Answer:
The molecular formula of the compounds indicates the amount of atoms that the substance contains. For example, water whose formula is H2O contains 2 atoms of hydrogen and 1 of oxygen. The representative unit of a molecular compound is a molecule while ions are represented by a formula unit.
Mole can be a unit 6.02 × 10²³. The units may be electrons, atoms, ions, or molecules, depending on the nature of the substance
Explanation:
The molecular formula of the compounds indicates the amount of atoms that the substance contains. For example, water whose formula is [tex]H2O[/tex] contains 2 atoms of hydrogen and 1 of oxygen. The representative unit of a molecular compound is a molecule while ions are represented by a formula unit.
Mole can be a unit 6.02 × 10²³. The units may be electrons, atoms, ions, or molecules, depending on the nature of the substance.
Therefore, water whose formula is [tex]H2O[/tex] contains 2 atoms of hydrogen and 1 of oxygen. The representative unit of a molecular compound is a molecule while ions are represented by a formula unit.
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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.
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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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.
The thermosphere is: 1. the layer of atmosphere closest to the Earth’s surface where weather occurs. 2. supports long distance communication because it reflects outgoing radio waves back to Earth without the use of satellites. 3. the layer where auroras form when electrically charged particles from the sun collide with gas molecules releasing energy visible as light of different colors
The thermosphere is a layer of Earth's atmosphere located 80-700km above sea level. It supports long-distance communication by reflecting radio waves back to Earth and is also where auroras form due to the collision of charged particles and gas molecules.
Explanation:The thermosphere is the fourth layer of Earth's atmosphere, located above the mesosphere and extend from about 80 to 700 kilometers above sea level. This layer holds a unique property due to the presence of electrically charged particles, or ions which allows it to support long-distance communication by reflecting radio waves back to Earth, bypassing the need for satellites. This is also the layer in which auroras form. Auroras occur when these charged particles from the sun collide with gas molecules in the Earth's atmosphere, releasing energy that manifests as visible, colorful light.
The Earth's atmosphere is divided into five main layers: The troposphere, the stratosphere, the mesosphere, the thermosphere and the exosphere. The Troposphere being the one closest to the Earth's surface and is where weather is generally observed. This layer expands to a height of roughly 12 km above the sea level and makes up around 80% of the atmosphere's mass. The Thermosphere is not the layer of atmosphere closest to the Earth’s surface where weather typically occurs, this property represents the troposphere.
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The third option is correct.
3. The thermosphere is the layer where auroras form when electrically charged particles from the sun collide with gas molecules releasing energy visible as light of different colors.
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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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.
Consider the following equilibrium:
O2(g)+ 2F2(g) ↔ 2OF2(g); Kp = 2.3×10-15
Which of the following statements is true?
a. If the reaction mixture initially contains only OF2(g), then at equilibrium, the reaction mixture will consist of essentially only O2(g) and F2(g).
b. For this equilibrium, Kc=Kp.
c. If the reaction mixture initially contains only OF2(g), then the total pressure at equilibrium will be less than the total initial pressure.
d. If the reaciton mixture initially contains only O2(g) and F2(g), then at equilibrium, the reaction mixture will consist of essentially only OF2(g).
e. If the reaction mixture initially contains only O2(g) and F2(g), then the total pressure at equilibrium will be greater than the total initail pressure.
Answer:
a. If the reaction mixture initially contains only OF2(g), then at equilibrium, the reaction mixture will consist of essentially only O2(g) and F2(g).
Explanation:
The answer is a) because the value for Kp is really close to zero (having x10⁻¹⁵), this means that at equilibrium O₂ and F₂ will be significantly more present than OF₂.
Final answer:
For the equilibrium O2(g) + 2F2(g) ↔ 2OF2(g) with Kp = 2.3×10^-15, the reaction favors the reactants, making the correct answer that a mixture initially containing only OF2(g) will consist of essentially only O2(g) and F2(g) at equilibrium.
Explanation:
The question considers the equilibrium O2(g) + 2F2(g) ↔ 2OF2(g); Kp = 2.3×10-15 and asks which statement is true. Given the extremely low value of Kp, this indicates a strong preference for reactants at equilibrium. Therefore, the correct answer is: If the reaction mixture initially contains only OF2(g), then at equilibrium, the reaction mixture will consist of essentially only O2(g) and F2(g). This is because a very small Kp value means the equilibrium lies heavily on the side of the reactants, making statement (a) true.
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
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.
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.
There are some data that suggest that zinc lozenges can significantly shorten the duration of a cold. If the solubility of zinc acetate, Zn(CH3COO)2, is 43.0 g/L, what is the solubility product Ksp of this compound?
