All categories

2 years ago

What are the period and group trends in electronegativities

1 answer:

8 0

Electronegativity of an element decreases as we move down a group on the periodic table and electronegativity increases while moving from left to right across a period on the periodic table.

Explanation:

The electronegativity increases as we move from left to right across a period because from left to right across a period, the nuclear charge is increasing Hence the attraction for the valence electrons also increases.
As we move down a group, the atoms of each element have an increasing number of energy levels. The distance between the nucleus and valence electron shell increases and reduces the attraction for valence electrons. Hence electronegativity decreases as we move from top to bottom down a group.

You might be interested in

Name two factors that influence the viscosity of a lava flow

mr_godi [17]

Three factors that affect magma viscosity are temperature, composition, and presence of dissolved gases.

6 0

2 years ago

Annual precipitation rates can be collected for several years added together and divided by the number of years to obtain an ave

larisa [96]

Answer:

true

Explanation:

3 0

3 years ago

Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti

gayaneshka [121]

Answer : The value of $K$ for this reaction is, $2.6\times 10^{15}$

Explanation :

The given chemical reaction is:

$CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)$

Now we have to calculate value of $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]$

where,

$\Delta G^o$ = Gibbs free energy of reaction = ?

n = number of moles

$\Delta G^0_{(HCH_3CO_2(g))}$ = -389.8 kJ/mol

$\Delta G^0_{(CH_3OH(g))}$ = -161.96 kJ/mol

$\Delta G^0_{(CO(g))}$ = -137.2 kJ/mol

Now put all the given values in this expression, we get:

$\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]$

$\Delta G^o=-89.4kJ/mol$

The relation between the equilibrium constant and standard Gibbs, free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol

R = gas constant = 8.314 J/L.atm

T = temperature = $30.0^oC=273+30.0=303K$

$K$ = equilibrium constant = ?

Now put all the given values in this expression, we get:

$-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K$

$K=2.6\times 10^{15}$

Thus, the value of $K$ for this reaction is, $2.6\times 10^{15}$

4 0

3 years ago

Why does hard water taste better than distilled water?

Maurinko [17]

Answer:

because hard water is salty

8 0

3 years ago

According to scientific evidence, earth’s earliest atmosphere lacked oxygen. Over time, oxygen was added to the atmosphere.

Marianna [84]

B.
The cynobacteria were already there without the oxygen, so that rules out A, and a lot of prokaryotes were anaerobic, so that rules out C. Finally, Photosynthesis does not require oxygen. Instead, Oxygen is a waste product of it. Therefore, it cannot be D. So, we are only left with B
Hoped this helped :D

4 0

2 years ago