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Zarrin [17]
3 years ago
10

Write a balanced equation for the double-replacement precipitation reaction described, using the smallest possible integer coeff

icients. a precipitate forms when aqueous solutions of chromium(iii) iodide and potassium hydroxide are combined.
Chemistry
1 answer:
sergey [27]3 years ago
8 0
CrI_{3}_{(aq)} + 3KOH_{(aq)} --\ \textgreater \   Cr(OH)_{3} _{(s)} +3KI_{(aq)}
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If you are 5 foot 10 inches, how tall are you in meters? Meters ( write you answer 4 significant figures e.g. 2.852)
horsena [70]
5 foot 10 inch = 1.55448 meter
3 0
3 years ago
What percent of AgNO3 is silver?
Inga [223]

Answer:

63.499%

Explanation:

8 0
2 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
Why is a cis-1,3-disubstituted cyclohexane more stable thanits<br> trans isomer?
anzhelika [568]

Answer:

Explanation has been given below

Explanation:

  • In diaxial conformation of cis-1,3-disubstituted cyclohexane, 4 gauche-butane interactions along with syn-diaxial interaction are present. Hence it readily gets converted to diequitorial conformation where no such gauche-butane interaction is present
  • In two possible conformations of trans-1,3-disubstituted cyclohexane, 2 gauche-butane interactions are present in each of them.
  • Hence cis-1,3-disubstituted cyclohexane exists almost exclusively in diequitorial form. But trans-1,3-disubstituted cyclohexane has no such option.
  • Trans-1,3-disubstituted cyclohexane experiences gauche butane interaction in each of the two conformations.
  • Therefore cis-1,3-disubstituted cyclohexane is more stable than trans conformation

5 0
3 years ago
The density of a metal is 9.80 g/mL. What is the mass of a sample of metal when dropped in 28.9 mL of water, the volume increase
yan [13]

The correct answer is 2.70 × 10² g or 270 g.  

It is given, that the density of a metal is 9.80 g/ml.  

Let the mass of a sample of metal be x.  

The sample of metal is dropped in 28.9 ml of water, due to which the volume of the water increases to 56.4 ml.  

In order to calculate the mass of a metal, there is a need to use the formula, mass = density * volume

Mass = (9.80 g/ml) (56.4 ml - 28.9 ml)

= (9.80 g/ml) (27.5 ml)

= 2.70 × 10² g or 270 g


6 0
3 years ago
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