How is Hess's law used to measure enthalpy of a desired reaction?A. The enthalpy is obtained from the enthalpy of an intermediate
step.
B. The enthalpy is determined from the enthalpy of similar reactions.
C. The enthalpy from the final equation in a series of reactions is
used
D. Intermediate equations with known enthalpies are added together.

Answers

Answer 1

Answer:

Hess's law is used to measure the enthalpy of a desired chemical reaction because: D. Intermediate equations with known enthalpies are added together.

What is Hess's Law?

Hess's Law is also known as Hess's law of constant heat summation (enthalpy) and it was named after a Swiss-born Russian chemist called Germain Hess.

Hess's Law states that the energy change (enthalpy) experienced in a desired chemical reaction is equal to the sum of the energy changes (enthalpies) in each chemical reactions that it is made up of or contains.

Read more on Hess's Law here: brainly.com/question/9328637

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Develop expressions for the mole fraction of reacting species functions of the reaction coordinate for: A system initially containing 2 mol NH3 and 5 mol O2 and undergoing the reaction: 4NH3 (g) + 5O2 (g) ® 4NO (g) + 6 H20 (g) A system initially containing 3 mol NO2, 4 mol NH3, and 1 mol N2 and undergoing the reaction: 6NO2 (g) + 8NH3 (g) ® 7N2 (g) +12H2O (g)

Answers

Answer:

Individual mole fractions of all the species of the all reaction is as follows.

(a)

$y_{NH_(3)}=(2+(-4)\epsilon)/(7+\epsilon)$

$y_{O_(2)}=(5+(-5)\epsilon)/(7+\epsilon)$

$y_(NO)=(0+(4)\epsilon)/(7+\epsilon)=(4\epsilon)/(7+\epsilon)$

$y_{H_(2)O}=(0+6\epsilon)/(7+\epsilon)$

(b)

$y_{H_(2)S}=(3+(-2)\epsilon)/(8- \epsilon)$

$y_{O_(2)}=(5+(-3)\epsilon)/(8- \epsilon)$

$y_{H_(2)O}=(2\epsilon)/(8- \epsilon)$

$y_{SO_(2)}=(2\epsilon)/(8- \epsilon)$

(c)

$y_{NO_(2)}=(3+(-6)\epsilon)/(8+5\epsilon)$

$y_{NH_(3)}=(4+(-8)\epsilon)/(8+5\epsilon)$

$y_{N-{2}}=(1+7\epsilon)/(8+5 \epsilon)$

$y_{H_(2)O}=(12\epsilon)/(8+5\epsilon)$

Explanation:

(a)

Initial number of moles of $NH_(3)$ and $O_(2)$ are 2 mol and 5 mol respectively.

The given chemical reaction is as follows.

$4NH_(3)(g)+5O_(2)(g)\rightarrow 4NO(g)+6H_(2)O$

The stoichiometric numbers are as follows.

$v_{NH_(3)}=-4$

$v_{O_(2)}=-5$

$v_(NO)=4$

$v_{H_(2)O}=6$

The total number of moles initially present -7

$\Sigma v_(i)\epsilon = (-4-5+4+6)= 1\epsilon$

The expression for the mole fraction of species"i" is as follows.

$y_(i)=\frac{(n_{i_(o)})+(v_(i)\epsilon) }{n_(o)+v\epsilon}$

The individual mole fractions of all the species are as follows.

$y_{NH_(3)}=(2+(-4)\epsilon)/(7+\epsilon)$

$y_{O_(2)}=(5+(-5)\epsilon)/(7+\epsilon)$

$y_(NO)=(0+(4)\epsilon)/(7+\epsilon)=(4\epsilon)/(7+\epsilon)$

$y_{H_(2)O}=(0+6\epsilon)/(7+\epsilon)$

(b)

Initial number of moles of $H_(2)S$ and $O_(2)$ are 3 mol and 5 mol respectively.

The given chemical reaction is as follows.

$2H_(2)S(g)+3O_(2)(g)\rightarrow 2H_(2)O(g)+2SO_(2)$

The stoichiometric numbers are as follows.

$v_{H_(2)S}=-2$

$v_{O_(2)}=-3$

$v_{H_(2)O}=2$

$v_{SO_(2)=2$

The total number of moles initially present -8

$\Sigma v_(i)\epsilon = (-2-3+2+2)= - \epsilon$

The expression for the mole fraction of species"i" is as follows.

