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musickatia [10]
3 years ago
9

Determine the overall charge on each complex.a) tetrachlorocuprate(i)b) pentaamminechlorocobalt(iii)c) diaquadichloroethylenedia

minecobalt(iii)
Chemistry
1 answer:
balu736 [363]3 years ago
4 0
Complex a: <span>tetrachlorocuprate(i)
In present complex chlorine is negatively charged ligand (-1) and oxidation state of copper is +1. Therefore, total charge on complex = 4(-1) + 1 = -3

Complex b: </span><span>pentaamminechlorocobalt(iii)
In present complex, ammine is a neutral ligand (charge = 0), chlorine is negatively charged ligand (charge = -1) and oxidation state of Co is +3. Therefore, total charge on complex is 5(0) + (-1) + (+3) = +2

</span><span>Complex c: diaquadichloroethylenediaminecobalt(iii)
</span>In present complex, aqua is a neutral ligand (charge = 0), chlorine is a negatively charged ligand (charge = -1), ethylenediamine is a neutral ligand (charge = 0) and oxidation state of cobalt is +3. Therefore, total charge on complex is 2(0) + 2(-1) + 2(0) + 3 = +1. 
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The second-order diffraction for a gold crystal is at an angle of 22.20o for X-rays of 154 pm. What is the spacing between the c
Alenkinab [10]

<u>Answer:</u> The spacing between the crystal planes is 4.07\times 10^{-10}m

<u>Explanation:</u>

To calculate the spacing between the crystal planes, we use the equation given by Bragg, which is:

n\lambda =2d\sin \theta

where,

n = order of diffraction = 2

\lambda = wavelength of the light = 154pm=1.54\times 10^{-10}m     (Conversion factor:  1m=10^{12}pm )

d = spacing between the crystal planes = ?

\theta = angle of diffraction = 22.20°

Putting values in above equation, we get:

2\times 1.54\times 10^{-10}=2d\sin (22.20)\\\\d=\frac{2\times 1.54\times 10^{-10}}{2\times \sin (22.20)}\\\\d=4.07\times 10^{-10}m

Hence, the spacing between the crystal planes is 4.07\times 10^{-10}m

4 0
3 years ago
Suppose an ice cube weighing 36.0 g at a temperature of 10°C is placed in 360 g water at a temperature of 20°C. Calculate the te
Scilla [17]

Answer:

10.44 °C

Explanation:

When the thermal equilibrium is reached, both of the substances have the same final temperature (T). The liquid water will lose heat, and the ice cube will absorb this heat. The temperature of the ice will increase until it reaches 0°C, at this temperature, it will change of phase for liquid, absorbing heat, but without a change in the temperature. Then the temperature will increase until the equilibrium.

By the energy conservation, the total amount of heat must be equal to 0:

Qice + Qmelting + Qliquid1 + Qliquid2 = 0

Liquid 1 is the ice after melting, and liquid 2 the liquid that was already at the flask. When there's a change of temperature:

Q = n*c*ΔT, where n is the number of moles, c is the heat capacity and ΔT is the temperature change (final - initial). The temperature variation in °C is equal in K, so the temperature may be used in °C.

The melting heat is:

Q = n*Hfus, Hfus = 6007 J/mol

The molar mass of the water is 18 g/mol, so the number of moles of the water and the ice are:

nwater = nliquid1 = 360/18 = 20 moles

nice = 36/18 = 2 moles

Qice + Qmelting + Qliquid1 + Qliquid2 = 0

2*38*(0 - (-10)) + 2*6007 + 2*75*(T - 0) + 20*75*(T - 20) = 0

760 + 12014 + 150T + 1500T - 30000 = 0

1650T = 17226

T = 10.44 °C

4 0
3 years ago
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If the equation is complete the products would be manganese chloride and oxygen gas would be given off.

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These will cancel out making it plain MnSO4

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How many hydrogen atoms are involved in this reaction? 3
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