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4 years ago

When group 2A element from ion they

1 answer:

4 0

Well, knowing that group 2A elements are metals, and metals lose electrons to gain a full electron shell, we can eliminate B and D. Knowing that these are 2 electrons, that the group 2A element provides to another atom, to form an ion.

The solution is C. Lose 2 electrons.

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A liquid that evaporates at a slow rate exhibits __________.

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Strong internolecurar forces (A) hope it helps

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4 years ago

Calculate the standard enthalpy change for the reaction at 25 ∘ 25 ∘ C. Standard enthalpy of formation values can be found in th

WINSTONCH [101]

<u>Answer:</u> The standard enthalpy change of the reaction is coming out to be -16.3 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

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:

$Mg(OH)_2(s)+2HCl(g)\rightarrow MgCl_2(s)+2H_2O(g)$

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

$\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]$

We are given:

$\Delta H_f_{(Mg(OH)_2(s))}=-924.5kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(MgCl_2(s))}=-641.8kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ$

Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ

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4 years ago

At the candy store, gourmet jelly beans sell for $1.15 per ounce. There are 28 students in Alfred's class. If each student eats

saw5 [17]

Answer:

Total cost = $40.25 (Approx)

Explanation:

Given:

Per ounce = $1.15

Number of student = 28

Each student eat = 0.035 kg

Find:

Total cost

Computation:

Total weight of candy = 28 × 0.035 kg

Total weight of candy = 0.98 kg

1 ounce = 0.028 kg (approx).

Total weight of candy = 0.98 kg / 0.028

Total weight of candy = 35 ounce (Approx)

Total cost = 35 × $1.15

Total cost = $40.25 (Approx)

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3 years ago

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3 years ago