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2 years ago

The indicator propyl red has ka = 3.3 × 10–6 . estimate the approximate ph range over which this indicator would change color.

1 answer:

7 0

We let the chemical equation for the weak acid indicator propyl red HPr be
HPr + H2O ↔ H3O+ + Pr-
Therefore, the acid dissociation constant Ka is
Ka = [H3O+][Pr-] / [HPr]
The color of this indicator turns from red to yellow or the other way around at its turning point at which
[HPr] = [Pr-]
Substituting this to the equation for Ka, we now have
Ka = [H3O+]
The pH of the solution at its turning point is
pH = -log[H3O+] = -log(Ka) = -log 3.3×10^-6 = 5.48
<span>On this account, the pH range is pH 5.0 to 6.0.</span>

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In an experiment, the density of an unknown liquid was calculated to be 0.78 g/ml. if the accepted value is 0.75 g/ml, calculate

Vikki [24]

Percentage error is the relative error your measured value is from the true or accepted value. The formula for percentage error is written below:

Percentage error = |True Value - Measured Value|/True Value * 100
Percentage error = |0.75 g/mL - 0.78 g/mL|/0.75 g/mL * 100
Percentage error = 4%

6 0

1 year ago

In the reaction N2 + 3H2 ⇌ 2NH3, an experiment finds equilibrium concentrations of [N2] = 0.1 M, [H2] = 0.05 M, and [NH3] = 0.00

jasenka [17]

The equilibrium constant Kc for this reaction is calculated as follows

from the equation N2 + 3H2 =2 NH3

qc = (NH3)2/{(N2)(H2)^3}

Qc is therefore = ( 0.001)2 /{(0.1) (0.05)^3} = 0.08

3 0

2 years ago

Read 2 more answers

Describe how you would prepare exactly 100 mL of 0.109 M picolinate buffer, pH 5.61. Possible starting materials are pure picoli

Pepsi [2]

Answer:

1.342g of picolinic acid and 6.743mL of 1.0M NaOH diluting the mixture to 100.0mL

Explanation:

<em>The pKa of the picolinic acid is 5.4.</em>

Using Henderson-Hasselbalch formula for picolinic-picolinate buffer:

pH = pKa + log [Picolinate] / [Picolinic]

<em>Where [] could be taken as moles of each species</em>

<em />

5.61 = 5.4 + log [Picolinate] / [Picolinic]

0.21 = log [Picolinate] / [Picolinic]

1.62181 = [Picolinate] / [Picolinic] <em>(1)</em>

Now, both picolinate and picolinic acid will be:

0.100L * (0.109mol / L) =

0.0109 moles = [Picolinate] + [Picolinic] <em>(2)</em>

First, as we will start with picolinic acid, we need add:

0.0109 moles picolinic acid * (123.10g/mol) = 1.342g of picolinic acid

Now, replacing (2) in (1):

1.62181 = 0.0109 moles - [Picolinic] / [Picolinic]

1.62181 [Picolinic] = 0.0109 moles - [Picolinic]

2.62181 [Picolinic] = 0.0109 moles

[Picolinic] = 4.157x10⁻³ moles

And:

[Picolinate] = 0.0109 - 4.157x10⁻³ moles =

<h3>6.743x10⁻³ moles</h3><h3 />

To obtain these moles of picolinate ion we need to make the reaction of the picolinic acid with NaOH:

Picolinic acid + NaOH → Picolinate + Water

<em>That means to obtain 6.743x10⁻³ moles of picolinate ion we need to add 6.743x10⁻³ moles of NaOH</em>

<em />

6.743x10⁻³ moles of NaOH that is 1.0M are, in mL:

6.743x10⁻³ moles * (1L / 1mol) = 6.743x10⁻³L * 1000 =

<h3>6.743mL of the 1.0M NaOH must be added</h3><h3 />

Thus, we obtain the desire moles of picolinate and picolinic acid to obtain the buffer we want, the last step is:

<h3>Dilute the mixture to 100mL, the volume we need to prepare</h3>

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2 years ago

Is iron bromide magnetic if no why

Nataly_w [17]

Iron bromide isn't considered magnetic because all iron compounds are not magnetic

4 0

1 year ago

Read 2 more answers

Gold is one of the densest substances known, with a density of 19.3 g/cm3. If the gold in the crown was mixed with a less-valuab

Andrews [41]

Answer:

Explanation:

Density of gold is 19.3 g / cm³

Density of copper is 8.96 g / cm³

Density of bronze is 8.7 g / cm³

Hence when the gold and copper or bronze are mixed , the density of gold will be reduced due to less density of copper and bronze in comparison to that of gold.

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2 years ago