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3 years ago

In which of the following are the symbol and name for the ion given correctly

Chemistry

2 answers:

Alinara [238K]3 years ago

6 0

Answer:

This question is incomplete but the completed question is below

In which of the following are the symbol and name for the ion given correctly?

a. Fe2+: ferrous ion; Fe3+: ferric ion

b. Sn2+: stannic ion; Sn4+: stannous ion

c. Co2+: cobalt(II) ion; Co3+: cobaltous ion

d. Pb2+: lead ion; Pb4+: lead(IV) ion

The correct option is a

Explanation:

In this question, the knowledge of ion naming is been tested. Below, the ions and there correct names will be presented

Pb2+ is lead (II) ion

Pb4+ is lead (IV) ion

Co2+ is cobalt (II) ion and it is also called cobaltous ion

Co3+ is cobalt (III) ion

Sn2+ is tin (II) ion and it is also called stannous ion

Sn4+ is tin (IV) ion and it is also called stannic ion

Fe2+ is iron (II) ion and it is also called ferrous ion

Fe3+ is iron (III) ion and it is also called ferric ion

From the above, we can deduce that option a is the one that has it's symbol correctly attached to it's name

Ivenika [448]3 years ago

3 0

I don't see the symbols or the ions. sorry. can you type them out so I can help

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9. If 28.56 g of K2O is produced when 25.00 g K is reactedaccording to the following equation, what is the percent yieldof the r

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Answer

b. 95%

Explanation

Given:

Mass of K₂O produced (actual yield) = 28.56 g

Mass of K that reacted = 25.00 g

Equation: 4K(s) + O₂(g) → 2K₂0(s)

What to find:

The percent yield of K₂O.

Step-by-step solution:

The first step is to calculate the theoretical yield of K₂O produced.

From the balanced equation, 4 mol K produced 2 mol K₂O

Molar mass of K₂O = 94.20 g/mol)

Molar mass of K = 39.10 g/mol)

This means 4 mol x 39.10 g/mol = 156.40 g K produced 2 mol x 94.20 g/mol = 188.40 g K₂O

So 25.00 g K will produce:

$\frac{25.00\text{ }g\text{ }K\times188.40\text{ }g\text{ }K₂O}{156.40\text{ }g\text{ }K}=30.1151\text{ }g\text{ }K₂O$

Actual yield of K₂O = 28.56 g

Theoretical yield of k₂O = 30.1151 g

The percent yield for the reaction can now be calculated using the formula below:

$\begin{gathered} Percent\text{ }yield=\frac{Actual\text{ }yield}{Theoretical\text{ }yield}\times100\% \\ \\ Percent\text{ }yield=\frac{28.56\text{ }g}{30.1151\text{ }g}\times100\% \\ \\ Percent\text{ }yield=0.9484\times100\% \\ \\ Percent\text{ }yield=94.84\%\approx95\% \end{gathered}$

Therefore, the percent yield for the reaction is 95%.

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