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steposvetlana [31]
2 years ago
10

What occurs along a convergent plate boundary ?

Chemistry
2 answers:
vichka [17]2 years ago
5 0

Answer:

What happens along a convergent plate boundary?

If two tectonic plates collide, they form a convergent plate boundary. Usually, one of the converging plates will move beneath the other, a process known as subduction. ... The new magma (molten rock) rises and may erupt violently to form volcanoes, often building arcs of islands along the convergent boundary

Explanation:

Margaret [11]2 years ago
3 0

Answer:

Two plates pull towards each other

Explanation:

Along a convergent plate boundary, two plates moves towards each other as the move in the same direction.

This results in different forms of plate interactions depending on the plate types.

  • At an ocean - ocean and continental - ocean convergent front, subduction of the oceanic plate occurs. This is because the oceanic plate below is denser than the asthenosphere.
  • At a continental - continental convergent front, the continental crust is pulls upward and build up as a mountain.
  • The subduction produces trenches and some volcanic islands.
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What is the maximum amount of water (in grams) that can be removed from 15ml of toluene by the addition?
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Complete Question

Magnesium sulfate forms a hydrate with the formula MgSO_4. 7H_20. What is the maximum amount of water (in grams) that can be removed from 15 ml of toluene by the addition of 200 mg of anhydrous magnesium sulfate? The molar mass of MgSO_4 is 120.4 g/mol; H20 = 18 g/mol.

Answer:

The value  is  z =  0.2093 \  g of  H_2O

Explanation:

From the question we are told that

   The volume of toluene is  V = 15 mL

    The mass of  anhydrous magnesium sulfate is  m =  200m g  = 200 *10^{-3} \  g

   The formula of the hydrate is   MgSO_4. 7H_20

    The molar mass of   MgSO_4  is  z =120.4 \ g/mol

From the formula given we see that

  1 mole of  Mg SO_4 wil remove  7 moles of H_2O to for the given formula

Hence

  120.4 g (1 mole) will remove  7 moles (7 * 18 g = 126 g  ) of  H_2O to for the given formula

Therefore 1 g of  Mg SO_4  x g  of  H_2O  

So

     x  =  \frac{x]126 *  1}{ 120.4 }

=>     x  =  1.0465 \  g

From our calculation we obtained that

  1 g of Mg SO_4 will remove  x  =  1.0465 \  g  of  H_2O  

Then  

   200 *10^{-3} \  g of Mg SO_4 will remove z g of  x  =  1.0465 \  g  of  H_2O  

So

   z =  200 *10^{-3} *  1.0465

=>z =  200 *10^{-3} *  1.0465

=>z =  0.2093 \  g of  H_2O  

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