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ICE Princess25 [194]
3 years ago
6

How many grams of oxygen (O2), x, are needed to react with the 700 g of iron to produce 1000 g of iron (III) oxide, Fe2O3?

Physics
1 answer:
Murljashka [212]3 years ago
6 0

As we know that, molecular mass of ferric oxide, Fe2O3, is 159.69 grams.
Out of which, iron contributes 111.69 g (2 X 55.845 g) and oxygen contributes
48 g (3 X 16 g).   
 Each gram of iron (III) oxide contains 111.69/159.69 g of iron and 48/159.69
g of oxygen.   
 To produce 1000 g iron (III) oxide we need,    
 Iron = 111.69*1000/159.69 = 699.42 g   
 Oxygen = 48*1000/159.69 = 300.58 g
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Answer:

The tension on the string is  T  =  43.302 \ N

Explanation:

From the question we are told that

    The mass of the rock is m_r = 5.00 \ kg =  5000 \ g

       The density of the rock is \rho  =  4300 \ kg/m^3 =  4.3 g/dm^3

       

Generally the volume of the rock is mathematically evaluated as

          V    =  \frac{m_r}{\rho}

substituting values

        V    =  \frac{5000}{4.3}

       V    =  1162.7 \  dm^3

The volume of the rock immersed in water is

      V_w = \frac{V}{2}  

substituting values

     V_w = \frac{1162.7 }{2}

     V_w = 581.4 \ dm^3

mass of water been displaced by the this volume is

     m_w  = V_w     According to Archimedes principle

=>   m_w =  581.4 \ g

     m_w =  0.5814 \ kg

The weight of the water displace is  

      W _w =  m_w  * g

      W _w =  0.5814  * 9.8

      W _w = 5.698 \ N

The actual weight of the rock is  

      W_r  =  m_r * g

     W_r  =  5.0 *  9.8

     W_r  =  49.0 \ N

The tension on the string is

       T  = W_r - W_w

substituting values

       T  = 49.0 -  5.698

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Answer:

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