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zaharov [31]
3 years ago
10

1. (8 pts) Consider if the initial unknown acid solution used in this experiment started out more dilute. Predict any difference

s you would expect to see in the titration curve ‐ clearly identify what changes, if any, are expected for:
 initial pH
 pH at the half‐equivalence point
 NaOH volume needed to reach the equivalence point
 pH at the equivalence poin
Chemistry
1 answer:
malfutka [58]3 years ago
5 0

Answer:

The answers are in the explanation

Explanation:

- Initial pH: An acid solution more dilute has a higher pH because concentration of H⁺ decreases.

- pH at the half‐equivalence point: In a titration curve. The pH at the half-equivalence point will be higher because the initial pH is higher and the equivalence point pH is the same.

- NaOH volume needed to reach the equivalence point: As the diulte solution has a higher pH, the NaOH volume you need is lower than original solution.

- pH at the equivalence point: The pH at the equivalence point will be always the same (pH = 7,0). Because is the pH where the total H⁺ of the acid were consumed.

I hope it helps!

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8 0
3 years ago
Kathleen makes iced coffee by adding ice to her room temperature coffee in a glass. What statement explains what is taking place
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Hope this helps! :)

3 0
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5 0
3 years ago
How much water (H2O ) would form if 4.04 g of hydrogen (H2) reacted with 31.98 g of oxygen (O2 )?
Norma-Jean [14]

Answer:

Mass = 36 g

Explanation:

Given data:

Mass of water formed = ?

Mass of hydrogen = 4.04 g

Mass of oxygen = 31.98 g

Solution:

Chemical equation:

2H₂ + O₂   →   2H₂O

Number of moles of hydrogen:

Number of moles = mass/molar mass

Number of moles = 4.04 g/ 2 g/mol

Number of moles = 2.02 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 31.98 g/ 32 g/mol

Number of moles = 1.0 mol

Now we will compare the moles of water with hydrogen and oxygen.

                O₂         :         H₂O

                 1           :           2

                H₂         :         H₂O

                 2          :          2

               2.02       :      2.02

Number of moles of water formed by oxygen are less thus oxygen will limiting reactant.

Mass of water:

Mass = number of moles × molar mass

Mass = 2 mol × 18 g/mol

Mass = 36 g

8 0
2 years ago
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