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3 years ago

When nitrogen, oxygen, fluorine, sodium, magnesium and aluminum ionize, they all will have:

Chemistry

1 answer:

Ray Of Light [21]3 years ago

5 0

Answer:

the same electron configuration (isoelectronic) as neon.

Explanation:

The electron configuration refers to the distribution of electrons of an atom or molecule in atomic or molecular orbitals. Electronic configuration is also known as an electronic structure.

When nitrogen, oxygen, fluorine, sodium, magnesium and aluminum ionize, they all will have the same electron configuration (isoelectronic) as neon.

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Water molecules have unique physical and chemical properties that enable them to perform important functions in living organisms

Serhud [2]

Answer: cohesion

Explanation:

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2 years ago

5. Compare the threats faced by the Everglades and the Louisiana wetan

SOVA2 [1]

The Everglades and the Louisiana wetan are the same

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3 years ago

A recipe calls for 1.2 cups of oil. How many liters of oil is this?

Nadusha1986 [10]

.3 liters... im pretty sure this is correct!!

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3 years ago

Read 2 more answers

Given the following at 25C calculate delta Hf for HCN (g) at 25C. 2NH3 (g) +3O2 (g) + 2CH4 (g) ---> 2HCN (g) + 6H2O (g) delta

AysviL [449]

Answer: The $\Delta H_f$ for HCN (g) in the reaction is 135.1 kJ/mol.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$2NH_3(g)+3O_2(g)+2CH_4(g)\rightarrow 2HCN(g)+6H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(HCN)})+(6\times \Delta H_f_{(H_2O)})]-[(2\times \Delta H_f_{(NH_3)})+(3\times \Delta H_f_{(O_2)})+(2\times \Delta H_f_{(CH_4)})]$

We are given:

$\Delta H_f_{(H_2O)}=-241.8kJ/mol\\\Delta H_f_{(NH_3)}=-80.3kJ/mol\\\Delta H_f_{(CH_4)}=-74.6kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=-870.8kJ$

Putting values in above equation, we get:

$-870.8=[(2\times \Delta H_f_{(HCN)})+(6\times (-241.8))]-[(2\times (-80.3))+(3\times (0))+(2\times (-74.6))]\\\\\Delta H_f_{(HCN)}=135.1kJ$

Hence, the $\Delta H_f$ for HCN (g) in the reaction is 135.1 kJ/mol.

8 0

3 years ago

3.0 cm x 4.0 cm x 1.0 cm [?]cm^3

CaHeK987 [17]

Explanation:

Hi there!!

you asked to multiply these all right,

you can simply multiply it ,

=3cm × 4 cm × 1cm

= 12cm^2×1cm (4×3=12)

= 12cm^3 (12×1=12)

Therefore, theanswer is 12 cm^3.

Hope it helps..

4 0

3 years ago