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vladimir2022 [97]
3 years ago
13

A beaker contains a 25 ml solution of an unknown monoprotic acid that reacts in a 1:1 stoichiometric ratio with naoh. titrate th

e solution with naoh to determine the concentration of the acid. perform a titration by setting the concentration of the naoh solution and adding it to the acid solution using the different add base buttons. the equivalence point of the titration is passed when the solution color changes. the unknown sample can be titrated multiple times by pressing the retitrate button and starting over. enter the concentration of the unknown acid solution.
Chemistry
1 answer:
yuradex [85]3 years ago
5 0

<u>Given:</u>

Volume of the unknown monoprotic acid (HA) = 25 ml

<u>To determine: </u>

The concentration of the acid HA

<u>Explanation:</u>

The titration reaction can be represented as-

HA + NaOH → Na⁺A⁻ + H₂O

As per stoichiometry: 1 mole of HA reacts with 1 mole of NaOH

At equivalence point-

moles of HA = moles of NaOH

For a known concentration and volume of added NaOH we have:

moles of NaOH = M(NaOH) * V(NaOH)

Thus, the concentration of the unknown 25 ml (0.025 L) of HA would be-

Molarity of HA = moles of HA/Vol of HA

Molarity of HA = M(NaOH)*V(NaOH)/0.025 L


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The solubility of acetanilide is 12.8 g in 100 mL of ethanol at 0 ∘C, and 46.4 g in 100 mL of ethanol at 60 ∘C. What is the maxi
weqwewe [10]

Answer: 72.41% and 26.90% respectively.

Explanation:

At 60°C, you can dissolve 46.4g of acetanilide in 100mL of ethanol. If you lower the temperature, at 0°C, you can dissolve just 12.8g, which means (46.4g-12.8g)=33.6g of acetanilide must have precipitated from the solution.

We can calculate recovery as:

\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{33.6\ g}{46.4\ g}*100=72.41\%

So the answer to the first question is 72.41%.

For the second part just use the same formula, the mass of the precipitate is the final mass minus the initial mass, (171mg-125mg)=46mg.

\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{46\ mg}{171\ mg}*100=26.90\%

So the answer to the second question is 26.90%.

3 0
3 years ago
Two types of heterogeneous mixtures are suspensions and colloids.<br>true or false<br><br>​
scoray [572]
<h2>Answer:True</h2>

Explanation:

Heterogeneous mixture is a mixture with non-uniform composition.

The properties of the mixture like concentration may change for different parts of the mixture.

Colloids contain solute particles of size 2nm-500nm.The presence of these particles makes the mixture heterogeneous.

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If volumes are additive and 253 mL of 0.19 M potassium bromide is mixed with 441 mL of a potassium dichromate solution to give a
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Answer:

The concentration of the Potassium Dichromate solution is 0.611 M

Explanation:

First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:

KBr (aq) → K+ (aq) + Br- (aq)

K2Cr2O7 (aq) → 2K+ (aq) + Cr2O7 2- (aq)

According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.

Having said that, we calculate the moles of potassium ions coming from the KBr solution:

0.19 M KBr: this means that we have 0.19 moles of KBr in 1000 mL solution. So:

1000 mL solution ----- 0.19 moles of KBr

253 mL solution ----- x = 0.04807 moles of KBr

As we said before, 1 mole of KBr will contribute with 1 mole of K+, so at the moment we have 0.04807 moles of K+.

Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):

1000 mL solution ----- 0.846 moles K+

694 mL solution ----- x = 0.587124 moles K+

This is the final quantity of potassium ion moles we have present once we mixed the KBr and K2Cr2O7 solutions. Because we already know the amount of K+ moles that were added with the KBr solution (0.04807 moles), we can calculate the contribution corresponding to K2Cr2O7:

0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7

If we go back and take a look a the chemical reactions, we can see that 1 mole of K2Cr2O7 dissociates into 2 moles of K+ ions, so:

2 K+ moles ----- 1 K2Cr2O7 mole

0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles

Now this quantity of potassium dichromate moles came from the respective  solution, that is 441 mL, so we calculate the amount of them that would be present in 1000 mL to determine de molar concentration:

441 mL ----- 0.269527 K2Cr2O7 moles

1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M

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