The mole fraction of nitrogen in the gas mixture can be calculated using the ideal gas law and the given pressure, volume, and temperature. The calculation suggests a mole fraction of 1.509, which might indicate that we've made an erroneous assumption about the mixture's composition.
Explanation:First, in order to find the mole fraction of nitrogen in the mixture, we should first find the number of moles of the total mixture. To do this, we can use the ideal gas law, which states that the volume, pressure, and temperature of a gas can be related to the number of moles of the gas. This law can be written as: PV=nRT.
To find the pressure in the correct units, we convert mmHg to atm by dividing by 760 (since 1 atm = 760 mmHg). The pressure is therefore 870.2 / 760 = 1.145 atm. Similarly, we should convert the temperature to Kelvin by adding 273 to the Celsius temperature: 31.2 + 273 = 304.2 K.
So, our equation becomes: 1.145 atm * 15.1 L = n * 0.0821 atm*L/mol/K * 304.2 K. Solving for n, we find that the total number of moles in the mixture is n = 0.570 moles.
Given that molecular nitrogen (N₂) has a molar mass of 28.01 g/mol, we can calculate the number of moles of nitrogen in the 24.1-g mixture. The mass of nitrogen divided by its molar mass gives the number of moles, which is 24.1 g / 28.01 g/mol = 0.860 moles. Therefore, the mole fraction of nitrogen (X_N2) in the mixture can be calculated as the number of moles of nitrogen divided by the total number of moles, or 0.860 / 0.570 = 1.509.
Therefore, the mole fraction of nitrogen in the mixture is 1.509. This value may seem unusual as mole fractions are typically less than 1, but this discrepancy could be due to the initial assumption that the entire 24.1 g consisted of nitrogen. In a real situation, you would need to know the mass of each component in the mixture to accurately calculate the mole fractions.
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The mole fraction of nitrogen in the gas mixture is calculated to be approximately 43.8 percent using the Ideal Gas Law and the given molar masses of nitrogen and carbon dioxide.
Explanation:To find the mole fraction of nitrogen in this mixture, we first need to use the Ideal Gas Law (PV=nRT) to find the total moles (n) of the gases in the mixture. Given the volume (V) of the gas mixture is 15.1 L, the pressure (P) is 870.2 mmHg or 1.145 atm, and the temperature (T) is 31.2oC or 304.35 K, with R being the ideal gas constant which equals 0.0821 L·atm/K·mol, the total number of moles is found to be 0.573 mol.
Then, we need to know that the molar masses of nitrogen (N₂) and carbon dioxide (CO₂) are 28.01 g/mol and 44.01 g/mol respectively. So, we can set up the equation x(28.01) + (1-x)(44.01) = 24.1 (where x is the mole fraction of nitrogen in the mixture) to solve for x. We find that x equals 0.438.
So, the mole fraction of nitrogen in this mixture is approximately 0.438 or 43.8 percent.
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Which statement is TRUE in describing what occurs when a solid melts to a liquid? The process is endothermic and the heat of fusion is negative. The process is endothermic and the heat of fusion is positive. The process is exothermic and the heat of fusion is negative. The process is exothermic and the heat of fusion is positive. not enough information
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
The balanced equation for the reaction of bromate ion with bromide in acidic solution is [tex]BrO^- + 5Br^- + 6H^+ \rightarrow 3Br_2 + 3H_2O[/tex]At a particular instant in time, the rate of disappearance of Br– is 2.0 x 10⁻³ mol/L • s. What is the rate of appearance of Br₂ at the same instant?
The rate of appearance of Br₂ at the same instant is 6.0 x 10⁻³ mol/L • s.
Explanation: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]
As more nitrogen (or any other inert gas) is added to a flame, the flame temperature drops and the oxidation reactions cannot proceed fast enough to keep going and sustain the flame. This point is known as the:
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.
In a voltaic cell, electrons flow from the ________ to the ________. In a voltaic cell, electrons flow from the ________ to the ________.
a. anode, salt bridge.
b. salt bridge, cathode.
c. anode, cathode.
d. salt bride, anode.
e. cathode, anode
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
Predict the sign of the entropy change of the system for each of the following reactions. (a) N2(g)+3H2(g)→2NH3(g) (b) CaCO3(s)→CaO(s)+CO2(g) (c) 3C2H2(g)→C6H6(g) (d) Al2O3(s)+3H2(g)→2Al(s)+3H2O(g) socratic.org
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.
