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2 years ago

For each metal complex give the coordination number for the metal species. m(nh34cl2

1 answer:

8 0

The coordination number is 6. The outer Cl- ion isn't included, so you count 4 x NH3 and 2 x Cl- are within Co's coordination sphere.

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When air pressure is high what is the weather like

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high-pressure systems normally associate with dry weather and mostly clear skies. This usually brings some light winds of cool, dry air, and brings fair weather.

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3 years ago

To calculate the percentage of people who have a particular allele population studies are conducted to collect data.true or fals

kondaur [170]

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2 years ago

Explain how the copper could be in the lake sample near the picnic area but not have been detected by this test.

Masja [62]

Answer:

May be the instrument is incorrect or may be error in it.

Explanation:

The copper have not been detected by this test because the test may be not for the detection of copper, may be it is used for identification of another minerals. If there is copper in the lake sample but can't be detected in the test so it means that the instrument which is used for detection is not the right one or having error in that instrument. Every mineral has a specific type of instrument that detect its presence, if we use incorrect instrument for the mineral then we can't detect the presence of that specific mineral.

8 0

2 years ago

Review the equation below.

olchik [2.2K]

2KClO3 --> 2KCl + 3O2

3 moles of oxygen are produced when 2 mol of potassium chlorate (KClO3) decompose.

3 0

3 years ago

Read 2 more answers

Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.57×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.46×10−19, what is the equilib

Dvinal [7]

Answer: The value of $K_c$ for the final reaction is $7.16\times 10^{25}$

Explanation:

The given chemical equations follows:

Equation 1: $H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^(aq.);K_1$

Equation 2: $HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^(aq.);K_2$

The net equation follows:

$S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_c$

As, the net reaction is the result of the addition of reverse of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of inverse of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=\frac{1}{K_1}\times \frac{1}{K_2}$

We are given:

$K_1=9.57\times 10^{-8}$

$K_2=1.46\times 10^{-19}$

Putting values in above equation, we get:

$K_c=\frac{1}{(9.57\times 10^{-8})}\times \frac{1}{(1.46\times 10^{-19})}=7.16\times 10^{25}$

Hence, the value of $K_c$ for the final reaction is $7.16\times 10^{25}$

5 0

3 years ago