$y_(i)=\frac{(n_{i_(o)})+(v_(i)\epsilon) }{n_(o)+v\epsilon}$

The individual mole fractions of all the species are as follows.

$y_{H_(2)S}=(3+(-2)\epsilon)/(8- \epsilon)$

$y_{O_(2)}=(5+(-3)\epsilon)/(8- \epsilon)$

$y_{H_(2)O}=(2\epsilon)/(8- \epsilon)$

$y_{SO_(2)}=(2\epsilon)/(8- \epsilon)$

(c)

Initial number of moles of $NO_(2)$ , $NH_(3)$ and $N_(2)$ are 3 mol,4 mol and 1 mol respectively.

The given chemical reaction is as follows.

$6NO_(2)(g)+8NH_(3)(g)\rightarrow 7N_(2)(g)+12H_(2)O$

The stoichiometric numbers are as follows.

$v_{NO_(2)}=-6$

$v_{NH_(3)}=-8$

$v_{N_(2)}=7$

$v_{H_(2)O}=12$

The total number of moles initially present -8

$\Sigma v_(i)\epsilon = (-6-8+7+12)= 5 \epsilon$

The expression for the mole fraction of species"i" is as follows.

$y_(i)=\frac{(n_{i_(o)})+(v_(i)\epsilon) }{n_(o)+v\epsilon}$

The individual mole fractions of all the species are as follows.

$y_{NO_(2)}=(3+(-6)\epsilon)/(8+5\epsilon)$

$y_{NH_(3)}=(4+(-8)\epsilon)/(8+5\epsilon)$

$y_{N-{2}}=(1+7\epsilon)/(8+5 \epsilon)$

$y_{H_(2)O}=(12\epsilon)/(8+5\epsilon)$

Final answer:

Expressions for the mole fractions of reacting species are determined using stoichiometry and the initial molar amounts, taking into account the stoichiometric coefficients of the chemical reactions.

Explanation:

To develop expressions for the mole fraction of reacting species as functions of the reaction coordinate for the given systems, we will examine each reaction individually. For the reaction 4NH3 (g) + 5O2 (g) ® 4NO (g) + 6 H2O (g), we can use stoichiometry to correlate the molar amounts of each species with reaction progress. Given the initial amounts, we will track how the molar amount changes for each mole of NH3 reacted.

Starting with 2 mol NH3 and 5 mol O2, the mole ratio from NH3 to NO and H2O is 1:1 and 1:1.5, respectively. The mole ratio from NH3 to O2 is 4:5. If x moles of NH3 react, the mole fractions for each species at any point in the reaction can be expressed as follows:

Mole fraction of NH3: (2 - x)/(Total moles)
Mole fraction of O2: (5 - 5x/4)/(Total moles)
Mole fraction of NO: (4x)/(Total moles)
Mole fraction of H2O: (6x)/(Total moles)

Note that 'Total moles' is the sum of the ongoing moles of all species. The mole fractions must always add up to 1 at any point during the reaction.

For the second reaction 6NO2 (g) + 8NH3 (g) ® 7N2 (g) +12H2O (g), with initial amounts of 3 mol NO2, 4 mol NH3, and 1 mol N2, similar steps are taken. For every mole of NH3 reacted, the corresponding changes in molar amounts can be calculated from the stoichiometry of the balanced equation.

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How many moles of carbondioxide are produced when 0.2mol of sodium carbonate react with excess hydrovhloric acid

Answers

Answer:

0.2 moles of CO₂ are produced

Explanation:

Given data:

Moles of CO₂ produced = ?

Moles of Na₂CO₃ react = 0.2 mol

Solution:

Chemical equation:

Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O

Now we will compare the moles of CO₂ with Na₂CO₃ .

Na₂CO₃ : CO₂

1 : 1

0.2 : 0.2

Thus, 0.2 moles of CO₂ are produced.

Increasing which factor will cause the gravitational force between two objects to decrease?weights of the objects
distance between the objects
acceleration of the objects
masses of the objects

Answers

Increasing distance between the objects factor will cause the gravitational force between two objects to decrease. Therefore, option B is correct.

What causes gravitational force to decrease?

The gravitational force grows in proportion to the size of the masses . The gravitational force weakens rapidly as the distance between masses grows. Unless at least one of the objects has a lot of mass, detecting gravitational force is extremely difficult.

