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3 years ago

A solution of which of the following coordination compounds will form a precipitate when treated with aqueous AgNO3?

Chemistry

1 answer:

lidiya [134]3 years ago

3 0

Answer:

[Cr(NH3)6.]C13

Explanation:

Alfred Werner's coordination theory (1893) recognized two kinds of valency;

Primary valency which are nondirectional and secondary valency which are directional.

Hence, the number of counter ions precipitated from a complex depends on the primary valency of the central metal ion in the complex.

We must note that it is only these counter ions that occur outside the coordination sphere that can be precipitated by AgNO3.

If we consider the options carefully, only [Cr(NH3)6.]C13 possess counter ions outside the coordination sphere which can be precipitated when treated with aqueous AgNO3.

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2 years ago

Perform the calculation and record the answer with the correct number of significant figures.

kotykmax [81]

The question is incomplete, here is the complete question:

A. (6.5-6.10)/3.19

B. (34.123 + 9.60) / (98.7654 - 9.249)

Answer:

For A: The answer becomes 0.1

For B: The answer becomes 0.4884

Explanation:

Significant figures are defined as the figures present in a number that expresses the magnitude of a quantity to a specific degree of accuracy.

Rules for the identification of significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.
All non-zero numbers are always significant. For example: 664, 6.64 and 66.4 all have three significant figures.
All zeros between the integers are always significant. For example: 5018, 5.018 and 50.18 all have four significant figures.
All zeros preceding the first integers are never significant. For example: 0.00058 has two significant figures.
All zeros after the decimal point are always significant. For example: 2.500, 25.00 and 250.0 all have four significant figures.
All zeroes used solely for spacing the decimal point are not significant. For example: 10000 has one significant figure.

Rule applied for addition and subtraction:

The least precise number present after the decimal point determines the number of significant figures in the answer.

Rule applied for multiplication and division:

In case of multiplication and division, the number of significant digits is taken from the value which has least precise significant digits

For A: (6.5-6.10)/3.19

This a a problem of subtraction and division.

First, the subtraction is carried out.

$\Rightarow \frac{6.5-6.10}{3.19}=\frac{0.4}{3.19}$

Here, the least precise number after decimal was 1.

$\Rightarrow \frac{0.4}{3.19}=0.125$

Here, the least precise number of significant digit is 1. So, the answer becomes 0.1

For B: (34.123 + 9.60) / (98.7654 - 9.249)

This a a problem of subtraction, addition and division.

First, the subtraction and addition is carried out.

$\Rightarow \frac{34.123+9.60}{98.7654-9.249}=\frac{43.723}{89.5164}=\frac{43.72}{89.516}$

Here, the least precise number after decimal in addition are 2 and in subtraction are 3

$\Rightarrow \frac{43.72}{89.516}=0.48840$

Here, the least precise number of significant digit are 4. So, the answer becomes 0.4884

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3 years ago