Aerosol cans have a label that warns the user not to use them above a certain temperature and not to dispose of them by incineration. Even an empty can contains residual gaseous propellant. For example, the residual pressure in a can is 1.11 atm when it is sitting on a shelf at 23 ∘C. If the can is placed on top of the furnace where the temperature reaches the boiling point of water, what is the pressure inside the can?

Answer:

The number of moles of hydrochloric acid are 0.861

Explanation:

In first solution, the [HCl] is 11.7 M, which it means that 11.7 moles are present in 1 liter.

So we took 25 mL and we have to know how many moles, do we have now.

1000 mL ____ 11.7 moles

25 mL _____ (25 . 11.7)/1000 = 0.2925 moles

This are the moles, we add to the solution where the [HCl] is 3.25 M

In 1000 mL __ we have __ 3.25moles

175 mL ____ we have __ (175 . 3.25)/1000 = 0.56875 moles

Total moles: 0.2925 + 0.56875 = 0.861 moles

The number of moles of hydrochloric acid from 175.00 mL of 3.25 M hydrochloric acid solution is 0.56875 mol.

To determine the number of moles of hydrochloric acid (HCl) in the 175.00 mL of 3.25 M hydrochloric acid solution, we can use the formula for molarity (M), which is:

[tex]\[ M = \frac{\text{moles of solute (mol)}}{\text{volume of solution (L)}} \][/tex]

 Rearranging the formula to solve for the moles of solute, we get:

[tex]\[ \text{moles of solute (mol)} = M \times \text{volume of solution (L)} \][/tex]

 Given that the molarity (M) of the hydrochloric acid solution is 3.25 M and the volume is 175.00 mL, we first need to convert the volume from milliliters to liters because molarity is defined in terms of moles per liter.

[tex]\[ \text{Volume in liters (L)} = \frac{175.00 \text{ mL}}{1000 \text{ mL/L}} = 0.175 \text{ L} \][/tex]

Now, we can calculate the moles of HCl:

[tex]\[ \text{moles of HCl} = 3.25 \text{ M} \times 0.175 \text{ L} \][/tex]

[tex]\[ \text{moles of HCl} = 0.56875 \text{ mol} \][/tex]

 Therefore, the number of moles of hydrochloric acid from 175.00 mL of 3.25 M hydrochloric acid solution is 0.56875 mol.

Answer:

The value of gold in the oceans is 4.07x10¹⁴ $

Explanation:

A problem with relation between the units.

This are the data, 3800 m (depth of the ocean)

The total surface 3.63×10⁸ km² (total surface)

5.8×10⁻⁹g/L (gold concentration)

$1595 (value of each troy oz)

31.1 g (mass of each troy oz in grams)

19.3 g/cm³ (density of gold)

As we have a depth (a kind of height) and the total surface we can know the volume that the ocean occupies. This height is in m, the surface in km².

We should convert eveything in dm to work with concentration.

3800 m to cm = 3800 . 10 = 3800 → 3.8 x10⁴ dm

3.63×10⁸ km² to dm² = 3.63×10⁸ . 1x10⁸ = 3.63×10¹⁶ dm²

(1 km² = 1x10⁸dm)

3.8 x10⁴ dm  . 3.63×10¹⁶ dm² = 1.37 x10²¹ dm³

This is the volume that the ocean occupies. By using concentration, we can know the mass of gold in all the ocean.

1L = 1 dm³

1L _____ 5.8×10⁻⁹g

1.37 x10²¹ L ____ 7.9 x10¹² g

So 1 troy oz pays $1595 and 1 troy oz is 31.1 grams, so 31.1 grams pay $1595.

The final rule of three will be

31.1 g __ pay ___ $1595

7.9 x10¹² g ___ pay (8 x10¹¹ g . $1595) / 31.1 g = 4.07x10¹⁴ $

This detailed answer explains how to calculate the value of gold in Earth's oceans based on surface area, depth, and gold concentration. The value of gold in the oceans is found to be around $8.1 trillion.

Earth's oceans have a total surface area of 3.63×[tex]10^8 km^2[/tex] and an average depth of 3800 m. The oceans have an average concentration of dissolved gold of 5.8×[tex]10^{-9}[/tex] g/L. Given the price of gold as $1595 per troy oz, we can calculate the value of gold in the oceans.

To calculate the value of gold in the oceans, we first determine the total mass of gold present in oceans, which is approximately 1.4 × [tex]10^{14}[/tex] g. By using the density of gold as 19.3 [tex]g/cm^{3}[/tex] and the conversion factor of 1 troy oz = 31.1 g, we can find the total value of gold in the oceans.

The value of gold in the oceans is found to be around $8.1 trillion based on the given data points and calculations.

Answer: The percent yield of silver is 86.96 %

Explanation:

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

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]     .....(1)

Given mass of copper = 40.0 g

Molar mass of copper = 63.55 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of copper}=\frac{40.0g}{63.55g/mol}=0.629mol[/tex]

The given chemical equation follows:

[tex]Cu+2AgNO_3\rightarrow Cu(NO_3)_2+2Ag[/tex]

By Stoichiometry of the reaction:

1 mole of copper produces 2 moles of silver

So, 0.629 moles of copper will produce = [tex]\frac{2}{1}\times 0.629=1.258mol[/tex] of silver

Now, calculating the mass of silver from equation 1, we get:

Molar mass of silver = 107.9 g/mol

Moles of silver = 1.258 moles

Putting values in equation 1, we get:

[tex]1.258mol=\frac{\text{Mass of silver}}{107.9g/mol}\\\\\text{Mass of silver}=(1.285mol\times 107.9g/mol)=135.7g[/tex]

To calculate the percentage yield of silver, we use the equation:

[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]

Experimental yield of silver = 118 g

Theoretical yield of silver = 135.7 g

Putting values in above equation, we get:

[tex]\%\text{ yield of silver}=\frac{118g}{135.7g}\times 100\\\\\% \text{yield of silver}=86.96\%[/tex]

Hence, the percent yield of silver is 86.96 %

The percent yield of silver when 40.0 g of copper react with silver nitrate, producing 118 g of silver, is calculated to be 86.9%.

