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Vilka [71]
3 years ago
8

From the list of the instructions and tips below, select the statement that would not lead to a successful recrystallization.

Chemistry
1 answer:
viva [34]3 years ago
8 0
What instructions and tips?
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What is the correct answer?!????
ddd [48]

Answer:

B

Explanation:

Dalton worked with mainly about the chemistry of atoms.

how do atoms combine to form various molecules.

—rather than the details of the physical, internal structure of atoms, although he never denied the possibility of atoms' having a substructure.

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A patient can't move his legs after a motorcycle accident. The doctor suspects nerve damage. What part of the peripheral nervous
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Motor neuron! it included volentoury movement.
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What is the precipitate for CuSO4+NaOH<br> (Balancing Equations)
jeka57 [31]

Answer:

CuSO4 + 2NaOH → Cu(OH)2 + Na2SO4

Explanation:

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4 years ago
Calculate the volume of a 0.323-mol sample of a gas at 265 K and<br> 0.900 atm.
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Answer:

V = 7.8 L

Explanation:

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5 0
3 years ago
Calculate the standard enthalpy of formation of NOCl(g) at 25 ºC, knowing that the standard enthalpy of formation of NO(g) at th
stepan [7]

Answer:

The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol

Explanation:

The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:

Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants

In this case, you have:  2 NOCl(g) → 2 NO(g) + Cl₂(g)

So, ΔH=2*H_{NO} +H_{Cl_{2} }-2*H_{NOCl}

Knowing:

  • ΔH= 75.5 kJ/mol
  • H_{NO}= 90.25 kJ/mol
  • H_{Cl_{2} }= 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound  the chlorine Cl₂)
  • H_{NOCl}=?

Replacing:

75.5 kJ/mol=2* 90.25 kJ/mol + 0 - H_{NOCl}

Solving

-H_{NOCl}=75.5 kJ/mol - 2*90.25 kJ/mol

-H_{NOCl}=-105 kJ/mol

H_{NOCl}=105 kJ/mol

<u><em>The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol</em></u>

8 0
3 years ago
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