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otez555 [7]
3 years ago
7

Mohamedaidh00 answer me fast​

Chemistry
2 answers:
kolezko [41]3 years ago
6 0
I don’t understand?
KengaRu [80]3 years ago
5 0

Answer:

why can't u leave me........

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Placing a sample of iron (II) oxide into a graduated cylinder makes the water volume increase 12.0 mL. The weight of the sample
vivado [14]

Answer:6.38

Explanation:D=76.6/12

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A cylinder is labeled \"PENTANE.\" When the gas inside the cylinder is monochlorinated, five isomers of formula C5H11Cl result.
melisa1 [442]

Answer:

a mixture of two these

Explanation:

The number of isomeric monochlorides depends on the structure and number of equivalent hydrogen atoms in each isomer of pentane.

n-pentane has three different kinds of equivalent hydrogen atoms leading to three isomeric monochlorides formed.

Isopentane has four different types of equivalent hydrogen atoms hence four isomeric monochlorides are formed.

Lastly, neopentane has only one type of equivalent hydrogen atoms that yields one mono chlorination product.

Hence the cylinder must contain a mixture of isopentane and neopentane which yields four and one isomeric monochlorides giving a total of five identifiable monochloride products as stated in the question.  

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3 years ago
Every atom has atleast two shells ?
VladimirAG [237]
Nope. Hydrogen has one.
4 0
3 years ago
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Which element above has 8 protons in its nucleus?
Nataliya [291]
A. Oxygen. Oxygen has an atomic mass of 16. The atomic mass of an atom is the combined weight o the protons and neutrons. Since Oxygen's atomic mass is 16, it has 8 protons and 8 neutrons.
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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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