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algol
2 months ago
13

In a titration 5.0 mL of a 2.0 M NaOH aq solution exactly neutralizes 10.0 of an HCL aq solution what is the concentration of th

e HCL(aq)solution
Chemistry
1 answer:
VMariaS [2.9K]2 months ago
8 0
The concentration of the HCl solution can be determined as follows:

The reaction equation is written as

NaOH + HCl = NaCl + H2O

Next, the moles of NaOH are calculated: moles = molarity x volume /1000

= 5 x 2/1000 = 0.01 moles

Using the mole ratio of NaOH to HCl, which is 1:1, the moles of HCl is also equal to 0.01 moles

The concentration is given by: concentration = moles/volume x 1000

= 0.01/10 x 1000 = 1M

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93.2 mL of a 2.03 M potassium fluoride (KF) solution
alisha [2963]

Response:

1.98 M

Clarification:

Provided data

  • Starting volume (V₁): 93.2 mL
  • Starting concentration (C₁): 2.03 M
  • Water volume added: 3.92 L

Step 1: Convert V₁ to liters

Using the relationship 1 L = 1000 mL.

93.2mL \times \frac{1L}{1000mL} = 0.0932 L

Step 2: Calculate the final volume (V₂)

The final volume is the total of the initial volume and the added water volume.

V_2 = 0.0932L + 3.92 L = 4.01L

Step 3: Calculate the final concentration (C₂)

Utilizing the dilution rule.

C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{2.03 M \times 3.92L}{4.01L} = 1.98 M

3 0
2 months ago
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In an acid-base neutralization reaction 43.74 ml of 0.500 m potassium hydroxide reacts with 50.00 ml of sulfuric acid solution.
lorasvet [2795]
The neutralization reaction that occurs between potassium hydroxide and sulfuric acid can be represented as follows
2KOH + H2SO4 ---> K2SO4 + 2H2O

The quantity of moles of KOH is calculated using (43.74 x 0.500)/ 1000 = 0.02187 moles

Given that the stoichiometric ratio of KOH to H2SO4 is 2:1, the moles of H2SO4 can be determined as 0.02187/2 = 0.01094 moles

To find the concentration (molarity), use the formula (0.01094/50) x 1000 = 0.2188M
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2 months ago
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Which statement best describes how an ionic bond forms?
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