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2 years ago

The electronic configuration of an atom of element X is 2,8,2 what will be the valency of the element x

Chemistry

1 answer:

KIM [24]2 years ago

6 0

The element is Magnesium

It has total 12 electrons

it has electronic configuration

$\\ \sf\longmapsto 2,8,2$

$\\ \sf\longmapsto 1s^22s^22p^63s^2$

In valence cell it has 2 electrons and therefore its valency is 2

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Which statement best describes the reaction pathway graph for an exothermic reaction but not an endothermic reaction? It has a h

ruslelena [56]

Answer: "The reactants are higher in energy than the products"

Explanation:

The exothermic reactions are characterized by the release of heat to the surroundings. The reactants lose heat that is delivered to the surroundings which implies that the products will be lower in energy than the reactants.

The hills that you can see in a reaction energy diagram are not related with the final change of energy. The hills are an indication of the activation energy needed to start the reaction, but they do not measure the change of energy from the products to the reactants.

The enthalpy that is a state variable that identifies the content of heat. Then the change of enthalpy for the exothermic reactions is negative, meaning that the energy of the products is lower than the energy of the reactants.

3 0

3 years ago

Read 2 more answers

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

A 50.0 g sample of liquid water at 25.0 degree C is mixed with 29.0 g of water at 45 degree C. The final temperature of the wate

kotegsom [21]

Answer: The final temperature of water is 32.3°C

Explanation:

When two solutions are mixed, the amount of heat released by solution 1 (liquid water) will be equal to the amount of heat absorbed by solution 2 (liquid water)

$Heat_{\text{absorbed}}=Heat_{\text{released}}$