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3 years ago

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

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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,

V₁ + V₂ = V₃ = 20L ......................................(1)

Using

C₁V₁ + C₂V₂ = C₃V₃

40×V₁ + 10×V₂ = 28×20

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

V₁ = 12 L

So,

V₂ = 20 - V₁ = 8L

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.

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