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grigory [225]
3 years ago
15

You will observe a weak acid-strong base titration in this experiment. Select all statements that are true about weak acid-stron

g base titrations.
A. Weak acid-strong base titrations always start at a higher pH than strong acid-strong base titrations, no matter the initial concentration.
B. The pH is less than 7 at the equivalence point.
C. The pH is greater than 7 at the equivalence point.
D. Half way to the equivalence point, a buffer region is observed.
Chemistry
1 answer:
DiKsa [7]3 years ago
6 0

Answer:

The pH is greater than 7 at the equivalence point.

Explanation:

Equivalence point is the point where the acid reacts with the base as stipulated in the equation of the reaction.

When a weak acid and a strong base are titrated, the pH of the solution at equivalence point is actually found to be around about pH ~ 9.

Hence, for a weak acid and strong base titration, The pH is greater than 7 at the equivalence point.

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g Oxidation refers to ________ Select one: a. an increase in oxidation number. b. a decrease in oxidation number. c. an increase
mihalych1998 [28]

Answer:

The correct alternative is "Option a".

Explanation:

Oxidation has become a mechanism whereby the physicochemical properties transform attributed to the formation of O₂.

  • The connection involving magnesium as well as O₂ requires the oxidation of the component named magnesium.
  • Even before exposed to the air, silicon is oxidized as well as generates silicon dioxide.

Other possibilities are not connected to the scenario in question. So Choice A is the best option.

4 0
3 years ago
How would an element on the left side of row 2 of the periodic table differ from an element in the middle of the same row?
maw [93]

Explanation:

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7 0
3 years ago
Read 2 more answers
If 56.8 ml of bacl2 solution is needed to precipitate all the sulfate ion in a 554 mg sample of na2so4 (forming baso4), what is
IrinaVladis [17]
<span>0.0687 m The balanced equation is BaCl2 + Na2SO4 ==> BaSO4 + 2 NaCl Looking at the equation, it indicates that there's a 1 to 1 ratio of BaCl2 and Na2SO4 in the reaction. So the number of moles of each will be equal. Now calculate the number of moles of Na2SO4 we had. Start by looking up atomic weights. Atomic weight sodium = 22.989769 Atomic weight sulfur = 32.065 Atomic weight oxygen = 15.999 Molar mass Na2SO4 = 2 * 22.989769 + 32.065 + 4 * 15.999 = 142.040538 g/mol Moles Na2SO4 = 0.554 g / 142.040538 g/mol = 0.003900295 mol Molarity is defined as moles per liter, so let's do the division. 0.003900295 mol / 0.0568 l = 0.068667165 mol/l = 0.068667165 m Rounding to 3 significant figures gives 0.0687 m</span>
3 0
4 years ago
if 100. mL of 0.800 M Na2SO4 is added to 200. mL of 1.20 M NaCl, what is the concentration of Na+ ions in the final solution? As
Black_prince [1.1K]
Compounds Na₂SO₄ and NaCl are mixed together are we are asked to find the concentration of Na⁺ in the mixture 
Na₂SO₄ ---> 2 Na⁺ + SO₄³⁻

1 mol of Na₂SO₄ gives out 2 mol of Na⁺ ions 

the number of Na₂SO₄ moles added - 0.800 M/1000 * 100 ml
                                                         = 0.08 mol
therefore number of Na⁺ ions from Na₂SO₄ = 0.08 * 2 = 0.16 mol

NaCl ----> Na⁺ + Cl⁻ 
1 mol of NaCl gives 1 mol of Na⁺ ions
number of NaCl moles added = 1.20 M/1000 * 200 ml
                                               = 0.24 mol
number of Na⁺ ions from NaCl = 0.24 mol

total number of Na⁺ ions in the mixture = 0.16 mol + 0.24 mol = 0.4 mol
as stated the volumes are additive, 
therefore total volume  = 100 ml + 200 ml = 300 ml 
the concentration of Na⁺ ions = number of moles / volume 
                                              = 0.4 mol/ 0.3 dm³
concentration of Na⁺ = 1.33 mol/dm³
5 0
3 years ago
Calculate the molarity of the solution formed when 0.72 moles of CaBr2 is dissolved in 1.50 L water.
xxTIMURxx [149]

Answer:

The molarity of the formed CaBr2 solution is 0.48 M

Explanation:

Step 1: Data given

Number of moles CaBr2 = 0.72 moles

Volume of water = 1.50 L

Step 2: Calculate the molarity of the solution

Molarity of CaBr2 solution = moles CaBr2 / volume water

Molarity of CaBr2 solution = 0.72 moles / 1.50 L

Molarity of CaBr2 solution = 0.48 mol / = 0.48 M

The molarity of the formed CaBr2 solution is 0.48 M

4 0
3 years ago
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