When 40.0 g of copper (Cu) are reacted with silver nitrate (AgNO3) to produce 118 g of silver (Ag), the percent yield of silver can be calculated using the actual yield and the theoretical yield. The balanced chemical equation for the reaction is: Cu (s) + 2 AgNO3(aq) → Cu(NO3)2 (aq) + 2 Ag (s).

First, we calculate the theoretical yield of silver that should have been produced from 40.0 g of copper. The molar mass of copper is 63.55 g/mol, so 40.0 g of copper is equal to 40.0 g / 63.55 g/mol = 0.629 mol of copper. According to the balanced equation, 1 mol of copper will produce 2 mol of silver. Hence, 0.629 mol of copper will produce 0.629 × 2 = 1.258 mol of silver. With the molar mass of silver being 107.9 g/mol, the theoretical yield of silver is 1.258 mol × 107.9 g/mol = 135.8 g.

To find the percent yield, divide the actual yield (118 g of Ag) by the theoretical yield and multiply by 100:
Percent yield = (118 g / 135.8 g) × 100 = 86.9%

Answer:

A

Explanation:

the rate of evaporation equals the rate of condensation.

The values of ΔH and ΔE for the combustion of one mole of butane are 2658 kJ and 2655 kJ, respectively.

Part A:

The values of ΔH and ΔE for the combustion of one mole of butane can be determined using the first law of thermodynamics:

ΔH = q + w

where ΔH is the change in enthalpy, q is the heat released or absorbed, and w is the work done during the reaction.

From the given information, we know that 1 mole of butane produces 2658 kJ of heat and does 3 kJ of work. Therefore, ΔH = 2658 kJ and ΔE = ΔH - w = 2658 kJ - 3 kJ = 2655 kJ.

Part B:

Expressing the answers using four significant figures, we have ΔH = 2658 kJ and ΔE = 2655 kJ.

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The change in enthalpy (∆H) for the combustion of one mole of butane is -2658 kJ and the change in internal energy (∆E) is -2655 kJ, both values are expressed with four significant figures.

In thermodynamics, the heat transfer at constant pressure is defined as the change in enthalpy (∆H), while the change in internal energy (∆E) is given by heat transfer minus work done. For the combustion of one mole of butane, given that the heat produced is 2658 kJ (which is released so it is -2658 kJ), and the work done is 3 kJ (the system does work so it's -3kJ).

For Part A, we can calculate these values as follows: ∆H = q_p = -2658 kJ and ∆E = ∆H - work = -2658 kJ - (-3 kJ) = -2655 kJ

For part B, using four significant figures, the values will be ∆H = -2658 kJ and ∆E = -2655 kJ

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Answer:

90 mg of H₂O

Explanation:

The reaction that takes place is:

2HCl + Ba(OH)₂ → BaCl₂(aq) + 2H₂O

With the information given by the problem and the definition of molarity (M=n/V), we can calculate the moles of HCl:

20.00 mL * 0.250 M = 5 mmol HCl

Now we use the stoichiometric ratio to convert moles of HCl to moles of H₂O and then to mass of H₂O:

5 mmol HCl * [tex]\frac{2mmolH_{2}O}{2mmolHCl} *\frac{18mg}{1mmolH_{2}O}[/tex] = 90 mg H₂O

Final answer:

To determine the mass of H2O produced from a 20.00 mL sample of 0.250 M HCl solution, calculate the moles of HCl, use the stoichiometric ratios from the balanced equation to find moles of H2O, and then convert moles of H2O to grams using its molar mass. The mass of H2O produced is 0.090075 grams.

Explanation:

Solution stoichiometry is an essential part of chemistry that deals with the calculations involving volumes of solutions of reacting substances. To determine the mass of H2O produced when 20.00 mL of a 0.250 M HCl solution reacts with excess Ba(OH)2, we must first write the balanced chemical equation for the reaction:

HCl(aq) + Ba(OH)2(s) → BaCl2(aq) + 2 H2O(l)

From the balanced equation, we see that one mole of HCl produces one mole of H2O. Using the molarity of HCl, we can calculate the number of moles of HCl in the 20.00 mL sample:

moles of HCl = Molarity of HCl × Volume of HCl (in liters) = 0.250 mol/L × 0.02000 L = 0.00500 mol

Since the molar ratio of HCl to H2O is 1:1, the moles of H2O produced will also be 0.00500 mol. We can then find the mass of H2O using the molar mass of water (approximately 18.015 g/mol):

mass of H2O = moles of H2O × molar mass of H2O = 0.00500 mol × 18.015 g/mol = 0.090075 g

Therefore, the mass of H2O produced is 0.090075 grams.

Answer:

Correct answer is structural formula expanded.

Explanation:

A. is wrong.

The molecular formula only show the number of atoms of each element present and the ratio in which they are present. It does not provide any information as regards the bonds whether in its expanded or condensed form.

