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3 years ago

How will overshooting the endpoint in the titration affect your final result?

1 answer:

5 0

The end result will be inaccurate as the concentration of the acid or base being calculated will be more concentrated than it actually is due to the endpoint being surpassed

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Classify the below solids as amorphous or crystalline. emerald glass MgCl, rubber

svet-max [94.6K]

Answer:

Here yo

Explanation:

6 0

2 years ago

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How many moles of S would I have if I had 11 grams? (Stoichiometry) HELP

zepelin [54]

<h3>Answer:</h3>

0.34 mol S

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

11 g S

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of S - 32.07 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 11 \ g \ S(\frac{1 \ mol \ S}{32.07 \ g \ S})$
Multiply/Divide: $\displaystyle 0.343 \ mol \ S$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

0.343 mol S ≈ 0.34 mol S

4 0

3 years ago

A sample of an ionic compound NaA, where A- is the anion of a

vitfil [10]

Answer:

a) Kb = 10^-9

b) pH = 3.02

Explanation:

a) pH 5.0 titration with a 100 mL sample containing 500 mL of 0.10 M HCl, or 0.05 moles of HCl. Therefore we have the following:

[NaA] and [A-] = 0.05/0.6 = 0.083 M

Kb = Kw/Ka = 10^-14/[H+] = 10^-14/10^-5 = 10^-9

b) For the stoichiometric point in the titration, 0.100 moles of NaA have to be found in a 1.1L solution, and this is equal to:

[A-] = [H+] = (0.1 L)*(1 M)/1.1 L = 0.091 M

pKb = 10^-9

Ka = 10^-5

HA = H+ + A-

Ka = 10^-5 = ([H+]*[A-])/[HA] = [H+]^2/(0.091 - [H+])

[H+]^2 + 10^5 * [H+] - 10^-5 * 0.091 = 0

Clearing [H+]:

[H+] = 0.00095 M

pH = -log([H+]) = -log(0.00095) = 3.02

6 0

3 years ago

You have 345 mL of a 2.5 M NaCl solution. If you boil the water until the volume of the solution is 250 mLwhat will the molarity

HACTEHA [7]

The molarity of the solution will be 3.45 M

<h3>Further explanation </h3>

Dilution is the process of adding a solvent to get a more dilute solution.

The moles(n) before and after dilution are the same.

Can be formulated :

M₁V₁=M₂V₂

M₁ = Molarity of the solution before dilution

V₁ = volume of the solution before dilution

M₂ = Molarity of the solution after dilution

V₂ = Molarity volume of the solution after dilution

The initial solution

345 mL of a 2.5 M NaCl solution.

M₁=2.5 M

V₁=345 ml

The solution after boiled :

V₂=250 ml

$\tt M_2=\dfrac{M_1.V_1}{V_2}\\\\M_2=\dfrac{2.5\times 345}{250}\\\\M_2=3.45~M$

5 0

3 years ago

Initial Concentration mol/L[A] Initial Concentration mol/L[B] Initial Rate mol/Ls 0.20 0.10 20 0.20 0.20 40 0.40 0.20 160 Given

zimovet [89]

The generalized rate expression may be written as:
r = k[A]ᵃ[B]ᵇ

We may determine the order with respect to B by observing the change in rate when the concentration of B is changed. This can be done by comparing the first two runs of the experiment, where the concentration of A is constant but the concentration of B is doubled. Upon doubling the concentration of B, we see that the rate also doubles. Therefore, the order with respect to concentration of B is 1.
The same can be done to determine the concentration with respect to A. The rate increases 4 times between the second and third trial in which the concentration of B is constant, but that of A is doubled. We find that the order with respect to is 2. The rate expression is:

r = k[A]²[B]

4 0

3 years ago

Read 2 more answers