When 1 mole of H2(g) reacts with F2(g) to form HF(g) according to the following equation, 542 kJ of energy are evolved H2(g) + F2(g2HF(g) Is this reaction endothermic or exothermic? What is the value of q?

Answers

Answer 1

Answer:

Answer: The reaction is exothermic. The value of q is -542 kJ.

Explanation:

Endothermic reactions are defined as the reactions in which energy of the product is greater than the energy of the reactants. The total energy is absorbed in the form of heat and for the reaction comes out to be positive.

Exothermic reactions are defined as the reactions in which energy of the product is lesser than the energy of the reactants. The total energy is released in the form of heat and for the reaction comes out to be negative.

Thus $H_2(g)+F_2(g)\rightarrow 2HF(g)$ evolves heat , it is exothermic in nature. The value of q is -542kJ.

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Carbon and oxygen combine to form the molecular compound CO2, while silicon and oxygen combine to form a covalent network solid with the formula unit SiO2. Explain the difference in bonding between the two group 4A elements and oxygen. g

Answers

Answer:

See explanation below.

Explanation:

Both carbon and silicon are members of group 4A(now group 14) i n the periodic table. Carbon is the first member of the group. CO2 is a gas while SiO2 is a solid. In SiO2, there are single bonds between silicon and oxygen and the geometry around the central atom is tetrahedral while in CO2, there are double carbon-oxygen bonds and the geometry around the central atom is linear. CO2 molecules are discrete and contain only weak vanderwaals forces.

Again, silicon bonds to oxygen via its 3p orbital while carbon bonds to oxygen via a 2p orbital. As a result of this, there will be less overlap between the pi orbitals of silicon and that of oxygen. This is why tetrahedral bonds are formed with oxygen leading to a covalent network solid rather than the formation of a silicon-oxygen pi bond. A covalent network solid is known to be made up of a network of atoms of the same or different elements connected to each other continuously throughout the structure by covalent bonds.

In SiO2, each silicon atom is surrounded by four oxygen atoms. Each corner is shared with another tetrahedron. SiO2 forms an infinite three dimensional structure and melts at a very high temperature.

Final answer:

Carbon and oxygen form a molecular compound CO2 with weaker covalent bonds, while silicon and oxygen form a covalent network solid SiO2 with stronger, three-dimensional covalent bonds.

Explanation:

The difference in bonding between carbon and oxygen compared to silicon and oxygen is due to the different nature of their chemical bonds. In the case of carbon and oxygen, they form a molecular compound CO2, where carbon and oxygen atoms share electrons to form covalent bonds. This is because carbon and oxygen have similar electronegativities, so they can share electrons equally. The covalent bonds in CO2 are relatively weak, allowing the compound to exist as a gas at room temperature and pressure.

On the other hand, silicon and oxygen form a covalent network solid with the formula unit SiO2, known as quartz. In this case, silicon and oxygen atoms are covalently bonded in a three-dimensional network structure, where each silicon atom is bonded to four oxygen atoms and each oxygen atom is bonded to two silicon atoms. This network structure gives SiO2 its high melting point and hardness, making it a solid at room temperature and pressure.

In summary, the difference in bonding between carbon and oxygen compared to silicon and oxygen is that carbon and oxygen form a molecular compound with weaker covalent bonds, while silicon and oxygen form a covalent network solid with stronger, three-dimensional covalent bonds.

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Saved Propane burns in air according to the equation C3Ha(g 502lg)3CO2) + 4H20(g) What volume of O2 in liters would be required if 15.0 L of propane burns, assuming that all of the gases are under the same conditions? Short Answer Toolbar navigation E I E B IUS EA This question will be sent to your Instructor for grading. 20 of 25 l Next > Prev nere to search

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Answer: 75 liters of $O_2$ in liters would be required if 15.0 L of propane burns, assuming that all of the gases are under the same conditions.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 Lat STP and contains avogadro's number $6.023* 10^(23)$ of particles.

To calculate the number of moles, we use the equation:

$\text{Number of moles of propane}=\frac{\text{Given volume}}{\text{Molar volume}}=(15.0L)/(22.4L)=0.67moles$

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

According to stoichiometry:

1 mole of propane combines with = 5 moles of oxygen

Thus 0.67 moles of propane combine with = $(5)/(1)* 0.67=3.35moles$

Volume of $O_2=moles* {\text {Molar Volume}}=3.35* 22.4L=75L$

Thus 75 liters of $O_2$ in liters would be required if 15.0 L of propane burns, assuming that all of the gases are under the same conditions.

Which of the following describes the change in atomic mass and atomic number during this reaction?

Answers

The mass number tells us the number (the sum of nucleons) of protons and neutrons in the nucleus of an atom. ... It is traditionally represented by the symbol Z. The atomic number uniquely identifies a chemical element. In an atom of neutral charge, atomic number is equal to the number of electrons.

50.00 mL of 0.10 M HNO 2 (nitrous acid, K a = 4.5 × 10 −4) is titrated with a 0.10 M KOH solution. After 25.00 mL of the KOH solution is added, the pH in the titration flask will be a. 2.17
b. 3.35
c. 2.41
d. 1.48
e. 7.00

Answers

Answer:

b. 3.35

Explanation:

To calculate the pH of a solution containing both acid and its salt (produced as a result of titration) we need to use Henderson’s equation i.e.

pH = pKa + log ([salt]/[acid]) (Eq. 01)

Where

pKa = -log(Ka) (Eq. 02)

[salt] = Molar concentration of salt produced as a result of titration

[acid] = Molar concentration of acid left in the solution after titration

Let’s now calculate the molar concentration of HNO2 and KOH considering following chemical reaction:

HNO2 + KOH ⇆ H2O + KNO2 (Eq. 03)

This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.

Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles

Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles

As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.

Therefore

Amount of salt produced i.e [salt] = 0.0025 moles (Eq. 04)

Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)

Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:

pH= -log(4.5x10 -4) + log (0.0025/0.0025)

Solving above we get

pH = 3.35

Final answer:

The pH value in the titration flask after 25.00 mL of the 0.10 M KOH solution is added to 50.00 mL of 0.10 M HNO2 solution is 3.35.

Explanation:

The subject of this question is titration, which is a method used in chemistry to measure the concentration of an unknown solution. Given 50.00 mL of 0.10 M HNO2 (nitrous acid, Ka = 4.5 × 10-4), titrated with 0.10 M KOH (potassium hydroxide), we need to calculate the pH after 25.00 mL of the KOH solution is added.

First, we need to find the moles of the HNO2 and the KOH. Moles equals Molarity times Volume. So, for HNO2, it is 0.10 M * 0.050 L which equals 0.005 moles. For KOH, it is 0.10 M * 0.025 L which equals 0.0025 moles.

Then, subtract the moles of OH- from the moles of HNO2 to determine the concentration of HNO2 left, which is 0.005 moles - 0.0025 moles = 0.0025 moles. Divide this by the total volume of the solution (50.00 mL + 25.00 mL = 75.00 mL or 0.075 L to determine the new concentration of HNO2, 0.0025 moles / 0.075 L = 0.033 M. Then use the given Ka value with the equation [H+] = sqrt(Ka * [HNO2]) to get [H+].

To find acids' pH, we use the formula pH = -log[H+]. Use the calculated [H+] to find the pH.

Upon performing these calculations, the resulting pH value should be approximately 3.35 after 25.00 mL of the KOH solution is added, so the answer is (b) 3.35.

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What the first song come to your mind?

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Answer: Hey Brother by Avicii

Let her go by Passenger

Be sure to answer all parts. The first step in HNO3 production is the catalyzed oxidation of NH3. Without a catalyst, a different reaction predominates: 4 NH3(g) +3 O2 (g) ⇌ 2 N2(g) + 6 H2O(g) When 0.0150 mol of NH3(g) and 0.0150 mol of O2(g) are placed in a 1.00−L container at a certain temperature, the N2 concentration at equilibrium is 1.96 × 10−3 M. Calculate Kc.

Answers

Answer: The value of $K_c$ for the reaction is $6.005* 10^(-6)$

Explanation:

We are given:

Initial moles of $NH_3=0.0150mol$

Initial moles of $O_2=0.0150mol$

Volume of the container = 1.00 L

Molarity of the solution = $\frac{\text{Number of moles}}{\text{Volume of container}}$

$[NH_3]_i=(0.0150)/(1.00)=0.0150M$

$[O_2]_i=(0.0150)/(1.00)=0.0150M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(g)$

Initial: 0.0150 0.0150

At eqllm: 0.0150-4x 0.0150-3x 2x 6x

The expression of $K_c$ for above equation follows:

$K_c=([N_2]^2[H_2O]^6)/([NH_3]^4[O_2]^3)$ .......(1)

We are given:

Equilibrium concentration of $N_2=1.96* 10^(-3)$

Equating the equilibrium concentrations of nitrogen, we get:

$2x=1.96* 10^(-3)\n\nx=0.98* 10^(-3)M$

Calculating the equilibrium concentrations:

Concentration of $NH_3=(0.0150-4x)=0.0150-4(0.00098)=0.01108M$

Concentration of $O_2=(0.0150-3x)=0.0150-3(0.00098)=0.01206M$

Concentration of $N_2=2x=2(0.00098)=0.00196M$

Concentration of $H_2O=6x=6(0.00098)=0.00588M$

Putting values in expression 1, we get:

$K_c=((0.00196)^2* (0.00588)^6)/((0.01108)^4* (0.01206)^3)\n\nK_c=6.005* 10^(-6)$

Hence, the value of $K_c$ for the reaction is $6.005* 10^(-6)$

Final answer:

To calculate the equilibrium constant, Kc, for the reaction that produces HNO3 from NH3 and O2, you need to determine the equilibrium concentrations of NH3 and O2. The given information includes the initial moles and concentration of NH3 and O2, as well as the equilibrium concentration of N2. Using the stoichiometry of the reaction and the given data, you can calculate the equilibrium concentrations and substitute them into the Kc expression to determine the numerical value of Kc.

Explanation:

The question asks to calculate the equilibrium constant, Kc, for the reaction that produces HNO3 from NH3 and O2. The reaction equation is 4 NH3(g) + 3 O2(g) ⇌ 2 N2(g) + 6 H2O(g). The given information is that 0.0150 mol of NH3(g) and 0.0150 mol of O2(g) are placed in a 1.00-L container, and the N2 concentration at equilibrium is 1.96 × 10−3 M. To solve for Kc, we need to calculate the equilibrium concentrations of NH3 and O2.

Using the stoichiometry of the reaction, we can determine that the equilibrium concentration of NH3 is (0.0150 - 2*1.96 × 10−3) M and the equilibrium concentration of O2 is (0.0150 - 3*1.96 × 10−3) M. Substituting these values into the equilibrium expression for Kc, we can calculate the value of Kc.

In this case, the equilibrium constant, Kc, can be calculated as [N2]^2 / ([NH3]^4 * [O2]^3). Substitute the given equilibrium concentration of N2 and the calculated equilibrium concentrations of NH3 and O2 into the Kc expression to determine the numerical value of Kc.

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