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4 years ago

8

All elements have same number of valence electrons and similar properties

1 answer:

Yanka [14]4 years ago

6 0

Answer:

Groups

Explanation:

Groups(row going up and down) All have the same amount of valence electrons) Ex: First group the alkali metals all have 1 valence electron and are all very highly reactive.

The only ones that this rule does not work for is the smaller column in the middle called transitional metals and the amount of valence electron that each one has differs

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In general, the number of valence electrons is the same within a column and increases from left to right within a row. Group 1 elements have just one valence electron and group 18 elements have eight, except for helium, which has only two electrons total.

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3 years ago

The charge on a magnesium ion is +2. When it formed an ionic compound with another element, it lost

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3 years ago

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Explanation:

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3 years ago

The ideal gas heat capacity of nitrogen varies with temperature. It is given by:

hammer [34]

Answer:

A) 1059 J/mol

B) 17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

$C_p=C_v+R$

We can determine $C_v$ from above if we make $C_v$ the subject of the formula as:

$C_v$ $=C_p-R$

$C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314$

$C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4$

A).

The formula for calculating change in internal energy is given as:

$dU=C_vdT$

If we integrate above data into the equation; it implies that:

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 1059J/mol$

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 17,920 J/mol$

3 0

4 years ago