Answer:
[tex]K_{sp}[/tex] of [tex]Zn(CH_{3}COO)_{2}[/tex] is 0.0513
Explanation:
Solubility equilibrium of [tex]Zn(CH_{3}COO)_{2}[/tex]:
[tex]Zn(CH_{3}COO)_{2}\rightleftharpoons Zn^{2+}+2CH_{3}COO^{-}[/tex]
Solubility product of [tex]Zn(CH_{3}COO)_{2}[/tex] ([tex]K_{sp}[/tex]) is written as- [tex]K_{sp}=[Zn^{2+}][CH_{3}COO^{-}]^{2}[/tex]
Where [tex][Zn^{2+}][/tex] and [tex][CH_{3}COO^{-}][/tex] represents equilibrium concentration (in molarity) of [tex]Zn^{2+}[/tex] and [tex]CH_{3}COO^{-}[/tex] respectively.
Molar mass of [tex]Zn(CH_{3}COO)_{2}[/tex] = 183.48 g/mol
So, solubility of [tex]Zn(CH_{3}COO)_{2}[/tex] = [tex]\frac{43.0}{183.48}M[/tex] = 0.234M
1 mol of [tex]Zn(CH_{3}COO)_{2}[/tex] gives 1 mol of [tex]Zn^{2+}[/tex] and 2 moles of [tex]CH_{3}COO^{-}[/tex] upon dissociation.
so, [tex][Zn^{2+}][/tex] = 0.234 M and [tex][CH_{3}COO^{-}][/tex] = [tex](2\times 0.234)M=0.468M[/tex]
so, [tex]K_{sp}=(0.234)\times (0.468)^{2}=0.0513[/tex]
Based on the data provided, the Ksp of zinc acetate is 0.051 M^{2}.
What is the solubility product of Zinc acetate?The solubility product, Ksp, of zinc acetate is derubed from the equation for the dissolution of Zinc acetate given below:
Zn(CH_{3}COO)_{2} <------> Zn^{2+} + 2 CH_{3}COO^{-}The solubility product, is given below:
Ksp = [Zn^{2+] × [CH_{3}COO^{-}]^{2}Molar concentration of the zinc acetate = mass concentration/molar mass
Molar mass of Zinc acetate = 183.48 g/molMolar concentration of Zinc acetate = 43.0/183.48
Molar concentration of Zinc acetate = 0.234 M
From the equation of the reaction:
1 mole of Zn(CH_{3}COO)_{2} produces 1 mole Zn^{2+} and 2 CH_{3}COO^{-}
Hence;
[Zn^{2+] = 0.234[CH_{3}COO^{-}]^{2} = 0.234 × 2 = 0.468Ksp = 0.234 × 0.468^{2}
Ksp = 0.051 M^{2}
Therefore, the Ksp of zinc acetate is 0.051 M^{2}.
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A sample of impure tin of mass 0.526 g is dissolved in strong acid to give a solution of Sn2 . The solution is then titrated with a 0.0448 M solution of NO3−, which is reduced to NO(g). The equivalence point is reached upon the addition of 3.67×10−2 L of the NO3− solution.
Find the percent by mass of tin in the original sample, assuming that it contains no other reducing agents.
Answer:
55.7%
Explanation:
The reaction that takes place is:
3Sn²⁺ + 2NO₃⁻ + 8H⁺ → 2NO + 3Sn⁺⁴ + 4H₂OWith the volume and concentration of NO₃⁻ solution, we can calculate the moles of Sn²⁺ that reacted:
3.67x10⁻² L * 0.0448 M = 1.64x10⁻³ mol NO₃⁻1.64x10⁻³ mol NO₃⁻ * [tex]\frac{3molSn^{2+}}{2molNO_{3}^{-}}[/tex] = 2.47x10⁻³mol Sn²⁺Now we convert moles of Sn to mass, using its atomic weight:
2.47x10⁻³mol Sn²⁺ * 118.71 g/mol = 0.293 g SnFinally we calculate the percent by mass of Sn:
0.293 g / 0.526 g * 100% = 55.7%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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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.
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]
Mothballs are comprised primarily of naphthalene (C10H8). When 1.025 g of Naphthalene burns in a bomb calorimeter, the temperature rises from 24.25 degrees Celsius to 32.33 degrees Celsius. Find the change in energy for the combustion of a mole of naphthalene. The heat capacity of the bomb calorimeter is 5.11 kJ/degree Celsius.
Answer:
The change in energy for the combustion of a mole of naphthalene is 79 kJ
Explanation:
An excersise where you have to apply the Heat Capacity formula
C = Q . ΔT
where C is Heat Capacity and Q is Heat
ΔT means (T° final - T° initial)
5.11 kJ/°C = Q (32.33°C - 24.25°C)
5.11 kJ/°C = Q (32.33°C - 24.25°C)
5.11 kJ/°C = Q . 8.08°C
5.11 kJ / 8.08°C = Q
0.632 kJ = Q
This heat is released by 1.025 grams of naphtalene.
