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2 years ago

Please help me with this

Chemistry

2 answers:

ratelena [41]2 years ago

5 0

Explanation:

Magnesium representation : 1s²2s²2p⁶3s².

therefore, to become Mg2+ ion Magnesium losses electron from its 3s orbital to form 1s²2s²2p⁶.

therefore option C is correct.

hope this helps you.

yaroslaw [1]2 years ago

3 0

Answer:

3s orbital

Explanation:

3s is the outermost orbital of magnesium and it loses electron from it, it doesnt even have 4f, 5p or 3d orbitals.

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Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2O(g) (Delta

shusha [124]

Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

Explanation:

The balanced chemical reaction is,

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]$

Putting the values we get :

$\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]$

$\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]$

$\Delta H=-5314.8kJ$

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat = $\frac{5314.8}{2}\times 1=2657.4kJ$

Thus enthalpy of combustion per mole of butane is -2657.4 kJ

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3 years ago

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ryzh [129]

Answer:

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3 years ago

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Calculate the molarity of a solution consisting of 25.0 g of KOH in 3.00 L of solution.

melomori [17]

Answer:

0.15M

Explanation:

The equation for molarity is M= n/L. Where "M" is Molarity, "n" is the number of moles of solute, and "L" is the total liters in solution.

You need to calculate the number of moles from the given grams. The molar mass of KOH is (39.098+ 16 +1.008)= 56.106g. To calculate the mols of KOH, $\frac{25.0g}{1}$ × $\frac{1 mol}{56.106g}$ = 0.44558... mol, you see that the grams unit cancel out leaving you with mol as the unit.

The volume is given in L already so no need to do any conversion. M= $\frac{0.4558mol}{3.00L}$ = 0.1485M ≈ 0.15M

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2 years ago

Need help with chemistry hw please help

Andru [333]

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3 years ago

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spin [16.1K]

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3 years ago