]

Answer:

Transition elements are elements which have partially filled d-orbitals and form at least one or more stable ions.

6 0

3 years ago

Compute the percent ionic character of the interatomic bonds for the following compounds : a. TiO2 b. ZnTe c. CsCld. InSb e. MgC

sesenic [268]

Answer:

a. 63.2%

b. 11.7%

c. 73.3%

d. 0.995%

e. 55.5%

Explanation:

An ionic compound is a compound that is formed by ions, so one of the elements must donate electrons (which is the cation, the positive ion), and the other will receive these electrons (which is the anion, the negative ion).

The power of an element has to attract the electrons is called electronegativity, and so, as higher is the difference of electronegative of the elements, it is more probable that one of them will "still" the electrons and will form an ionic compound. The percent of this ionic character can be found by the Pauling's equation:

$%IC = (1 - e^{-0.25*(x_A - x_B)^2})$ *100%

Where $x_A - x_B$ is the electronegativity difference of the elements. Thus, consulting an electronegativity table:

a. $x_{Ti}$ = 1.5

$x_{O}$ = 3.5

$%IC = (1 - e^{-0.25*(3.5 - 1.5)^2})$ *100%

%IC = 63.2%

b. $x_{Zn}$ = 1.6

$x_{Te}$ = 2.1

$%IC = (1 - e^{-0.25*(2.1 - 1.6)^2})$ *100%

%IC = 11.7%

c. $x_{Cs}$ = 0.7

$x_{Cl}$ = 3.0

$%IC = (1 - e^{-0.25*(3.0 - 0.7)^2})$ *100%

%IC = 73.3%

d. $x_{In}$ = 1.7

$x_{Sb}$ = 1.9

$%IC = (1 - e^{-0.25*(1.9 - 1.7)^2})$ *100%

%IC = 0.995 %

e. $x_{Mg}$ = 1.2

$x_{Cl}$ = 3.0

$%IC = (1 - e^{-0.25*(3.0 - 1.2)^2})$ *100%

%IC = 55.5%

4 0

3 years ago

Pls help.............

Alina [70]

C is the correct answer

6 0

3 years ago

5. What is the density of water in g/mL? Why?

goldfiish [28.3K]

Answer:

1g/ml @ 4 degrees C by definition

Explanation:

7 0

3 years ago

What volume of 0.500 M HNO3(aq) must completely react to neutralize 100.0 milliliters of 0.100 M KOH(aq)?

asambeis [7]

KOH+ HNO3--> KNO3+ H2O<span>
From this balanced equation, we know that 1 mol HNO3= 1 mol KOH (keep in mind this because it will be used later).

We also know that 0.100 M KOH aqueous solution (soln)= 0.100 mol KOH/ 1 L of KOH soln (this one is based on the definition of molarity).

First, we should find the mole of KOH:
100.0 mL KOH soln* (1 L KOH soln/ 1,000 mL KOH soln)* (0.100 mol KOH/ 1L KOH soln)= 1.00*10^(-2) mol KOH.

Now, let's find the volume of HNO3 soln:
1.00*10^(-2) mol KOH* (1 mol HNO3/ 1 mol KOH)* (1 L HNO3 soln/ 0.500 mol HNO3)* (1,000 mL HNO3 soln/ 1 L HNO3 soln)= 20.0 mL HNO3 soln.

The final answer is </span>(2) 20.0 mL.<span>

Also, this problem can also be done by using dimensional analysis.

Hope this would help~ </span>

6 0

3 years ago