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2 years ago

Solid samples of the element phosphorus can be white, black, or red in color. The variations in color are due to different

Chemistry

2 answers:

kirza4 [7]2 years ago

7 0

Answer:

Due to molecular structures.

Explanation:

The different forms of phosphorus (white, black or red) are allotropes of phosphorus.

They differ in molecular structures.

White phosphorus: It has four phosphorus which form a tetrahedral shape.

Black phosphorus : It has orthorhombic structure and due to interlinked six membered rings, it is least reactive.

Red phosphorus: It is formed on heating white phosphorus. It is initially amorphus and turns crystalline on further heating.

choli [55]2 years ago

4 0

Solid samples of the element phosphorus can be white, black, or red in color. The variations in color are due to different Molecular structures.

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2 years ago

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Andre45 [30]

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2 years ago

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The number of p atoms in 1.0 g of ba3(po4)2 is:

marta [7]

First, we determine the number of moles of barium phosphate. This is done using:

Moles = mass / Mr
Moles = 1 / 602
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Now, we see from the formula of the compound that each mole of barium phosphate has 2 moles of phosphorus (P). Therefore, the moles of phosphorus are:
0.002 * 2 = 0.004

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0.004 * 6.02 x 10²³

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2 years ago

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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Vitek1552 [10]

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