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alexdok [17]
1 year ago
7

consider a different titrant for this exercise. suppose ca(oh)2 were used as the titrant, instead of naoh. this will make the ti

trant twice as concentrated in hydroxide ion. the analyte will still be hc2h3o2. (a) what is the stoichiometry of hc2h3o2 to ca(oh)2?
Chemistry
1 answer:
Ivan1 year ago
4 0

The titrant for this exercise. suppose Ca(OH)₂ were used as the titrant, instead of NaOH. This will make the titrant twice as concentrated in hydroxide ion. the analyte will still be HC₂H₃O₂. the stoichiometry ratio of  HC₂H₃O₂ to Ca(OH)₂ is 1 : 2.

The balanced reaction of the given condition as follow :

 Ca(OH)₂   +      2HC₂H₃O₂      ------>     Ca(C₂H₃O₂)₂     +    2H₂O

from the equation it is clear that stoichiometry of Ca(OH)₂ is 1 and the stoichiometry of HC₂H₃O₂    is 2. therefore the stoichiometry ratio of  HC₂H₃O₂ to Ca(OH)₂ is 1 : 2.

Thus, The titrant for this exercise. suppose Ca(OH)₂ were used as the titrant, instead of NaOH. This will make the titrant twice as concentrated in hydroxide ion. the analyte will still be HC₂H₃O₂. the stoichiometry ratio of  HC₂H₃O₂ to Ca(OH)₂ is 1 : 2.

To learn more about stoichiometry here

brainly.com/question/13145466

#SPJ4

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artcher [175]

Answer:

1. V₁ = 2.0 mL

2. V₁ = 2.5 mL

Explanation:

<em>You are provided with a stock solution with a concentration of 1.0 × 10⁻⁵ M. You will be using this to make two standard solutions via serial dilution.</em>

To calculate the volume required (V₁) in each dilution we will use the dilution rule.

C₁ . V₁ = C₂ . V₂

where,

C are the concentrations

V are the volumes

1 refers to the initial state

2 refers to the final state

<em>1. Perform calculations to determine the volume of the 1.0 × 10⁻⁵ M stock solution needed to prepare 10.0 mL of a 2.0 × 10⁻⁶ M solution.</em>

C₁ . V₁ = C₂ . V₂

(1.0 × 10⁻⁵ M) . V₁ = (2.0 × 10⁻⁶ M) . 10.0 mL

V₁ = 2.0 mL

<em>2. Perform calculations to determine the volume of the 2.0 × 10⁻⁶ M solution needed to prepare 10.0 mL of a 5.0 × 10⁻⁷ M solution.</em>

C₁ . V₁ = C₂ . V₂

(2.0 × 10⁻⁶ M) . V₁ = (5.0 × 10⁻⁷ M) . 10.0 mL

V₁ = 2.5 mL

8 0
3 years ago
How many moles of HNO3 are present if 4.90×10−2 mol of Ba(OH)2 was needed to neutralize the acid solution?
Amiraneli [1.4K]

Answer:

0.098 moles

Explanation:

Let y represent the number of moles present

1 mole of Ba(OH)₂ contains 2 moles of OH- ions.

Hence, 0.049 moles of Ba(OH)2 contains y moles of OH- ions.

To get the y moles, we then do cross multiplication

1 mole *  y mole = 2 moles * 0.049 mole

y mole = 2 * 0.049 / 1

y mole =  0.098 moles of OH- ions.

1 mole of OH- can neutralize 1 mole of H+

Therefore, 0.098 moles of HNO₃ are present.

3 0
3 years ago
A compound such as CaCl, would have an overall charge of?
Rina8888 [55]

Calcium Chloride would have a charge of +2

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7 0
3 years ago
A certain ionic compound X has a solubility in water of 40.3 g/L at 20. degrees C. Calculate the mass X of required to prepare 5
tino4ka555 [31]

Answer:

20.1 g

Explanation:

The solubility indicates how much of the solute the solvent can dissolve. A solution is saturated when the solvent dissolved the maximum that it can do, so, if more solute is added, it will precipitate. The solubility varies with the temperature. Generally, it increases when the temperature increases.

So, if the solubility is 40.3 g/L, and the volume is 500 mL = 0.5 L, the mass of the solute is:

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40.3 g/L = m/0.5L

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Answer:

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