B. is wrong

While the structural formula will show the inter linkage between the atoms, the condensed structural formula won't provide complete information as regards these interlinkages.

D. is wrong

As established in A above, the molecular formula only provides the number of atoms and their ratios. It does not elucidate the types of bonds present therein.

E. is wrong

The isometric formula, although will elucidate to an extent does not serve the purpose of providing bonding details. It only provides comparisons between isomers.

Final answer:

A formula that displays all bonds in a molecule is called a structural formula, which may be written as an expanded structure or, more commonly, as a condensed structural formula. So the correct option is b.

Explanation:

A formula that shows the arrangement of all bonds in a molecule is called a structural formula. A structural formula can come in various forms such as an expanded structure, which shows all the carbon and hydrogen atoms as well as the bonds attaching them. However, as molecules increase in size, structural formulas can become complex. To simplify this, chemists often use a condensed structural formula, which provides a shorthand representation of the molecule by listing the atoms bonded to each carbon atom directly next to it. This helps to visualize the molecule's structure in a more compact form.

Answer:

[tex]\DeltaG=158235.4 J[/tex]

Explanation:

The free-energy equation for a redox reaction is:

[tex]\DeltaG=-n*F*E[/tex]

Where:

In this case:

[tex]n=2 mol[/tex]

[tex]F=96485 C/mol[/tex]

[tex]E=0.82 V[/tex]

So:

[tex]\DeltaG=-2mol*96485 C/mol*0.82 V[/tex]

[tex]\DeltaG=158235.4 J[/tex]

Final answer:

To determine the enthalpy of neutralization, we can use the equation q = mcΔT to calculate the heat absorbed or released by the solution. Then, we can use the molar ratios from the balanced equation to calculate the enthalpy of neutralization.

Explanation:

From the given information, we can calculate the enthalpy of neutralization using the equation q = mcΔT. The heat absorbed or released by the reaction is equal to the heat absorbed or released by the solution. First, we need to calculate the heat absorbed or released by the solution using the formula q = mcΔT, where q is the heat, m is the mass of the solution, c is the specific heat capacity of the solution, and ΔT is the change in temperature. Then, we can use the molar ratios from the balanced equation to calculate the enthalpy of neutralization.

Explanation:

It is known that relation between partial pressure, mole fraction and pressure is as follows.

      Partial pressure of gas = mole fraction of gas × Pressure of gas

Therefore, putting the given values into the above formula as follows.

   Partial pressure of gas = mole fraction of gas × Pressure of gas

                                          = [tex]0.21 \times 1.13 atm[/tex]

                                           = 0.237 atm

According to Henry's law,

       Concentration of oxygen = Henry's law constant × partial pressure of oxygen

             = [tex]1.3 \times 10^{-3} M/atm \times 0.2373 atm[/tex]

             = [tex]3.08 \times 10^{-4}[/tex] M

Therefore, calculate moles of oxygen in 5.00 L present as follows.

   Moles of oxygen in 5.00 L = volume × concentration

                                                 = [tex]5.00 \times 3.0849 \times 10^{-4}[/tex]

                                                 = [tex]1.542 \times 10^{-3}[/tex] mol

Now, we will calculate the mass of oxygen as follows.

        Mass of oxygen = moles × molar mass of oxygen

                                    = [tex]1.542 \times 10^{-3} mol \times 32 g/mol[/tex] mol    

                                    = 0.0494 g

or,                                 = 49.4 mg           (As 1 g = 1000 mg)

thus, we can conclude that the mass of given oxygen (in mg) is 49.4 mg.

The mass of oxygen dissolved in water has been 49.4 mg.

The partial pressure of oxygen in the air:

Partial pressure = Mole fraction [tex]\times[/tex] Pressure of gas

Partial pressure = 0.21 [tex]\times[/tex] 1.13 atm

Partial pressure of oxygen = 0.237 atm.

The concentration of oxygen can be given by Henry's law.

Concentration of Oxygen = Henry's constant ([tex]\rm k_H[/tex]) [tex]\times[/tex] Partial pressure

Concentration = 1.3 [tex]\rm \times\;10^-^3[/tex] M/atm [tex]\times[/tex] 0.237 atm

Concentration of oxygen = 3.08 [tex]\rm \times\;10^-^4[/tex] M

Moles can be given by:

Moles = Molarity [tex]\times[/tex] Volume

Moles of oxygen =  3.08 [tex]\rm \times\;10^-^4[/tex] M [tex]\times[/tex] 5

Moles of oxygen = 1.542  [tex]\rm \times\;10^-^3[/tex] mol

Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]

Weight of oxygen = Moles [tex]\times[/tex] Molecular weight

Weight of oxygen =  1.542  [tex]\rm \times\;10^-^3[/tex] mol [tex]\times[/tex] 32 g/mol

Weight of oxygen = 0.0494 grams

Weight of oxygen = 49.4 mg.

The mass of oxygen dissolved in water has been 49.4 mg.

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Answer:

The answer is A. Newton's First law

Explanation:

Newton's First Law states that an object will stay at rest if it's at rest and an object in motion will stay in motion unless another object comes.

Hope this helps : )

The relations among the forces acting on a body and the motion of the body is first formulated by physicist Isaac Newton. An object which is at rest resume at rest whereas an object in motion carry on with motion is given by Newton's First Law. The correct option is A.

The Newton's First Law is also called the Law of inertia which states that if a body at rest or moving at a constant speed, it will remain at rest or keep its motion in a straight line unless it is acted upon by a force.

