Chemistry College


1. Show a correct numerical setup for calculating the molarity of the sodium hydroxide solution.
2. Determine both the total volume of HCl(aq) and the total volume of NaOH(aq) used in the titration.

1. Show A Correct Numerical Setup For Calculating The Molarity Of The Sodium Hydroxide Solution.2. Determine

Answers

Answer 1

To calculate the molarity of the sodium hydroxide (NaOH) solution, we need to perform a titration with a standardized solution of hydrochloric acid (HCl). Here is the numerical setup for calculating the molarity of the NaOH solution:

Measure the volume of the HCl solution used in the titration. Let's say you used 25.0 mL of 0.100 M HCl.

Calculate the number of moles of HCl used in the titration: moles of HCl = M x V = 0.100 mol/L x 0.0250 L = 0.00250 mol.

Use the balanced chemical equation for the reaction between HCl and NaOH to determine the number of moles of NaOH that reacted with the HCl. The balanced chemical equation is:

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Since the stoichiometry of the reaction is 1:1 between HCl and NaOH, the number of moles of NaOH that reacted is also 0.00250 mol.

4. Determine the volume of NaOH used in the titration. Let's say you used 30.0 mL of NaOH solution.

Calculate the molarity of the NaOH solution: Molarity of NaOH = moles of NaOH / volume of NaOH solution (in L) = 0.00250 mol / 0.0300 L = 0.0833 mol/L.

To determine the total volume of HCl(aq) and NaOH(aq) used in the titration, simply add together the volumes of HCl and NaOH that were used. In this example, the total volume would be 25.0 mL + 30.0 mL = 55.0 mL.

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Related Questions

2 NaN3 → 2 Na + 3 N
Given 9.98 grams of N2, how many moles of NaN3 are produced?​

Answers

0.238 moles of NaN₃ are produced from 9.98 grams of N₂.

What is the moles of NaN₃ produced?​

The moles of he mass of NaN₃ produced

The balanced equation for the reaction is:

2 NaN₃ → 2 Na + 3 N₂

The molar ratio between NaN₃ and N₂ is 2:3, which means that for every 2 moles of NaN₃, 3 moles of N₂ are produced.

The mole ratio is used to determine how many moles of NaN₃ are produced from 9.98 grams of N₂.

First, we need to convert the mass of N₂ to moles:

moles of N₂ = mass of N2 / molar mass of N₂

moles of N₂ = 9.98 g / 28.02 g/mol

moles of N₂ = 0.356 mol

moles of NaN₃ = (2/3) * moles of N₂

moles of NaN₃ = (2/3) * 0.356 mol

moles of NaN₃ = 0.238 mol

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Does anyone know the answer to this question

Answers

Answer:

A

Explanation:

If Hydrogen is H₂  There will be two silver

and is Carbon is C There will only be one gray

and if Oxygen is O₃ There will be three red

What is the final temperature when 625 grams of water at 75.0 deg C loses 7.96 x 10^4 J? (hint: remember ΔT = Tfinal - Tinitial )

Answers

The final temperature of the water is 71.99°C.

The final temperature when 625 grams of water at 75.0°C loses 7.96 x 10⁴ J can be found using the specific heat capacity equation:

q = mcΔT

where q is the amount of heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

First, we need to determine the specific heat capacity of water, which is 4.18 J/g°C. Then we can rearrange the equation to solve for ΔT:

ΔT = q / (mc)

Substituting the given values, we get:

ΔT = (7.96 x 10⁴ J) / (625 g x 4.18 J/g°C)

ΔT = 3.01°C

Therefore, the final temperature is:

Tfinal = Tinitial - ΔT

Tfinal = 75.0°C - 3.01°C

Tfinal = 71.99°C

As a result, the water's ultimate temperature is 71.99°C.

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