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nasty-shy [4]
3 years ago
7

Give the systematic name for this coordination compound. k3 cr(cn)6

Chemistry
1 answer:
Helen [10]3 years ago
8 0

Potassium hexacyanochromate(III)

A systematic name is a name given in a systematic way to a chemical substance, out of a definite collection.

The systematic name for the coordination compound k3 cr(cn)6 is Potassium hexacyanochromate(III). This compound contains potassium (k3), six cn which is called hexacyano, and cr (chromium).

 






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Answer:

The mass of a sample of iron that has had 300 J applied to it and heats up from 20 degrees Celsius to 40 degrees Celsius is 32.61 grams.

Explanation:

Calorimetry is the measurement and calculation of the measurement of heat changes exchanged by a body or a system produced in physical and chemical processes.

The sensible heat of a body is the amount of heat received or transferred by a body to produce a change in temperature but without a change in physical state.

The sensible heat in a constant pressure is calculated by:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c, and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial)

In this case:

  • Q= 300 J
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Replacing:

300 J= 0.46 \frac{J}{g*C} * m* (40 - 20) C

Solving:

300 J= 0.46 \frac{J}{g*C} * m* 20 C

m=\frac{300 J}{0.46 \frac{J}{g*C} * 20 C}

m= 32.61 g

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Calculate the energy (in J/atom) for vacancy formation in silver, given that the equilibrium number of vacancies at 800 C is 3.6
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Answer:

the energy vacancies for formation in silver is \mathbf{Q_v = 3.069*10^{-4} \ J/atom}

Explanation:

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the equilibrium  number of vacancies at 800 °C

i.e T = 800°C     is  3.6 x 10¹⁷ cm3

Atomic weight of sliver = 107.9 g/mol

Density of silver = 9.5 g/cm³

Let's first determine the number of atoms in silver

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N =  \dfrac{N_A* \rho _{Ag}}{A_{Ag}}

where ;

N_A = avogadro's number = 6.023*10^{23} \ atoms/mol

\rho _{Ag} = Density of silver = 9.5 g/cm³

A_{Ag} = Atomic weight of sliver = 107.9 g/mol

N =  \dfrac{(6.023*10^{23} \ atoms/mol)*( 9.5 \ g/cm^3)}{(107.9 \ g/mol)}

N = 5.30 × 10²⁸ atoms/m³

However;

The equation for equilibrium number of vacancies can be represented by the equation:

N_v = N \ e^{^{-\dfrac{Q_v}{KT}}

From above; Considering the  natural logarithm on both sides; we have:

In \ N_v =In N - \dfrac{Q_v}{KT}

Making Q_v the subject of the formula; we have:

{Q_v =  - {KT}   In( \dfrac{ \ N_v }{ N})

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Temperature T = 800 °C = (800+ 273) K = 1073 K

Q _v =-( 8.62*10^{-5} \ eV/atom.K * 1073 \ K) \ In( \dfrac{3.6*10^{17}}{5.3 0*10^{28}})

\mathbf{Q_v = 2.38 \ eV/atom}

Where;

1 eV = 1.602176565 × 10⁻¹⁹ J

Then

Q_v =  (2.38 \ * 1.602176565 * 10^{-19} ) J/atom  }

\mathbf{Q_v = 3.069*10^{-4} \ J/atom}

Thus, the energy vacancies for formation in silver is \mathbf{Q_v = 3.069*10^{-4} \ J/atom}

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