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Leni [432]
3 years ago
14

Suppose that you are performing a titration on a monoprotic acid. Titration of this monoprotic acid required a volume of 0.1L of

the base being used. If the same volume of a diprotic acid would instead have been used (same volume of a diprotic acid versus the originally used monoprotic acid), and if we assume furthermore that the concentration of the diprotic acid and the monoprotic acids are equal, how does the volume of base required to reach the endpoint for the diprotic acid compare to the volume of base required to reach the endpoint for the monoprotic acid?
A) More base is likely required to reach the endpoint for the diprotic acid than for the monoprotic acid under these conditionsB) Less base is likely required to reach the endpoint for the diprotic acid than for the monoprotic acid. C) Exactly the same amount of base is required for both monoprotic and diprotic acids under the situations described here.D) A completely different indicator will need to be used here.
Chemistry
1 answer:
Alex73 [517]3 years ago
6 0

Answer: A) More base is likely required to reach the endpoint for the diprotic acid than for the monoprotic acid under these conditions

Explanation:

The monoprotic acid (HA) has a valency of 1 and diprotic acid  (H_2A) has a valency of 2.

As the concentration and volume of the diprotic acid and the monoprotic acids are equal.

The neutralization reaction for monoprotic acid is:

HA+BOH\rightarrow BA+H_2O

The neutralization reaction for diprotic acid is:

H_2A+2BOH\rightarrow B_2A+2H_2O

Thus more number of moles of base are required for neutralization of diprotic acid and thus the volume required will be more as concentration and volume of the diprotic acid and the monoprotic acids are equal.

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MrRa [10]
No, don't try, it will explode  close to 187 kPa
6 0
3 years ago
Determine the pH of a 2.8 ×10−4 M solution<br> of Ca(OH)2.
shepuryov [24]

Answer:

pH = 10.75

Explanation:

To solve this problem, we must find the molarity of [OH⁻]. With the molarity we can find the pOH = -log[OH⁻]

Using the equation:

pH = 14 - pOH

We can find the pH of the solution.

The molarity of Ca(OH)₂ is 2.8x10⁻⁴M, as there are 2 moles of OH⁻ in 1 mole of Ca(OH)₂, the molarity of [OH⁻] is 2*2.8x10⁻⁴M = 5.6x10⁻⁴M

pOH is

pOH = -log 5.6x10⁻⁴M

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pH = 14-pOH

<h3>pH = 10.75</h3>
3 0
3 years ago
5)
Drupady [299]

What amount of heat absorbs 50 g of steel (ce = 0.115 cal / g. ° C) that

does its temperature vary by 25 ° C?

Answer:

143.75cal

Explanation:

Given parameters:

Mass of steel  = 50g

Specific heat capacity of the steel  = 0.115cal/g°C

Temperature  = 25°C

Unknown:

Amount of heat = ?

Solution:

The amount of heat to cause this temperature change is dependent on mass and specific heat capacity of the substance.

  Amount of heat  = m C (ΔT)

m is the mass

c is the specific heat capacity

ΔT is the temperature change

 Now insert the parameters and solve;

  Amount of heat  = 50 x 0.115 x 25

  Amount of heat  = 143.75cal

4 0
3 years ago
A 0.5438 g of a C.H.O. compound was combusted in air to make 1.039 g of CO2 and 0.6369 g H20. What is the empirical formula? Bal
goblinko [34]

Answer:

C₃H₅O₂

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

Explanation:

The reaction can be expressed as:

CₓHₓOₓ + nO₂ → CO₂ + H₂O

Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so <u>the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂</u>:

1.039gCO_{2}*\frac{1molCO_{2}}{44gCO_{2}} *\frac{1molC}{1molCO_{2}} *\frac{12gC}{1molC} =0.2834gC

All of the hydrogens atoms in the compound ended up becoming H₂O, so <u>the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O</u>:

0.6369g*\frac{1molH_{2}O}{18gH_{2}O} *\frac{1molH}{1molH_{2}O} *\frac{1gH}{1molH} =0.0354gH

Because the compound is composed only by C, H and O, <u>the mass of O in the compound can be calculated by substraction</u>:

0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O

In order to determine the empirical formula, we calculate the moles of each component:

  • mol C = 0.2834 g C ÷ 12 g/mol = 0.0236 mol C
  • mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H
  • mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O

Then we divide those values by the lowest one:

0.0236 mol C ÷ 0.0141 = 1.67

0.0354 mol H ÷ 0.0141 = 2.51

0.0141 mol O ÷ 0.0141 = 1

If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.

  • The reaction is:

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

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Marta_Voda [28]
The answer to this is metal
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