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vitfil [10]
3 years ago
14

The highly exothermic thermite reaction, in which aluminum reduces iron(iii) oxide to elemental iron, has been used by railroad

repair crews to weld rails together. 2al(s) + fe2o3(s)  2fe(s) + al2o3(s) h = –850 kj what mass of iron is formed when 725 kj of heat are released?
Physics
1 answer:
trapecia [35]3 years ago
6 0

Answer:

To illustrate clearly, I will rewrite the reaction in a more understandable manner.

2 Al(s) +  Fe₂O₃ (s) ⇒ 2 Fe(s) + Al₂O₃(s)                Δhrxn = –850 kJ

This reaction has a negative sign for the change in enthalpy of reaction. The sign convention only means that the reaction releases energy to the surroundings. In other words, the reaction is exothermic. Focusing on only its magnitude, this means that 850 kJ of energy is needed for this reaction of 2 Aluminum moles and 1 mole of Fe₂O₃  to occur. 

Now, if you only had an energy of 725 kJ, then the reaction is incomplete but it will still form Iron (Fe). We use stoichiometric calculations as follows:

725 kJ * (2 mol Fe/850 kJ) = 1.7 moles of Fe

Knowing that the molar mass of Fe is 55.6 g/mol, then the mass of produced iron is

1.7 mol Fe * 55.6 g/mol = 94.85 g iron

Explanation:

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Answer:

0.536\sqrt{\frac{GM}{R}}

Explanation:

We are given that

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Mass of asteroid 2,m_2=1.97 M

Initial distance between their centers,d=13.63 R

Radius of each asteroid=R

d'=R+R=2R

Initial velocity of both asteroids

u=0

We have to find the speed of second asteroid just before they collide.

According to law of conservation of momentum

(m_1+m_2)u=m_1v_1+m_2v_2

(M+1.97 M)\times 0=Mv_1+1.97Mv_2

Mv_1=-1.97 Mv_2

v_1=-1.97v_2

According to law of conservation of energy

Gm_1m_2(\frac{1}{d'}-\frac{1}{d})=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2

GM(1.97M)(\frac{1}{2R}-\frac{1}{13.63R})=\frac{1}{2}M(-1.97v_2)^2+\frac{1}{2}(1.97M)v^2_2

1.97M^2G(\frac{13.63-2}{27.26R})=\frac{1}{2}Mv^2_2(3.8809+1.97)

1.97MG(\frac{11.63}{27.26 R})=\frac{1}{2}(5.8509)v^2_2

v^2_2=\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}

v_2=\sqrt{\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}}

v_2=0.536\sqrt{\frac{GM}{R}}

Hence, the speed of second asteroid =0.536\sqrt{\frac{GM}{R}}

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Answer:

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Answer:

750 people

Explanation:

From the question,

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