As long as the forces are not unbalanced, which means as long as the forces are balanced, the first law of motion is obeyed. When we shake the branch of a mango tree, the mangoes fall, which is an example of the inertia of rest.

During the breaking of a bus or train immediately, the passengers who are sitting lean forward. This denotes the inertia of motion.

Thus the correct option is A.

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The work done on a piston going from 0.130 liters to 0.600 liters under an external pressure of 8.00 atm is 380.882 Joules. This is calculated using the formula for work done under constant pressure, which is W = PΔV.

The subject of this question is regarding the computation of work done on a piston due to change in volume under constant external pressure, and this is a concept in Physics. To compute the work done when a gas expands or compresses, we can use the formula W = PΔV, where W is work done, P is the external pressure, and ΔV is the change in volume.

Here, the external pressure (P) is 8.00 ATM. But to obtain the work done in Joules, we first need to convert this pressure from ATM to Pa (Pascals): 1 ATM = 101325 Pa, thus 8.00 ATM = 8.00 * 101325 = 810600 Pa.

The change in volume (ΔV) is the final volume minus the initial volume, which is 0.600 liters - 0.130 liters = 0.470 liters. But again to match units, we should convert this volume from liters to cubic meters: 1 liter = 0.001 cubic meters, so 0.470 liters = 0.470 * 0.001 cubic meters = 0.00047 m^3.

Therefore, substituting the values into the formula, we get: Work W = PΔV = 810600 Pa * 0.00047 m^3 = 380.882 Joules.

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In order to calculate how much work has been done, it is crucial to convert the provided values to SI units before using the formula for work done by a gas at constant pressure. In this particular case, the amount of work done is approximately 380.9 joules.

The work done by a gas when it expands or contracts at constant pressure can be calculated using the formula W = PΔV, where W is the work done, P is the pressure, and ΔV is the change in volume. However, the pressure is given in atm and the volume in liters, we need to convert these to SI units. The conversion factors are 1 atm = 101.3 kPa = 101,300 Pa and 1 liter = 1 x 10-3 m3. Using these conversion factors, the pressure is 8.00 atm x 101,300 Pa/atm = 810,400 Pa and the change in volume is 0.600 liters - 0.130 liters = 0.470 liters = 0.470 x 10-3 m3. Substituting these values into the formula gives W = (810,400 Pa)(0.470 x 10-3 m3) = 380.9 J.

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The empirical formula of the hydrocarbon is C2H3. The molar mass of the hydrocarbon is about nine times the molar mass of the empirical formula, thus the molecular formula of the hydrocarbon is C18H27.

First off, we need to find the empirical formula using the given data. The molar mass for carbon (C) is 12 g/mol and for oxygen (O) is 16 g/mol. Since the combustion of the hydrocarbon produced 440 mg of CO2, which contains 12/44 of carbon, we multiply 440 mg by 12/44 to get 120 mg of carbon.

The molar mass for hydrogen (H) is 1 g/mol. The combustion of the hydrocarbon also produced 135 mg of H2O, which contains 2/18 of hydrogen, so we multiply 135 mg by 2/18 to get 15 mg of hydrogen.

The empirical formula mass of carbon (120 mg/12 g/mol = 10 mmol) and hydrogen (15 mg/1 g/mol= 15 mmol) gives us CH1.5. But because we can't have fractional moles of atoms in a molecule, we multiply by 2 to get C2H3.

Therefore, the empirical formula is C2H3.

The molecular formula is a multiple of the empirical formula. The molar mass of C2H3 is 29 g/mol. As the molar mass of the hydrocarbon is given as 270 g/mol, we find that this is approximately 270/29 = around 9 times the empirical formula mass. So we multiply C2H3 by this to give the molecular formula: C18H27.

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The molecular formula is 18 times the empirical formula:

[tex]\[ \text{Molecular formula} = (\text{CH}_3)_{18} = \text{C}_{18}\text{H}_{54}[/tex]

To determine the molecular formula of the hydrocarbon, we need to first find the empirical formula, which is the simplest whole-number ratio of atoms in the compound. Then, using the molar mass, we can find the molecular formula.

 Step 1: Calculate the moles of carbon and hydrogen in the sample.

 From the given information, we have:

- Mass of CO2 produced = 440 mg = 0.440 g

- Mass of H2O produced = 135 mg = 0.135 g

Using the molar masses of CO2 (44.01 g/mol) and H2O (18.015 g/mol), we can calculate the moles of carbon and hydrogen:

Moles of carbon (from CO2):

[tex]\[ n_C = \frac{\text{mass of CO2}}{\text{molar mass of CO2}} = \frac{0.440 \text{ g}}{44.01 \text{ g/mol}} \approx 0.0100 \text{ mol} \][/tex]

Moles of hydrogen (from H2O):

[tex]\[ n_H = \frac{\text{mass of H2O} \times 2}{\text{molar mass of H2O}} = \frac{0.135 \text{ g} \times 2}{18.015 \text{ g/mol}} \approx 0.0150 \text{ mol} \][/tex]

 Step 2: Determine the simplest whole-number ratio of moles of carbon to hydrogen.

 The ratio of moles of hydrogen to moles of carbon is:

[tex]\[ \frac{n_H}{n_C} = \frac{0.0150}{0.0100} = \frac{3}{2} \][/tex]

This means there are 3 moles of hydrogen for every 2 moles of carbon. Therefore, the empirical formula of the hydrocarbon is CH3 (methane).

