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ANTONII [103]
3 years ago
9

I need help please j

Chemistry
1 answer:
Sidana [21]3 years ago
3 0
Reactants Hydrogen: 5
Products Hydrogen: 5

Reactants Carbon: 3
Products Carbon: 3

Reactants Oxygen: 4
Products Oxygen: 5
You might be interested in
The maximum number of electrons in a single d subshell is:
Iteru [2.4K]

10 electrons

Explanation:

The maximum number of electrons in a single d-subshell is 10 electrons.

The d-sub-orbital used to denote azimuthal or secondary quantum numbers.

The maximum number of electrons in the orbitals of sublevels are:

    two electrons in the s-sublevel, it has one orbital

   

   six electrons in the p-sublevel, it has three orbital

   ten electrons in the d- sublevel, it has five orbitals

   

   fourteen electrons in the f-sublevel, it has seven orbitals

The maximum number of electrons in an orbital is two.

learn more:

Atomic orbitals brainly.com/question/1832385

#learnwithBrainly

   

4 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
How many grams are in 0.572 moles of glucose, C6H12O6
LUCKY_DIMON [66]

Answer:

103.00

Explanation:

1 mole of Glucose

6 C = 6 * 12 = 72 grams

12H = 12 * 1 = 12 grams

6O = 6 * 16 =96 grams

Total          = 180 grams

0.572 moles of Glucose

1 mol of glucose = 180 grams

0.572 mols of glucose = x

x = 0.572 * 180

x = 103.00 grams

6 0
3 years ago
A chemist has one solu6on that is 40% sulfuric acid and one that is 10% sulfuric acid. How much of each should she use to make 2
kipiarov [429]

Answer:

12 L of 40% sulfuric acid solution and 8 L of 10% sulfuric acid solution are needed to make 20 L of sulfuric acid solution.

Explanation:

For first solution of sulfuric acid :

C₁ = 40% , V₁ = ?

For second solution of sulfuric acid :

C₂ = 10% , V₂ = ?

For the resultant solution of sulfuric acid:

C₃ = 28% , V₃ = 20L

Also,

<u>V₁ + V₂ = V₃ = 20L</u> ......................................(1)

Using

<u>C₁V₁ + C₂V₂ = C₃V₃</u>

<u>40×V₁ + 10×V₂ = 28×20</u>

So,

40V₁ + 10V₂ = 560........................................(2)

Solving 1 and 2 as:

V₂ = 20 - V₁

Applying in 2

40V₁ + 10(20 - V₁)  = 560

40V₁ + 200 - 10V₁ = 560

30V₁ = 360

<u>V₁ = 12 L</u>

So,

<u>V₂ = 20 - V₁ = 8L</u>

<u><em>12 L of 40% sulfuric acid solution and 8 L of 10% sulfuric acid solution are needed to make 20 L of sulfuric acid solution.</em></u>

4 0
3 years ago
• Briefly discuss the cause of errors in the measurements
rewona [7]
(also called Observational Error) is the difference between a measured quantity and its true value. It includes random error
4 0
3 years ago
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