Explanation:Predict the sign of the entropy change of the system for each of the following reactions:
(a) N2(g) + 3H2(g) → 2NH3(g): The entropy change would be negative because the number of gas molecules is decreasing from 4 to 2.(b) CaCO3(s) → CaO(s) + CO2(g): The entropy change is positive since a solid reactant is producing a solid and a gas, increasing the number of molecules and the disorder.(c) 3C2H2(g) → C6H6(g): The entropy change would likely be negative because six molecules of gas are producing one molecule of gas, thus reducing the disorder.(d) Al2O3(s) + 3H2(g) → 2Al(s) + 3H2O(g): Though solids are forming, the entropy change is unclear without additional information such as the specific conditions of the reaction. Generally, producing water vapor from hydrogen gas could increase the entropy, but the transition from a gas to solid aluminum may decrease it.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.
Consider the following information. The lattice energy of CsCl is Δ H lattice = − 657 kJ/mol. The enthalpy of sublimation of Cs is Δ H sub = 76.5 kJ/mol. The first ionization energy of Cs is IE 1 = 376 kJ/mol. The electron affinity of Cl is Δ H EA = − 349 kJ/mol. The bond energy of Cl 2 is BE = 243 kJ/mol. Determine the enthalpy of formation, Δ H f , for CsCl ( s ) .
The enthalpy of formation, ΔHf, for CsCl, can be determined using the Born-Haber cycle by summing up the energy changes in various steps including the sublimation of Cs, ionization of Cs, dissociation of Cl2, and the formation of CsCl. The lattice energy of CsCl is an exothermic process and is equal to the negative of the enthalpy of formation.
Explanation:The enthalpy of formation, ΔHf, for CsCl(s) can be determined using the Born-Haber cycle. The cycle involves several steps including the sublimation of Cs(s), ionization of Cs(g), dissociation of Cl2(g), and the formation of CsCl(s). The lattice energy of CsCl is an exothermic process and is equal to the negative of the enthalpy of formation. By summing up the energy changes in all the steps, we can calculate the enthalpy of formation for CsCl.
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Final answer:
To find the enthalpy of formation for CsCl, the enthalpy changes from sublimation, ionization, bond dissociation, electron affinity, and lattice energy are combined resulting in an enthalpy of formation of -390.5 kJ/mol.
Explanation:
To determine the enthalpy of formation (ΔHf) of cesium chloride (CsCl), we must use the Born-Haber cycle which involves several energy changes related to the formation of ionic compounds. These are the steps to calculate ΔHf for CsCl(s):
The enthalpy of sublimation of Cs (ΔHsub)
The first ionization energy of Cs (IE1)
The bond energy of Cl2 (BE)
The electron affinity of Cl (ΔHEA)
The lattice energy of CsCl (ΔHlattice)
To calculate the enthalpy of formation, first, we need to break the Cl2 bond energy into two Cl atoms which takes 1/2 of the bond energy (1/2 x BE) for one mole of Cl atoms. The total enthalpy change of formation is:
ΔHf = ΔHsub + IE1 + (1/2 x BE) - ΔHEA + ΔHlattice
Substituting the given values:
ΔHf = 76.5 + 376 + (1/2 x 243) - (-349) - 657
=76.5 + 376 +121.5 + 349 - 657
= -657 + 266.5 kJ/mol
= -390.5 kJ/mol
Use bond energies to calculate the enthalpy change for the following reaction. H2(g) + CO2(g) ⟶ H2O(g) + CO(g)
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.
Explanation: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:
Bonds broken: H-H in H2 and O=O in CO2Bonds formed: O-H in H2O and C=O in COThe 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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A piston has an external pressure of 8.00 atm. How much work has been done in joules if the cylinder goes from a volume of 0.130 liters to 0.600 liters? Express your answer with the appropriate units.
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.
Explanation: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.
Explanation: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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Determine the enthalpy of neutralization in Joules/mmol for a solution resulting from 19 mL of 1.4 M NaOH solution and 19 mL of a HCl with the same molarity. If separately, each had a temperature of 27.3 degrees Celsius, and upon addition, the highest temperature reached by the solution was graphically determined to be 38 degrees Celsius. Round to the nearest whole number.
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.
Solution stoichiometry allows chemists to determine the volume of reactants or products involved in a chemical reaction when the chemical substances are dissolved in water. Instead of using mass to calculate the resulting amount of reactants or products, molarity and volume will be used to calculate the moles of substances. A 20.00 mL sample of a 0.250 M solution of HCl reacts with excess Ba(OH)2. What mass of H2O is produced in the reaction?