Step 3: Calculate the molar mass of the empirical formula.

 The molar mass of CH3 is:

[tex]\[ \text{Molar mass of CH}_3 = \text{molar mass of C} + 3 \times \text{molar mass of H} \][/tex]

[tex]\[ \text{Molar mass of CH}_3 = 12.01 \text{ g/mol} + 3 \times 1.008 \text{ g/mol} \approx 15.024 \text{ g/mol} \][/tex]

Step 4: Determine the molecular formula.

The given molar mass of the hydrocarbon sample is 270 g/mol. We can find the molecular formula by comparing the empirical formula molar mass to the given molar mass:

[tex]\[ \text{Molecular formula molar mass} = n \times \text{Empirical formula molar mass} \][/tex]

[tex]\[ 270 \text{ g/mol} = n \times 15.024 \text{ g/mol} \][/tex]

[tex]\[ n = \frac{270}{15.024} \approx 18 \][/tex]

 Therefore, the molecular formula is 18 times the empirical formula:

[tex]\[ \text{Molecular formula} = (\text{CH}_3)_{18} = \text{C}_{18}\text{H}_{54}[/tex]

The final answer is the molecular formula of the hydrocarbon, which is [tex]\(\boxed{\text{C}_{18}\text{H}_{54}}\).[/tex]

For the given reaction, the enthalpy change calculated with bond energies is 36 kJ/mol.

Given the following balanced reaction:

H₂(g) + CO₂(g) → H₂O(g) + CO(g)     (1)

The enthalpy change ([tex]\Delta H_{rxn}[/tex]) can be calculated with the next equation:

[tex] \Delta H_{rxn} = \Sigma (n*E_{r}) - \Sigma (m*E_{p}) [/tex]

Where:

n and m are the stoichiometric number of moles of the reactants and the products, respectively          

[tex]E_{r}[/tex] and [tex]E_{p}[/tex] are the bond energies of the reactants and the products, respectively

For reaction (1), the enthalpy change is given by:

[tex]\Delta H_{rxn} = \Sigma (n_{H_{2}}*E_{H-H} + n_{CO_{2}}*2*E_{C=O}) - \Sigma (m_{H_{2}O}*2*E_{H-O} + m_{CO}*E_{C\equiv O})[/tex]

[tex] \Delta H_{rxn} = \Sigma (1*436 kJ/mol + 1*2*799 kJ/mol) - \Sigma (1*2*463 kJ/mol} + 1*1072 kJ/mol) = 36 kJ/mol [/tex]                        

Therefore, the enthalpy change for the given reaction is 36 kJ/mol.

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The enthalpy change for the reaction H2(g) + CO2(g) ⟶ H2O(g) + CO(g) can be calculated by subtracting the sum of bond energies of bonds formed from the sum of bond energies of bonds broken in the reactants and products. Specific bond energies are needed from a bond enthalpy table to complete the calculation.

To calculate the enthalpy change for the reaction H2(g) + CO2(g) ⟶ H2O(g) + CO(g), you have to consider the bonds being broken in the reactants and the bonds being formed in the products. This involves using bond energies for each type of bond involved in the reaction:

The general approach to calculate the enthalpy change (ΔH) is:

ΔH = Sum of bond energies of bonds broken - Sum of bond energies of bonds formed

However, to provide a precise answer, the specific bond energies for each bond must be known, which are not provided in the question. Typically, you can find these values in a table of average bond enthalpies. For the sake of this example, if we had the bond energies for H-H, O=O, O-H, and C=O, we would use the following formula:

ΔH = (Bond energy of H-H + Bond energy of O=O) - (2 x Bond energy of O-H + Bond energy of C=O)

Since we don't have the actual values, you would need to look up the specific bond energies and plug them into the formula to get the enthalpy change for this reaction.

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Answer:

Empirical formula = S₁Cl₂ = SCl₂

Explanation:

The empirical formula of a compound is the simplest whole number ratio of each type of atom in a compound.

This compound contains 31.1% sulfur and 68.9% chlorine.

That is 100g of the compound contains 31.1 g of sulfur and 68.9 g of chlorine.

Convert the mass of each element to moles using the molar mass from the periodic table.

Moles of Sulfur = [tex]\frac{31.1}{32.065 } \\[/tex]

= 0.9699

Moles of Chlorine= [tex]\frac{68.9}{35.453 } \\[/tex]

= 1.9434

Divide each mole value by the smallest number of moles calculated.

Units of sulfur =  [tex]\frac{0.9699}{0.9699} \\[/tex]

= 1

Units of Chlorine =  [tex]\frac{1.9434}{0.9699} \\[/tex]

= 2

Empirical formula = S₁Cl₂ = SCl₂

Answer:

ΔH = -610.1 kJ

Explanation:

By the Hess Law, when a reaction is formed by various steps, the enthalpy change (ΔH) of the global reaction is the sum of the enthalpy change of the steps reactions. Besides, if it's necessary for a change in the reaction, ΔH will suffer the same change. If the reaction multiplied by a number, ΔH will be multiplied by the same number, and if the reaction is inverted, the signal of ΔH is inverted.

