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Cloud [144]
3 years ago
8

Which of the following will have a larger radius than zinc ? Gallium (Ga) Magnesium (Mg) Aluminum (Al) Strontium (St)

Chemistry
1 answer:
oksian1 [2.3K]3 years ago
7 0

Answer:

Strontium (St)

Explanation:

The atom that will have a larger radius than zinc is strontium, Sr.

Atomic radius is defined as the half of the inter-nuclear distance between two covalently bonded atoms of non-metallic elements or half of the distance between two nuclei in the solid state of metals.

  • Across the period atomic radii decreases progressively due to the increase in nuclear charge.
  • Down a group atomic radii increase progressively due to the successive shells of electrons being added.
  • Since strontium satisfies the criteria, it has the larger atomic radius.
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a. Work

Explanation:

If you apply a force over a given distance - you have done work. Work = Change in Energy. If an object's kinetic energy or gravitational potential energy changes, then work is done. The force can act in the same direction of motion.

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Which answer below correctly identifies the type of change and the explanation for the boiling of water?
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Does the tempurture of the vinegar affect the volcano flow
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2 years ago
Why do hotspots appear to move in relation to tectonic plates?
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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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