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dangina [55]
3 years ago
10

Which change increases the amount of iron produced in the equation fe3o4(s) 4h2(g) energy?

Chemistry
1 answer:
Debora [2.8K]3 years ago
6 0
FeSO4 (s) plus  4H2 (g) gives  3Fe(s)  plus  4H2O(l)
  from  the  equation  above   i  mole  of  FeSO4 reacted  with  4  mole  of H2  gas  to  give  3 moles  of  iron  and  4  moles  of  water.There  is  a relationship  between  volume  and  equilibrium  shift. Decrease  in  volume  favor  the  side  with  less  moles  of  a  gas. since  hydrogen  is  a  gas    a  decrease  in   volume  favor  the  forward  reaction  hence  more  Fe  is  produced.
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A solution is made by dissolving 23.5 grams of glucose (C6H12O6) in 0.245 kilograms of water. If the molal freezing point consta
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Answer:

- 0.99 °C ≅ - 1.0 °C.

Explanation:

  • We can solve this problem using the relation:

<em>ΔTf = (Kf)(m),</em>

where, ΔTf is the depression in the freezing point.

Kf is the molal freezing point depression constant of water = -1.86 °C/m,

m is the molality of the solution (m = moles of solute / kg of solvent = (23.5 g / 180.156 g/mol)/(0.245 kg) = 0.53 m.

<em>∴ ΔTf = (Kf)(m)</em> = (-1.86 °C/m)(0.53 m) =<em> - 0.99 °C ≅ - 1.0 °C.</em>

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4 years ago
How do newtons 3 laws relate ​
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5 0
4 years ago
What is the empirical formula of C3H6N2
Evgen [1.6K]

Answer:

The empirical formula of the compound C₃H₆N₂ is C₃H₆N₂

Explanation:

The empirical formula of a compound is the formula of the compound given in the (smallest) whole number ratio of the elements of the compound

The empirical formula of S₂O₂ is SO

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The given compound's molecular formula is C₃H₆N₂

The smallest whole number ratio of of the elements of the compound is 3:6:2, therefore, the empirical formula of the compound C₃H₆N₂ is C₃H₆N₂.

4 0
3 years ago
How many grams of NaCl are needed to prepare 50.0 grams of 35.0% of salt solution?
nevsk [136]

Answer:

17.5 g

Explanation:

Given data

  • Mass of solution to be prepared: 50.0 grams
  • Concentration of the salt solution: 35.0%

The concentration by mass of NaCl in the solution is 35.0%, that is, there are 35.0 grams of sodium chloride per 100 grams of solution. We will use this ratio to find the mass of sodium chloride required to prepare 50.0 grams of a 35.0% salt solution.

50.0gSolution \times \frac{35.0gNaCl}{100gSolution} = 17.5gNaCl

4 0
3 years ago
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