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Leto [7]
3 years ago
14

A student who is performing this experiment pours an 8.50 mL sample of the saturated borax solution into a 10 mL graduated cylin

der after the borax solution had cooled to a certain temperature T. The student rinses the sample into a small beaker using distilled water, and then titrates the solution with a 0.500 M HCl solution. 12.00 mL of the HCl solution is needed to reach the endpoint of the titration.
Calculate the value of Ksp for borax at temperature T. (Answer: 0.176 Show your work.)

Here is a suggested procedure for doing this calculation:

(a) Calculate the number of moles of HCl that were added during the titration.

(b) Use reaction (2) in the lab manual to relate the number of moles of HCl to the number of moles of tetraborate ion in the 8.50 mL sample.

(c) Calculate the concentration of tetraborate ions in the 8.50 mL sample.

(d) Use reaction (1) in the lab manual to relate the concentration of tetraborate ions to the concentration of sodium ions.

(e) Use the concentrations of tetraborate ions and sodium ions to calculate the equilibrium constant (Ksp) at temperature T for reaction (1) in the lab manual.
Chemistry
1 answer:
belka [17]3 years ago
6 0

Answer:

Ksp = 0.1762

Explanation:

Applying

a) moles of HCl added, n= CV=0.5×0.012 = 6×10-3mol

b) since 0.006mol is present in 0.012dm3 of HCl

It implies moles of borax

C) Concentration = 0.706M

Ksp = [0.5]^2[0.706]= 0.176

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Why should you always condition a buret before running a titration?.
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Answer:

Conditioning two or three times will insure that the concentration of titrant is not changed by a stray drop of water.

Explanation:

"Check the tip of the buret for an air bubble. To remove an air bubble, whack the side of the buret tip while solution is flowing".

5 0
2 years ago
Solid carbon (C) can burn in oxygen (O2). Select
babunello [35]

Answer:

OCO

Another way of writing CO₂

Explanation:

A reaction equation has <u>reactants on the left</u> and <u>products on the right</u>.

The reactants are carbon and oxygen. The product is carbon dioxide.

C + O₂ → CO₂

You might see the equation both ways.

C + O₂ → OCO

C + O₂ in the products would mean no reaction has occurred. The problem can <u>solid carbon can burn in oxygen</u>, so a reaction will occur. For no reaction, you would put "NR" in the products.

<u>OCO is the structural way of writing CO₂.</u> Both have one carbon atom (C) and two oxygen atoms (O).

C + 2O is not possible. Oxygen, if alone, has to be at least O₂ because it's a <u>diatomic molecule</u>.

3 0
3 years ago
Why are all of the substances on the periodic table classified as elements?
Readme [11.4K]

All elements are pure substances.

4 0
3 years ago
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Where would sound travel the slowest
viva [34]

Answer:

Wood

Explanation:

It is because sound is a longitudinal wave

6 0
3 years ago
2Mg + O2 → 2MgO<br><br> If you are burning 5.8332 g of Mg, how many grams of MgO will this make?
LekaFEV [45]
<h3>Answer:</h3>

9.6724 g MgO

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Stoichiometry</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2Mg + O₂ → 2MgO

[Given] 5.8332 g Mg

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

[RxN] 2 mol Mg = 2 mol MgO

Molar Mass of Mg - 24.31 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of MgO - 24.31 + 16.00 = 40.31 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up:                              \displaystyle 5.8332 \ g \ Mg(\frac{1 \ mol \ Mg}{24.31 \ g \ Mg})(\frac{2 \ mol \ MgO}{2 \ mol \ Mg})(\frac{40.31 \ g \ MgO}{1 \ mol \ MgO})
  2. Multiply/Divide:                                                                                               \displaystyle 9.67241 \ g \ MgO

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 5 sig figs.</em>

9.67241 g MgO ≈ 9.6724 g MgO

5 0
3 years ago
Read 2 more answers
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