P₄(s) + 6Cl₂(g) → 4 PCl₃(g) ΔH = -1225.6 kJ

P₄(s) + 5O₂(g) → P₄O₁₀(s) ΔH = -2967.3 kJ (inverted)

PCl₃(g) + Cl₂(g) → PCl₅(g) ΔH = -84.2 kJ (inverted and multiplied by 6)

PCl₃(g) + 1/2O₂(g) → Cl₃PO(g) ΔH = -285.7 kJ (multiplied by 10)

P₄(s) + 6Cl₂(g) → 4 PCl₃(g) ΔH = -1225.6 kJ

P₄O₁₀(s) → P₄(s) + 5O₂(g) ΔH = +2967.3 kJ

6PCl₅(g) → 6PCl₃(g) + 6Cl₂(g) ΔH = +505.2 kJ

10PCl₃(g) + 5O₂(g) → 10Cl₃PO(g) ΔH = -2857.0 kJ

----------------------------------------------------------------------------

The bolded substances will be eliminated because have the same amount in product and reactant:

P₄O₁₀(s) + 6PCl₅(g) → 10Cl₃PO(g)

ΔH = -1225.6 + 2967.3 + 505.2 -2857.0

ΔH = -610.1 kJ

Based on the data provided, the enthalpy change ΔH for the given reaction is -610.1 kJ.

The enthalpy change of a reaction is the energy evolved or absorbed when reactant molecules react to form products.

From Hess' Law of constant heat summation, the enthalpy change (ΔH) of the reaction is the sum of the enthalpy change of the several reaction steps. reactions.

Considering the sum of the intermediate reaction steps:

The reactions and enthalpy changes become:

Summing 1, 2, 3 and 4 gives:

Enthalpy change, ΔH is then calculated thus:

ΔH = -1225.6 + 2967.3 + 505.2 -2857.0

ΔH = -610.1 kJ

Therefore, the enthalpy change ΔH for the given reaction is -610.1 kJ.

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The reason and what will happen if one atom in the given type of covalent bond reaction can win the tug of war is explained below.

      In chemistry, tug of war is usually used in polar covalent bond. Polar covalent bond is the bond that occurs as a result of unequal sharing of electrons between 2 atoms.

    Now, the reason why polar covalent bond is referred to as tug of war is because in tug of war games, the person that is the strongest usually wins. Likewise in polar covalent bonds, between the 2 sharing atoms, the atom that has higher electronegativity will have a stronger pull for electrons and as a result of this, it will have the shared electrons closer to it.

     Therefore, the molecule of the atom with higher electronegativity will become  negative because it has drawn the electrons closer to itself while the molecule of the atom that has the lesser electronegativity will become positive.

      In conclusion, Yes one atom in polar covalent bond reaction can win the ''tug of war''.

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Answer:

d. -564 kJ

Explanation:

It is possible to find ΔE° using the formula:

ΔH° = ΔE° + PΔV

For the reaction:

2CO(g) + O₂(g) → 2CO₂(g)  ΔH°= -566kJ

As P is 1,00 atm and ΔV is -24,5L:

-566 kJ = ΔE° + 1,00atm×-24,5L×[tex]\frac{0,101kJ}{1atmL}[/tex]

-566 kJ = ΔE° - 2,47 kJ

ΔE° = -563,53 kJ

The ΔE° is:

d. -564 kJ

I hope it helps!

ΔE° for the reaction below if the process is carried out at a constant pressure of 1.00 atm and ΔV is d. -564 kJ

To find the ΔE° for the reaction, use the formula ΔE° = ΔH° - (Δn)RT and plug in the given values to calculate the energy change.

Therefore, the correct answer is d. -564 kJ.

Correct question is: Find ΔE° for the reaction below if the process is carried out at a constant pressure of 1.00 atm andΔV (the volume change) = -24.5 L. (1 L ∙ atm = 101 J)
2 CO(g) + O₂ (g) → 2 CO₂(g) ΔH° = -566. kJ
a. +2.47 kJ
b. -568 kJ
c. -2.47 kJ
d. -564 kJ

Answer:

763 g

Explanation:

Let's consider the following thermochemical equation.

SiO₂(s) + 4 HF(g) → SiF₄(g) + 2 H₂O(l)    ΔH°rxn = -184 kJ

From the enthalpy of the reaction (ΔH°rxn), we can affirm that 184 kJ are released when 2 moles of H₂O(l) are produced. Taking into account that the molar mass of H₂O is 18.01 g/mol, the mass of water formed to produce 3900 kJ of energy is:

[tex]-3900kJ.\frac{2molH_{2}O}{-184kJ} .\frac{18.01gH_{2}O}{1molH_{2}O} =763gH_{2}O[/tex]

Answer:

a. Negative

b. Positive

c. Negative

d. zero

Explanation:

Entropy is measure of disorder. Positive entropy implies that a system is becoming more disordered. The opposite is true.

(a) N2(g)+3H2(g) → 2NH3(g) Negative because the system is becoming less disordered since the number of gaseous moles is decreasing

(b) CaCO3(s)→CaO(s)+CO2(g)  Positive because a solid produces a gas which is more disorder therefore there is an increase in entropy

(c) 3C2H2(g)→C6H6(g) Negative because the number of moles of a gas decrease meaninng there is less disorder

(d) Al2O3(s)+3H2(g) → 2Al(s)+3H2O(g) zero because the gaseous moles do not change

The sign of the entropy change depends on the change in the number of gas molecules and the phases of the reactants and products. Generally, when gases are produced or the number of particles increases, entropy increases, and vice versa.

Predict the sign of the entropy change of the system for each of the following reactions:

Entropy, a measurement of disorder or randomness in a system, tends to increase when solids or liquids turn into gases, when the temperature increases, or when the number of individual particles in a system increases.

