Which one of the following forms an ionic bond with Chlorine?

A 5.0 milliliter sample of a substance has a mass of 12.5 grams. What is the mass of a 100 milliliter sample of the same substan

exis [7]

Answer:

It sould be 250 grams

Explanation:

4 0

3 years ago

This family (ethane, propane, butane, etc) of materials is likely to have what set of properties?

Ilya [14]

This family (ethane, propane, butane, etc) of materials is likely to have following set of properties.

The alkanes are non- polar solvents.
The alkanes are immiscible in water but freely miscible in other non-polar solvent .
The alkanes are consisting of weak dipole dipole bonds can not breaks the strong hydrogen bond.
The alkanes having only carbon (C) and hydrogen (H) atom which is bonded by a single bonds only.
The alkanes posses weak force of attraction that is weak van der waals force of attraction.

The ethane, propane, butane, belong to alkanes family.The alkanes are also considers as saturated hudrocarbons. Ethane is found in gaseous stae Ethane is the second alkane followed by propane followed by butane.

learn about butane

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7 0

2 years ago

Warm air and water both tend to rise while cooler air and water sink. When different parts of the oceans are heated unevenly, th

ale4655 [162]

Answer:

a

Explanation:

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

Consider the equilibrium: HCOOH(aq) + F-(aq) <----> HCOO-(aq) + HF (aq) Given that the Ka of HCOOH = 1.8 x 10-4 and the Ka

Aleks [24]

Hey there!:

K = Ka * Kb / Kw

Ka = 1.8*10⁻⁴

Kb = 10⁻¹⁴ / 6.8*10⁻⁴

K = 1.8*10⁻⁴ * ( 10⁻¹⁴/ 6.8*10⁻⁴ ) * ( 1 / 10⁻¹⁴ )

K = = 1.8 / 6.8

K = 0.265

Answer A

Therefore:

K is less than on the forward reaction is not favorable .

Hope That helps!

8 0

4 years ago