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Which determines the reactivity of an alkali metal?

adelina 88 [10]

Answer:

its ability to lose electron

3 0

3 years ago

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Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and

skelet666 [1.2K]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

Explanation :

The given balanced chemical reaction is,

$C_6H_6(l)\rightarrow C_6H_6(g)$

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

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

$\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]$

where,

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

n = number of moles

$\Delta H_f^0_{(C_6H_6(g))}$ = standard enthalpy of formation of gaseous benzene = 82.93 kJ/mol

$\Delta H_f^0_{(C_6H_6(l))}$ = standard enthalpy of formation of liquid benzene = 49.04 kJ/mol

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

$\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]$

$\Delta H^o=33.89kJ/mol=33890J/mol$

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

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

$\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0_{(C_6H_6(g))}$ = standard entropy of formation of gaseous benzene = 269.2 J/K.mol

$\Delta S^0_{(C_6H_6(l))}$ = standard entropy of formation of liquid benzene = 173.26 J/K.mol

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

$\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]$

$\Delta S^o=95.94J/K.mol$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is $25^oC\text{ or }298K$ .

$\Delta G^o=(33890J)-(298K\times 95.94J/K)$

$\Delta G^o=5299.88J/mol=5.299kJ/mol$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

6 0

3 years ago

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Compare and contrast physical property and chemical property

amm1812

Physical properties are properties that change form but not it's chemical compound

chemical properties are properties that change in chemical substance like burning tearing

3 0

3 years ago

4. Ammonia is produced by the chemical reaction of nitrogen and hydrogen. N2 (g) + 3H2(g) →→ 2NH3 (9) (A) Calculate the number o

Makovka662 [10]

Answer:

a) No. of moles of hydrogen needed = 5.4 mol

b) Grams of ammonia produced = 27.2 g

Explanation:

$N_2 (g) + 3H_2(g) \rightarrow 2NH_3 (g)$

a)

No. of moles of nitrogen = 1.80 mol

1 mole of nitrogen reacts with 3 moles of hydrogen

1.80 moles of nitrogen will react with

= 1.80 × 3 = 5.4 moles of hydrogen

b)

No. of moles of hydrogen = 2.4 mol

It is given that nitrogen is present in sufficient amount.

3 moles of hydrogen produce 2 moles of $NH_3$

2.4 moles of hydrogen will produce

= $\frac{2}{3} \times 2.4 = 1.6\ mol\ of\ NH_3$

Molar mass of ammonia = 17 g/mol

Mass in gram = No. of moles × Molar mass

Mass of ammonia in g = 1.6 × 17

= 27.2 g

7 0

3 years ago

I need help ASAP!

MaRussiya [10]

Calcium reacts gently with water to give hydrogen and calcium hydroxide, which is only slightly soluble, thus slows down the reaction.

It will be assumed that hydrochloric acid used is a dilute aqueous solution.

However, calcium reacts with hydrochloric acid to give calcium chloride which is readily soluble in water, and hydrogen, being a typical reaction of relatively active metals with acids.
Ca(s) + 2HCl(aq) -> CaCl2(aq) +H2(g) ↑ + heat

The clues that it is a chemical reaction could be:
- formation of a new substance, gaseous hydrogen
- disappearance of a metallic solid in the solution
- heat formed during the vigorous reaction.

As silver is below hydrogen in the electrochemical series, it will not be expected to react with dilute hydrocloric acid. (however, it dissolves in oxidizing acid such as nitric acid, but not displacing hydrogen as a product).

8 0

3 years ago