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ICE Princess25 [194]
3 years ago
9

Identify the charge carried by a phosphide ion

Chemistry
1 answer:
insens350 [35]3 years ago
5 0

Answer : The charge carried by a phosphide ion is, (-3)

Explanation :

In phosphide ion, the (-ide) indicates that the phosphide ion is the anion of phosphorous. So, we will have to add electrons.

As we know that the phosphorous belongs to group 5 and have 5 valence electrons.

The atomic number of phosphorous = 15

The electronic configuration of phosphorous is, 1s^22s^22p^63s^23p^3

In phosphorous, the 'p' sub-shell is half filled. For fully filled 'p' sub-shell, we will need '3' electrons. By adding 3 electrons in 'p' sub-shell of phosphorous, (-3) charge will be created on phosphorous and named as phosphide ion.

Phosphide ion is written as, P^{3-}

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An object has a mass of 3.50 grams and a density of 5.61 g/mL. What is the volume of this object? *
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<h2>0.62 mL</h2>

Explanation:

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We have the final answer as

<h3>0.62 mL</h3>

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You are given 25.00 mL of an acetic acid solution of unknown concentration. You find it requires 35.75 mL of a 0.1950 M NaOH sol
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Answer:

0.2788 M

1.674 %(m/V)

Explanation:

Step 1: Write the balanced equation

NaOH + CH₃COOH → CH₃COONa + H₂O

Step 2: Calculate the reacting moles of NaOH

0.03575 L \times \frac{0.1950mol}{L} = 6.971 \times 10^{-3} mol

Step 3: Calculate the reacting moles of CH₃COOH

The molar ratio of NaOH to CH₃COOH is 1:1.

6.971 \times 10^{-3} molNaOH \times \frac{1molCH_3COOH}{1molNaOH} = 6.971 \times 10^{-3} molCH_3COOH

Step 4: Calculate the molarity of the acetic acid solution

M = \frac{6.971 \times 10^{-3} mol}{0.02500L} =0.2788 M

Step 5: Calculate the mass of acetic acid

The molar mass of acetic acid is 60.05 g/mol.

6.971 \times 10^{-3} mol \times \frac{60.05g}{mol} =0.4186 g

Step 6: Calculate the percentage of acetic acid in the solution

\frac{0.4186g}{25.00mL}  \times 100\% = 1.674 \%(m/V)

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3 years ago
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