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The sulphide capacity (C<sub>s</sub>) of liquid slag of composition 55 wt% CaO, 20 wt% SiO<sub>2</sub>, 15 wt% Al<sub>2</sub> O<sub>3</sub> and 10 wt% MgO is given by the following equation<br>$$\log {C_s} = 3.44\left( {{X_{caO}} + 0.1\,{X_{MgO}} - 0.8\,{X_{A{l_2}{O_3}}} - {X_{si{o_2}}}} \right) - \left( {\frac{{9894}}{T}} \right) + 2.05$$<br>where, X is mole fraction of the respective components. Atomic weights of Ca, Mg, Si, Al and O are 40, 24, 28, 27, and 16 respectively. The value of C<sub>s</sub> at 1900 K is

Correct Answer: 0.0009

If a + b + c + d = 4, then find the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?

If a + b + c + d = 4, then the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?

Liquid steel contains initially 0.05 mass% P and this has to be reduced to 0.01 mass% using a basic slag. The equilibrium distribution ratio of P between slag and metal is $$Lp = \frac{{\left( {\% P} \right){\text{slag}}}}{{\left( {\% P} \right){\text{metal}}}}, = 80;$$     Assuming that in itially the slag does not contain any phosphorous (P), then the minimum weight of slag (ton) required per ton of steel is

The value of the expression $$\frac{{{{\left( {a - b} \right)}^2}}}{{\left( {b - c} \right)\left( {c - a} \right)}} + $$   $$\frac{{{{\left( {b - c} \right)}^2}}}{{\left( {a - b} \right)\left( {c - a} \right)}} + $$    $$\frac{{{{\left( {c - a} \right)}^2}}}{{\left( {a - b} \right)\left( {b - c} \right)}}$$   = ?

The Hamiltonian of a particle is given by $$H = \frac{{{p^2}}}{{2m}} + V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right) + \phi \left( { + \left| {\overrightarrow {\bf{r}} } \right|} \right)\overrightarrow {\bf{L}} .\overrightarrow {\bf{S}} ,$$        where $$\overrightarrow {\bf{S}} $$ is the spin, $$V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$  and $$\phi \left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$  are potential functions and $$\overrightarrow {\bf{L}} \left( { = \overrightarrow {\bf{r}} \times \overrightarrow {\bf{p}} } \right)$$   is the angular momentum. The Hamiltonian does not commute with

What is the mole fraction in instantaneous distillate if: Moles of liquid in residue at start W0= 1000 Moles of liquid in residue at end W= 500 Mole fraction in liquid at start X0 = 0.6 Mole fraction in liquid at end X= 0.5

What is the mole fraction in instantaneous distillate if: Moles of liquid in residue at start W0= 800 Moles of liquid in residue at end W= 500 Mole fraction in liquid at start X0 = 0.5 Mole fraction in liquid at end X= 0.1

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What is the mole fraction in instantaneous distillate if: Moles of liquid in residue at start W0= 1000 Moles of liquid in residue at end W= 500 Mole fraction in liquid at start X0 = 0.25 Mole fraction in liquid at end X= 0.1