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1 year ago

Determine the number of valence electrons for sodium through the use of this table.

Chemistry

1 answer:

Alecsey [184]1 year ago

4 0

Answer:

Sodium has one valence electron. Valence electrons are electrons found in the outermost shell of an atom. The shell number representing the valence shell will differ depending on the atom in question. For sodium, which is in the 3rd row of the periodic table, the valence electrons will be found in the 3rd shell.

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What number of moles of O2 will be produced by the decomposition of 4.4 moles of water?

adelina 88 [10]

2.2 moles of O2 will be produced by the decomposition of 4.4 moles of water.

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3 years ago

Why do all chemical changes occur

GenaCL600 [577]

Answer:

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2 years ago

Part a use these data to calculate the heat of hydrogenation of buta-1,3-diene to butane. c4h6(g)+2h2(g)→c4h10(g)

Reptile [31]

<u>Answer:</u> The heat of hydrogenation of the reaction is coming out to be 234.2 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:

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

For the given chemical reaction:

$C_4H_6(g)+2H_2(g)\rightarrow C_4H_{10}(g)$

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

$\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]$

We are given:

$\Delta H_{(C_4H_{10})}=-2877.6kJ/mol\\\Delta H_{(C_4H_6)}=-2540.2kJ/mol\\\Delta H_{(H_2)}=-285.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J$

Hence, the heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

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2 years ago

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lesya [120]

That is convection my friend

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2 years ago

Read 2 more answers

If an object has a density of 0.55 g/mL, what is its density in cg/L?

expeople1 [14]

"cg" is centigram, which is one-hundredth of a gram.

I will first convert from g to cg (multiply by 100), then from mL to L (multiply by 1000).

$\frac{0.55g}{mL}*\frac{100cg}{1g}*\frac{1000mL}{1L}=55,000\frac{cg}{L} \ or \ 5.5e4\frac{cg}{L}$

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3 years ago