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marin [14]
4 years ago
14

Chromium crystallizes with a body-centered cubic unit cell. The radius of a chromium atom is 125 pm . Calculate the density of s

olid crystalline chromium in grams per cubic centimeter.
Chemistry
1 answer:
mafiozo [28]4 years ago
8 0

Answer:

\rho=7.15\ g/cm^3

Explanation:

The expression for density is:

\rho=\frac {Z\times M}{N_a\times {{(Edge\ length)}^3}}

N_a=6.023\times 10^{23}\ {mol}^{-1}

M is molar mass of Chromium = 51.9961 g/mol

For body-centered cubic unit cell , Z= 2

\rho is the density  

Radius = 125 pm = 1.25\times 10^{-8}\ cm

Also, for BCC, Edge\ length=\frac{4}{\sqrt{3}}\times radius=\frac{4}{\sqrt{3}}\times 1.25\times 10^{-8}\ cm=2.89\times 10^{-8}\ cm

Thus,  

\rho=\frac{2\times \:51.9961}{6.023\times \:10^{23}\times \left(2.89\times 10^{-8}\right)^3}\ g/cm^3

\rho=7.15\ g/cm^3

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Answer:

Explanation:

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Cl⁻ is monovalent so

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.50 mole = .50 equivalent

Total equivalent = .50 of Mg²⁺ + .50 of Cl⁻

= 1 equivalent .

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A nuclear reactor core must stay at or below 95 °C to remain in good working condition. Cool water at a temperature of 10 °C is
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Answer:

\large \boxed{\text{67 000 g}}

Explanation:

This is a problem in calorimetry — the measurement of the quantities of heat that flow from one object to another.

It is based on the Law of Conservation of Energy — Energy can be transformed from one type to another, but it cannot be destroyed or created.

If heat flows out of the reactor (negative), the same amount of heat must flow into the water (positive).

Since there is no change in total energy,

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The symbol for the quantity of heat transferred is q, so we can rewrite the word equation as

q₁ + q₂  = 0

The formula for the heat absorbed or released by an object is

 q = mCΔT, where

 m = the mass of the sample

  C = the specific heat capacity of the sample, and

ΔT = T_f - T_i = the change in temperature

1. Equation

There are two heat flows in this problem,

heat released by reactor + heat absorbed by water = 0

               q₁                  +                        q₂                     = 0

               q₁                  +                 m₂C₂ΔT₂                 = 0

2. Data:

q₁ = -23 746 kJ

m₂ = ?; C₂ = 4.184 J°C⁻¹g⁻¹;  T_f = 95 °C; T_i = 10 °C

3. Calculations

(a) Convert kilojoules to joules

q_{1} = -\text{23 746 kJ} \times \dfrac{\text{1000 J}}{\text{1 kJ}} = -\text{23 746 000 J}

(b) ΔT  

ΔT₂ = T_f - T_i = 95 °C - 10 °C = 85 °C

(c) m₂

\begin{array}{rcl}q_{1} + q_{2} & = & 0\\\text{-23 746 000 J} + m_{2} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 85 \, ^{\circ}\text{C} & = & 0\\\text{-23 746 000 J} + 356m_{2} \text{J$\cdot$g}^{-1} & = & 0\\356m_{2} \text{g}^{-1} & = & 23746000\\m_2&=& \dfrac{23746000}{\text{356 g}^{-1}}\\\\ & = & \textbf{67000 g}\\\end{array}\\

\text{You must circulate $\large \boxed{\textbf{67 000 g}}$ of water each hour.}

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