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.
A 25.00 mL aliquot of concentrated hydrochloric acid (11.7M) is added to 175.00 mL of 3.25M hydrochloric acid. Determine the number of moles of hydrochloric acid from 175.00 mL of 3.25 M hydrochloric acid solution.
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.
Without any force to change it, an object at rest stays at rest and an object in motion stays in motion. This demonstrates which of Newton's Laws?
Newton's First Law
Newton's Second Law
Newton's Third Law
Answer
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.
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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.
What is Newton's First Law?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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Calculate the mass of oxygen (in mg) dissolved in a 5.00 L bucket of water exposed to a pressure of 1.13 atm of air. Assume the mole fraction of oxygen in air to be 0.21 given that kH for O2 is 1.3 × 10-3 M/ atm at this temperature.
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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a compound is found to contain 31.1% sulfur and 68.9% chlorine determine the empirical formula for the sulfur and chlorine sample
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
The action of some commercial drain cleaners is based on the following reaction: 2 NaOH(s) + 2 Al(s) + 6 H2O(l) → 2 NaAl(OH)4(s) + 3 H2(g) What is the volume of H2 gas formed at STP when 6.32 g of Al reacts with excess NaOH?
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.
Lighters are usually fueled by butane (C4H10). When 1 mole of butane burns at constant pressure, it produces 2658 kJ of heat and does 3 kJ of work.
Part A
What are the values of ΔH and ΔE for the combustion of one mole of butane?
Part B
Express your answer using four significant figures.
The values of ΔH and ΔE for the combustion of one mole of butane are 2658 kJ and 2655 kJ, respectively.
Explanation: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.
Explanation: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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According to the following thermochemical equation, what mass of H2O (in g) must form in order to produce 3900 kJ of energy?
SiO2(s) + 4 HF(g) → SiF4(g) + 2 H2O(l) ΔH°rxn = -184 kJ SiO2(s) + 4 HF(g) → SiF4(g) + 2 H2O(l) ΔH°rxn = -184 kJ 216 g 382 g 763 g 408 g 272 g
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]
A standardized solution that is 0.0500 0.0500 M in Na + Na+ is necessary for a flame photometric determination of the element. How many grams of primary-standard-grade sodium carbonate are necessary to prepare 800.0 800.0 mL of this solution?
Answer:
2.12 grams of primary-standard-grade sodium carbonate are necessary to prepare 800.0 mL of this solution.
Explanation:
Molarity of sodium ions = [tex][Na^+][/tex] = 0.0500 M
Moles of sodium ions = n
Volume of the solution = V = 800.0 mL = 0.800 L
[tex]Molarity=\frac{n}{V(L)}[/tex]
[tex][Na^+]=\frac{n}{V}[/tex]
[tex]n=[Na^+]\times V=0.0500 M\times 0.800 L=0.04 mol[/tex]
[tex]Na_2SO_4(aq)\rightarrow 2Na^+(aq)+CO_3^{2-}(aq)[/tex]
1 mole sodium carbonates gives 2 moles of sodium ion and 1 mole of carbonate ions.
Then 0.04 moles of sodium ions will be obtained from:
[tex]\frac{1}{2}\times 0.04 mol=0.02 mol[/tex] of sodium m carbonation.
Mass of 0.02 moles of sodium carbonate = 0.02 mol × 106 g/mol= 2.12 g
2.12 grams of primary-standard-grade sodium carbonate are necessary to prepare 800.0 mL of this solution.
An industrial chemist studying bleaching and sterilizing prepares a hypochlorite buffer using 0.350 M HClO and 0.350 M NaClO. (Ka for HClO = 2.9 × 10−8) Find the pH of 1.00 L of the solution after 0.030 mol of NaOH has been added.
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.
What is pH?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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19.9 g of aluminum and 235 g of chlorine gas react until all of the aluminum metal has been converted to AlCl₃. The balanced equation for the reaction is the following. [tex]2 Al(s) + 3Cl_2(g) \rightarrow 2AlCl_3(s)[/tex]What is the quantity of chlorine gas left, in grams, after the reaction has occurred, assuming that the reaction goes to completion? (The formula mass of aluminum metal, Al, is 26.98 g/mol, and the formula mass of chlorine gas, Cl₂, is 70.90 g/mol.)
