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3 years ago

Why are the alkali metals and the halogens very reactive

Chemistry

1 answer:

jeyben [28]3 years ago

4 0

Answer:

Answer in explanation

Explanation:

The reactivity or passiveness of an element depends solely on how close it is to attain a noble gas configuration. This means the closer an element is to attain a noble gas configuration, the greater its reactivity in both direction, positively or negatively.

Alkali metals belong to group 1 of the periodic table while halogens belong to group 17 of the periodic table. This means they are just one electron away from achieving the stability of a noble gas configuration. While alkali metals need to lose one electron to form a univalent positive ion, halogens news to gain one electron to form a univalent negative ion.

They tend to go about this vigorously and as such undergo several chemical reactions because of that single electron they neeed.

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A molecule made of hydrogen and carbon with Mr of 56 has the following composition: carbon 85.7%; hydrogen 14.3%. Calculate the

Veseljchak [2.6K]

Answer:

Assume that 100 grams of C2H4 is present. This means that there are 85.7 grams of carbon and 14.3 grams of hydrogen.

Convert these weights to moles of each element:

85.7 grams carbon/12 grams per mole = 7 moles of carbon.

14.3 grams hydrogen/1 gram per mole = 14 moles of hydrogen.

Divide by the lowest number of moles to obtain one mole of carbon and two moles of hydrogen.

Since we know that there cannot be a stable CH2 molecule, multiply by two and you have C2H4 which is ethylene - a known molecule.

The secret is to convert the percentages to moles and find the ration of the constituents.

5 0

2 years ago

Choose the aqueous solution that has the highest boiling point. These are all solutions of nonvolatile solutes and you should as

UNO [17]

Answer: 0.100 m $K_2SO_4$

Explanation:

Elevation in boiling point is given by:

$\Delta T_b=i\times K_b\times m$

$\Delta T_b=T_b-T_b^0$ = Elevation in boiling point

i= vant hoff factor

$K_b$ = boiling point constant

m= molality

1. For 0.100 m $K_2SO_{4}$

$K_2SO_4\rightarrow 2K^{+}+SO_4^{2-}$

, i= 3 as it is a electrolyte and dissociate to give 3 ions. and concentration of ions will be $3\times 0.100=0.300$

2. For 0.100 m $LiNO_3$

$LiNO_3\rightarrow Li^{+}+NO_3^{-}$

, i= 2 as it is a electrolyte and dissociate to give 2 ions, concentration of ions will be $2\times 0.100=0.200$

3. For 0.200 m $C_2H_8O_3$

, i= 1 as it is a non electrolyte and does not dissociate, concentration of ions will be $1\times 0.200=0.200$

4. For 0.060 m $Na_3PO_4$

$Na_3PO_4\rightarrow 3Na^{+}+PO_4^{3-}$

, i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be $4\times 0.060=0.24m$

Thus as concentration of solute is highest for $K_2SO_4$ , the elevation in boiling point is highest and thus has the highest boiling point.

3 0

3 years ago

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

Read 2 more answers

Perform the following

Marianna [84]

Answer:

167.980.

Explanation:

Hello!

In this case, since additions involving the display of the result with the appropriate number of significant figures involve the solution of the system without looking at them:

$156.325 +11.65498 =167.97998$

Considering that 156.325 is significant to the thousandths and 11.65498 to the hundred thousandths, we infer that the result should be taken to the thousandths as well as 156.325; thus, we obtain:

$167.980$

Because the second decimal nine is rounded to ten and therefore the 7 is taken to 8.

Best regards!

5 0

3 years ago

Problem PageQuestion Sulfuric acid is essential to dozens of important industries from steelmaking to plastics and pharmaceutica

Feliz [49]

The question is incomplete, here is the complete question:

Sulfuric acid is essential to dozens of important industries from steel making to plastics and pharmaceuticals. More sulfuric acid is made than any other industrial chemical, and world production exceeds 2.0×10¹¹ kg per year.

The first step in the synthesis of sulfuric acid is usually burning solid sulfur to make sulfur dioxide gas. Suppose an engineer studying this reaction introduces 1.8 kg of solid sulfur and 10.0 atm of oxygen gas at 650°C into an evacuated 50.0 L tank. The engineer believes Kp = 0.099 for the reaction at this temperature.

Calculate the mass of solid sulfur he expects to be consumed when the reaction reaches equilibrium. Round your answer to 2 significant digits.

Answer: The mass of solid sulfur that will be consumed is 19. grams

Explanation:

The chemical equation for the formation of sulfur dioxide gas follows:

$S(s)+O_2\rightarrow SO_2(g)$