Gravity is affected by object size and distance between objects. Mass is a unit of measurement for the amount of matter in an object.

The force of gravity is proportional to the masses of the two objects and inversely proportional to the square of the distance between them. This means that the force of gravity increases with mass but decreases as the distance between objects increases.

Thus, option B is correct.

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Answer:

Explanation:

What is the molarity of the following solutions?a. 19.5 g NaHCO3 in 460.0 ml solution
b. 26.0 g H2SO4 in 200.0 mL solution
c. 15.0 g NaCl dissolved to make 420.0 mL solution

Answers

Answer:

a) NaHCO3 = 0.504 M

b) H2SO4 = 1.325 M

c) NaCl = 0.610 M

Explanation:

Step 1: Data given

Moles = mass / molar mass

Molarity = moles / volume

a. 19.5 g NaHCO3 in 460.0 ml solution

Step 1: Data given

Mass NaHCO3 = 19.5 grams

Volume = 460.0 mL = 0.460 L

Molar mass NaHCO3 = 84.0 g/mol

Step 2: Calculate moles NaHCO3

Moles NaHCO3 = 19.5 grams / 84.0 g/mol

Moles NaHCO3 = 0.232 moles

Step 3: Calculate molarity

Molarity = 0.232 moles / 0.460 L

Molarity = 0.504 M

b. 26.0 g H2SO4 in 200.0 mL solution

Step 1: Data given

Mass H2SO4 = 26.0 grams

Volume = 200.0 mL = 0.200 L

Molar mass H2SO4 = 98.08 g/mol

Step 2: Calculate moles H2SO4

Moles H2SO4 = 26.0 grams / 98.08 g/mol

Moles H2SO4 = 0.265 moles

Step 3: Calculate molarity

Molarity = 0.265 moles / 0.200 L

Molarity =1.325 M

c. 15.0 g NaCl dissolved to make 420.0 mL solution

Step 1: Data given

Mass NaCl = 15.0 grams

Volume = 420.0 mL = 0.420 L

Molar mass NaCl = 58.44 g/mol

Step 2: Calculate moles NaCl

Moles NaCl = 15.0 grams / 58.44 g/mol

Moles NaCl = 0.256 moles

Step 3: Calculate molarity

Molarity = 0.256 moles / 0.420 L

Molarity =0.610 M

If you mix different amounts of two ideal gases that are originally at different temperatures, what must be true of the final state after the temperature stabilizes? (There may be more than one correct choice.) a) Both gases will reach the same final temperature.

b) The final rms molecular speed will be the same for both gases.

c) The final average kinetic energy of a molecule will be the same for both gases.

Answers

Answer:

a,c are correct

Explanation:

a) On mixing two gases the final temperature of both the gases becomes the same. The heat will flow from high temp. gas to lower temp gas till the temp of both gases become equal (Thermal equilibrium). This is correct.

b) The rms speed of the molecule is inversely proportional to its molar mass so the final rsm will not be the same. This is incorrect.

c) The average kinetic energy of the system will remain the same. Hence this is also correct.

Suppose a scientist made a claim that all spontaneous reactions are exothermic. Whic of the following would provide the strongest challenge to their claim? Suppose a scientist made a claim that all spontaneous reactions are exothermic. Which of the following would provide the strongest challenge to their claim? a. An exothermic reaction which is not spontaneous
b. An endothermic reaction that only proceeds when coupled to an exothermic reaction
c. An endothermic reaction that only proceeds when a catalytst is present
d. An endothermic reaction which is not spontaneous
e. All of the above

Answers

Answer: Option (c) is the correct answer.

Explanation:

It is given that the scientist is claiming that all the spontaneous reactions are exothermic in nature.

And, it is known that when a reaction is spontaneous in nature then $\Delta G$ is negative.

Now, the relation between Gibb's free energy, enthalpy and entropy is as follows.

$\Delta G$ = $\Delta H - T \Delta S$

So, when a catalyst is present in a chemical reaction then we do not need to give large amount of heat from outside. And, because of this the enthalpy of reaction will not be highly positive.

Hence, the value of $\Delta G$ will result in a negative value which means the reaction is spontaneous.

Thus, we can conclude that an endothermic reaction that only proceeds when a catalytst is present, would provide the strongest challenge to their claim.

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