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
Given the following data:
P4(s) + 6 Cl2(g) → 4 PCl3(g) ΔH = −1225.6 kJ
P4(s) + 5 O2(g) → P4O10(s) ΔH = −2967.3 kJ
PCl3(g) + Cl2(g) → PCl5(g) ΔH = −84.2 kJ
PCl3(g) + 1/2 O2(g) → Cl3PO(g) ΔH = −285.7 kJ
Calculate ΔH for the reaction P4O10(s) + 6 PCl5(g) → 10 Cl3PO(g).
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.
What is enthalpy change of a reaction?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:
P₄(s) + 6Cl₂(g) → 4 PCl₃(g) ΔH = -1225.6 kJP₄(s) + 5O₂(g) → P₄O₁₀(s) ΔH = -2967.3 kJ (reversing)PCl₃(g) + Cl₂(g) → PCl₅(g) ΔH = -84.2 kJ (reversed and multiplied by 6)PCl₃(g) + 1/2O₂(g) → Cl₃PO(g) ΔH = -285.7 kJ (multiplied by 10)The reactions and enthalpy changes become:
P₄(s) + 6Cl₂(g) → 4 PCl₃(g) ΔH = -1225.6 kJP₄O₁₀(s) → P₄(s) + 5 O₂(g) ΔH = +2967.3 kJ6 PCl₅(g) → 6 PCl₃(g) + 6 Cl₂(g) ΔH = +505.2 kJ10 PCl₃(g) + 5 O₂(g) → 10 Cl₃PO(g) ΔH = -2857.0 kJSumming 1, 2, 3 and 4 gives:
P₄O₁₀(s) + 6 PCl₅(g) → 10Cl₃PO(g)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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Calculate the standard free energy change as a pair of electrons is transferred from succinate to molecular oxygen in the mitochondrial respiratory chain. Oxidant Reductant n E∘′(V) Fumarate 2H 2e− ⇌ Succinate 2 0.03 12O₂ 2H 2e− ⇌ H2O 2 0.82 Express your answer to two significant figures and include the appropriate units.
Answer:
[tex]\DeltaG=158235.4 J[/tex]
Explanation:
The free-energy equation for a redox reaction is:
[tex]\DeltaG=-n*F*E[/tex]
Where:
n=is the number of electrons involved in the reactionF= is the constant of Faraday E=redox potentialIn 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]
A formula that shows the arrangement of all bonds in a molecule is called a(n) ________.
a. molecular formula expanded
b. structural formula condensed
c. structural formula condensed.
d. molecular formula .
e. isomeric formula
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.
Earth’s oceans have an average depth of 3800 m, a total surface area of 3.63×108km2, and an average concentration of dissolved gold of 5.8×10−9g/L.
Assuming the price of gold is $1595/troy oz, what is the value of gold in the oceans
(1 troy oz 5 31.1 g; d of gold 5 19.3 g/cm3)?
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.
The balanced combustion reaction for C 6 H 6 C6H6 is 2 C 6 H 6 ( l ) + 15 O 2 ( g ) ⟶ 12 CO 2 ( g ) + 6 H 2 O ( l ) + 6542 kJ 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ If 5.500 g C 6 H 6 5.500 g C6H6 is burned and the heat produced from the burning is added to 5691 g 5691 g of water at 21 ∘ 21 ∘ C, what is the final temperature of the water?
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].
Can one atom in this type of reaction win the "tug of war"? What might happen if it did?
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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A rigid stainless steel chamber contains 140 Torr of methane, CH4, and excess oxygen, O2, at 160.0 °C. A spark is ignited inside the chamber, completely combusting the methane. What is the change in total pressure within the chamber following the reaction? Assume a constant temperature throughout the process.
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.
Equilibrium is established between a liquid and its vapor when A. the rate of evaporation equals the rate of condensation. B. equal masses exist in the liquid and gas phases. C. equal concentrations (in molarity) exist in the liquid and gas phases. D. all the liquid has evaporated. E. the liquid ceases to evaporate and the gas ceases to condense.
Answer:
A
Explanation:
the rate of evaporation equals the rate of condensation.
What gas is produced when calcium metal is dropped in water
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:
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.
ΔE° for the reaction can be calculated using the formula ΔE° = ΔH° - (Δn)RT, where Δn is the change in moles of gas in the reaction.In this case, ΔH° = -566.0 kJ, Δn = 2 - (2+1) = -1, R = 0.0821 L∙atm/(mol∙K), and T = 298 K. Plugging in these values gives ΔE° = -566.0 kJ - (-1)(0.0821 L∙atm/(mol∙K))(298 K) = -564.2 kJ.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