Answer:

Lean Flammability limit ,or lean limit.

Explanation:

The point is know as Lean Flammability limit ,or lean limit. The lean limit is usually expressed in volume percent. It can be defined as the lower range of concentration of over which a flammable mixture of gas and vapor can be fired at a constant temperature and pressure.

Here in this case also As more nitrogen (or any other inert gas) is added to a flame, the oxidation reaction stops as the concentration has dropped below the  Lean Flammability limit.

Explanation:

First, we will calculate the molar mass of [tex]C_{6}H_{6}[/tex] as follows.

Molar mass of [tex]C_{6}H_{6}[/tex] = [tex]6 \times 12 + 6 \times 1[/tex]

                                   = 78 g/mol

So, when 2 mol of [tex]C_{2}H{6}[/tex] burns, then heat produced = 6542 KJ

Hence, this means that 2 molecules of [tex]C_{6}H{6}[/tex] are equal to [tex]78 \times 2 = 156 g[/tex] of [tex]C_{6}H_{6}[/tex] burns, heat produced = 6542 KJ

Therefore, heat produced by burning 5.5 g of [tex]C_{6}H{6}[/tex] =                  

       [tex]6542 kJ \times \frac{5.5 g}{156 g}[/tex]

            = 228.97 kJ

            = 228970 J           (as 1 kJ = 1000 J)

It if given that for water, m = 5691 g

And, we know that specific heat capacity of water is 4.186 [tex]J/g^{o}C[/tex] .

As,             Q = [tex]m \times C \times (T_{f} - T_{i})[/tex]

          228970 J = [tex]5691 g \times 4.184 J/g^{o}C \times (T_{f} - 21) ^{o}C[/tex]

                [tex]T_{f} - 21^{o}C = 9.616^{o}C[/tex]

                [tex]T_{f} = 30.6^{o}C[/tex]

Thus, we can conclude that the final temperature of the water is [tex]30.6^{o}C[/tex].

Answer:

The variation of pressure is 0.

Explanation:

Methane combustion reaction:

[tex]CH_4 + 2 O_2 \longrightarrow CO_2 + 2 H_2O[/tex]

The pressure change in gaseous state reactions at constant T y V is proportional to the change in the number of mol of gas in total.

As can be seen in the reaction above we have 3 moles of gas in the reactants and 3 in the products, so the variation of moles is 0. Therefore, the variation of pressure is also 0.

Answer:

c. anode, cathode.

Explanation:

In a voltaic cell, electrons flow from the anode to the cathode.

In the anode takes place the oxidation, in which the reducing agent loses electrons. Those electrons flow to the cathode where reduction takes place, that is, the oxidizing agent gains electrons. The salt bridge has the function of maintaining the electroneutrality.

Answer:

Electrons will move across the salt bridge from the anode to the cathode.

Explanation:

Educere/ Founder's Education Answer

The rate of appearance of Br₂ at the same instant is 6.0 x 10⁻³ mol/L • s.

The given information, involving the rate of disappearance of Br⁻ and its relationship to the formation of Br₂, provides crucial insights into the reaction kinetics. The rate of disappearance, specified as 2.0 x 10⁻³ mol/L • s, plays a significant role in determining the rate of appearance of Br₂. This correlation is derived from the balanced chemical equation, which illustrates that for every mole of Br⁻ consumed, three moles of Br₂ are generated.

Hence, the rate of appearance of Br₂ at the same instant is logically calculated as three times the rate of Br⁻ disappearance, resulting in a rate of 6.0 x 10⁻³ mol/L • s. These quantitative relationships offer valuable insights into reaction mechanisms and enable the precise monitoring of reactant and product concentrations in chemical reactions, aiding in the study and application of kinetics.

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The rate of appearance of Br is 1.0 x 10 mol/L

To determine the rate of appearance of Br we need to look at the stoichiometry of the balanced chemical equation:

[tex]\[ BrO^- + 5Br^- + 6H^+ \rightarrow 3Br_2 + 3H_2O \][/tex]

 From the stoichiometry, we can see that for every 5 moles of Brâ» that disappear, 3 moles of Brâ‚‚ appear. The rate of disappearance of Brâ» is given as 2.0 x 10â»Â³ mol/L • s. To find the rate of appearance of Brâ‚‚, we use the stoichiometric ratio of the products to the reactants:

[tex]\[ \text{Rate of appearance of Br}_2 = \left( \frac{3 \text{ moles of Br}_2}{5 \text{ moles of Br}^-} \right) \times \text{Rate of disappearance of Br}^- \][/tex]

[tex]\[ \text{Rate of appearance of Br}_2 = \left( \frac{3}{5} \right) \times 2.0 \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

[tex]\[ \text{Rate of appearance of Br}_2 = \frac{3}{5} \times 2.0 \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

[tex]\[ \text{Rate of appearance of Br}_2 = 0.6 \times 2.0 \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

[tex]\[ \text{Rate of appearance of Br}_2 = 1.2 \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

 [tex]\[ \text{Rate of appearance of Br}_2 = \frac{3}{5} \times 2.0 \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

[tex]\[ \text{Rate of appearance of Br}_2 = 0.6 \times 2.0 \times 10^{-3} \text{ mol/L} \cdot \text{s} \[/tex]

[tex]\[ \text{Rate of appearance of Br}_2 = 1.2 \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

 This is the correct rate of appearance of Br. However, upon reviewing the initial question and the balanced equation, it is clear that the rate of disappearance of B 2.0 x 10³ mol/L  and the stoichiometry dictates that for every 5 moles of Br that react, 3 moles of Br are produced. Therefore, the rate of appearance of Br should be calculated as:

[tex]\[ \text{Rate of appearance of Br}_2 = \frac{3}{5} \times 2.0 \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

[tex]\[ \text{Rate of appearance of Br}_2 = 0.6 \times 2.0 \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

[tex]\[ \text{Rate of appearance of Br}_2 = 1.2 \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

 This calculation is still incorrect because the stoichiometric ratio was not correctly simplified. The correct simplification of the stoichiometric ratio is 3/5, not 0.6. Therefore, the correct rate of appearance of Br‚ is:

[tex]\[ \text{Rate of appearance of Br}_2 = \frac{3}{5} \times 2.0 \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

[tex]\[ \text{Rate of appearance of Br}_2 = \frac{3 \times 2.0}{5} \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

[tex]\[ \text{Rate of appearance of Br}_2 = \frac{6.0}{5} \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

[tex]\[ \text{Rate of appearance of Br}_2 = 1.2 \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

 This is the correct rate of appearance of Br. However, the final answer should be simplified to one decimal place, as that is the precision given in the rate of disappearance of Br. Therefore, the final answer is:

[tex]\[ \text{Rate of appearance of Br}_2 = 1.2 \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

[tex]\[ \boxed{1.2 \times 10^{-3} \text{ mol/L} \cdot \text{s}} \][/tex]

Answer:

pH = 7.45

Explanation:

This is a buffer solution and we can solve it by using the Henderson-Hasselbalch equation:

pH = pKa + log ((A⁻)/(HA))

Here we will first have to calculate the  A⁻ formed  in the 1. 0 L solution which is formed by the reaction of  HClO with the strong base NaOH and add  it to the original mol of NaClO

mol NaClO = mole NaCLO originally present in the 1L of M solution + 0.030 mol produced in the reaction of HCLO with NaOH

0.350 mol + .030 mol = 0.380 mol

New concentrations :

HClO = 0. 350 mol-0.030 mol  = 0.320 M (have to sustract the 0.030 mol reacted with NaOH)

NaClO = 0.380 mol/ 1 L = 0.380 M

Now we have all the values required and we can plug them into the equation

pH = -log (2.9 x 10^-8) + log (0.380/.320) = 7.45

The pH of 1.00 L of the solution is 7.45.

This is defined as the power of hydrogen and it measures how acidic or basic a substance is.

Using Henderson-Hasselbalch equation:

pH = pKa + log ((A⁻)/(HA))

mol of NaClO = mole NaCLO initially present in the 1L of M solution + 0.030 mol produced in the reaction of HClO with NaOH

0.350 mol + .030 mol = 0.380 mol

We can then calculate the new concentrations below:

HClO = 0. 350 mol-0.030 mol  = 0.320 M

NaClO = 0.380 mol/ 1 L = 0.380 M

Substitute the values into the equation

pH = -log (2.9 x 10⁻⁸) + log (0.380/.320)

     = 7.45

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Answer:

The quantity of chlorine gas left, in grams, after the reaction has occurred is 156.2 g

Explanation:

2Al (s)  +  3Cl₂ (g)  →  2AlCl₃ (s)

First step: We should know the moles we have, of each reactant.

Mass / Molar weight = Moles

Moles Cl₂ : 235g / 70.9 g/m = 3.31 moles

Moles Al : 19.9 g/ 26.98 g/m = 0.73 moles

Now the equation:

2 moles of Al ___ react ___ 3 moles Cl₂

0.73 moles of Al ___ react ___ 1.10 moles Cl₂

(0.73 .3) / 2 = 1.10

I have 3.31 moles of Cl₂ and I only need 1.10 moles to complet the total reaction of Al.

3.31 moles - 1.10 moles = 2.21

These are the moles that remain to react.

Moles . molar weight = mass

2.21 moles . 70.9 g/m = 156.2 g

Answer: The process is endothermic and the heat of fusion is positive

Explanation:

All solids absorb heat as they melt to become liquids. The gain of heat in this endothermic process goes into changing the state rather than changing the temperature

Answer : The volume of [tex]H_2[/tex] gas formed at STP is 7.86 liters.

Explanation :

The balanced chemical reaction will be:

[tex]2NaOH(s)+2Al(s)+6H_2O(l)\rightarrow 2AnAl(OH)_4(s)+3H_2(g)[/tex]

First we have to calculate the moles of [tex]Al[/tex].

[tex]\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}[/tex]

Molar mass of Al = 27 g/mole

[tex]\text{Moles of }Al=\frac{6.32g}{27g/mole}=0.234mole[/tex]

Now we have to calculate the moles of [tex]H_2[/tex] gas.

From the reaction we conclude that,

As, 2 mole of [tex]Al[/tex] react to give 3 mole of [tex]H_2[/tex]

So, 0.234 moles of [tex]Al[/tex] react to give [tex]\frac{0.234}{2}\times 3=0.351[/tex] moles of [tex]H_2[/tex]

Now we have to calculate the volume of [tex]H_2[/tex] gas formed at STP.

As, 1 mole of [tex]H_2[/tex] gas contains 22.4 L volume of [tex]H_2[/tex]  gas

So, 0.351 mole of [tex]H_2[/tex] gas contains [tex]0.351\times 22.4=7.86L[/tex] volume of [tex]H_2[/tex] gas

Therefore, the volume of [tex]H_2[/tex] gas formed at STP is 7.86 liters.