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Alexandra [31]
3 years ago
13

Magnesium is an alkaline earth metal with an atomic number of 12. It’s in the third row in the periodic table. Which expression

shows one way to find the number of valence electrons in one atom of magnesium?
Chemistry
1 answer:
Leviafan [203]3 years ago
7 0

Answer:

group # = # valence electrons

group 2= 2 valence electrons

Explanation:

Mg is in group #2 and subsequently has 2 valence electrons (electrons on the outermost shell)

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the mass number of fluorine (F) is 19

C. 19

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The chemical equation for the reaction of baking soda (sodium bicarbonate, nahco3) and vinegar (acetic acid, ch3cooh) may be wri
satela [25.4K]
<h3><u>Answer;</u></h3>

Step 1;  NaHCO3(s) + CH3COOH(l)

Step 2 ; CO2(g)

<h3><u>Explanation;</u></h3>
  • The chemical equation for the reaction of baking soda (sodium bicarbonate, NaHCO3) and vinegar (acetic acid, CH3COOH) reaction occurs in two steps.

Step 1;  

  • A double displacement reaction in which acetic acid in the vinegar reacts with sodium bicarbonate to form sodium acetate and carbonic acid:
  • Equation;

NaHCO3(s)+ CH3COOH(l) → CH3COONa(aq) + H2CO3(l)

Step 2;

  • Carbonic acid is unstable and undergoes a decomposition reaction to produce the carbon dioxide gas:

H2CO3(l) → H2O(l) + CO2(g)

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3 years ago
The energy available to primary consumers in an energy pyramid is 800 kilocalories. Approximately how much energy is used by the
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Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang
Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

\Delta (H_{f})_{MnO_{2}} = -520.0 KJ/mole

\Delta (H_{f})_{Al_{2}O_{3}} = -1699.8 KJ/mole

The balanced chemical reaction is,

3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)

Formula used :

\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}

\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]

                        = (-3399.6) + (1560)

                        = -1839.6 KJ



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Column A. (4pts/.5 pts each) Order the layers from youngest to oldest. Write the letter for each layer in the correct order in t
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