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3 years ago

Which type of chemical model shows the bond angle and bond length bewteen atoms and compunds

Chemistry

1 answer:

Ivan3 years ago

5 0

A molecular structure shows the bond angle and the Bond length between different atoms in a compound.

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What would be the pH of a solution of a solution with a hydrogen ion concentration of 0.001 M?

son4ous [18]

0001 M HCl is the same as saying that 1 *10-4 moles of H+ ions have been added to solution. The -log[. 0001] =4, so the pH of the solution =4.

7 0

3 years ago

1. Draw the Dot Structure for Na!

stepladder [879]

Answer:

Na ·

Explanation:

Hello!

In this case, since the Lewis dot structure of any element illustrates the number of valence electrons at the outer shell; we can set up the electron configuration of sodium to obtain:

$Na^1^1\rightarrow 1s^2,2s^2,2p^6,3s^1$

We can see there is only one valence electron, for that reason the Lewis dot structure is:

Na ·

Whereas the dot represents the valence electron.

Best regards!

4 0

3 years ago

A factor that changes in an experiment from manipulation of the independent variable is the what?

melomori [17]

Manipulation of the independent variable should change the dependent variable, the value of which "depends" on the independent variable's change in value.

7 0

3 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

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

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$