Molar mass Naphtalene : 128.17 g/m
1.025 g / 128.17 g/m = 0.008 moles
This are the moles, so we have to divide heat/moles to get the change in energy for one mole.
0.632 kJ/0.008 mol = 79 kJ/m
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.
Arrange the colors of visible light, green, red, and blue, in order of increasing wavelength.
blue < green < red
Shortest wavelength red < green < blue Longest wavelength
green < blue < red
Answer: The increasing wavelength of colors:
Red > Green > Blue
Explanation:
Wavelength: This is the property of wave which includes the distance between two consecutive crests or trough. This is denoted by the Greek letter Lambda and it is found by dividing the velocity of the wave with its frequency.
Wavelength of colours are
Violet: 400 - 420 nm
Indigo: 420 - 440 nm
Blue: 440 - 490 nm
Green: 490 - 570 nm
Yellow: 570 - 585 nm
Orange: 585 - 620 nm
Red: 620 - 780 nm
In visible light, the wavelength increases from violet to red. For the colors mentioned, in order of increasing wavelength, it is blue < green < red and blue < yellow < red for the additional colors provided.
The question asks to arrange the colors of visible light (green, red, and blue) in order of increasing wavelength and then provides a similar task with the colors yellow, blue, and red. For visible light, wavelengths increase from violet through to red. Hence, using the mnemonic ROY G BIV (Red, Orange, Yellow, Green, Blue, Indigo, Violet), we can deduce that blue has a shorter wavelength than green, which in turn has a shorter wavelength than red.
Part A: For the colors yellow, blue, and red, in order of increasing wavelength, it would be: blue < yellow < red.
Part B: Since the frequency of light waves is inversely proportional to their wavelength, the order according to frequency, from lowest to highest, would inversely mirror the wavelengths: red < yellow < blue.
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.
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.
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.
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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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.
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].
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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When iron (III) oxide combines with sulfuric acid H2 S04 the product formed are water and iron (III) sulfate. What is the coefficient needed for water in the balanced equation
Answer:
coefficient 3 is needed t with water molecule to balance the equation.
Explanation:
Chemical equation:
Fe₂O₃ + H₂SO₄ → Fe₂(SO₄)₃ + H₂O
Balanced chemical equation:
Fe₂O₃ + 3H₂SO₄ → Fe₂(SO₄)₃ + 3H₂O
The equation is balanced. There are two iron three sulfate six hydrogen and three oxygen atoms are present on both side of equation.
The given equation completely follow the law of conservation of mass.
According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.
Coefficient with reactant and product:
Fe₂O₃ 1
H₂SO₄ 3
Fe₂(SO₄)₃ 1
H₂O 3
So coefficient 3 is needed t with water molecule to balance the equation.
What gas is produced when calcium metal is dropped in water
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
How many molecules of N2 are in a 400.0 mL container at 780 mm Hg and 135°C? Avogadro’s number = 6.022 x 1023A) 7.01 × 1021 moleculesB) 7.38 × 1021 moleculesC) 2.12 × 1022 moleculesD) 2.23 × 1022 molecules
Answer:
B
Explanation:
Firstly, we will need to calculate the number of moles. To do this, we make use of the ideal gas equation
PV = nRT
n = PV/RT
The parameters have the following values according to the question:
P = 780mmHg, we convert this to pascal.
760mHG = 101325pa
780mmHg = xpa
x = (780 * 101325)/760 = 103,991 Pa
V= 400ml = 0.4L
T = 135C = 135 + 273.15 = 408.15K
n = ?
R = 8314.463LPa/K.mol
Substituting these values into the equation yields the following:
n = (103991 * 0.4)/(8314.463 * 408.15)
= 0.012 moles
Now we know 1 mole contains 6.02 * 10^23 molecules, hence, 0.012moles will contain = 0.012 * 6.02 * 10^23 = 7.38 * 10^21 molecules
Answer: The number of nitrogen molecules in the container are [tex]7.38\times 10^{21}[/tex]
Explanation:
To calculate the moles of gas, we use the equation given by ideal gas which follows:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 780 mmHg
V = Volume of the gas = 400.0 mL = 0.4 L (Conversion factor: 1 L = 1000 mL)
T = Temperature of the gas = [tex]135^oC=[135+273]K=408K[/tex]
R = Gas constant = [tex]62.364\text{ L.mmHg }mol^{-1}K^{-1}[/tex]
n = number of moles of nitrogen gas = ?
Putting values in above equation, we get:
[tex]780mmHg\times 0.4L=n\times 62.364\text{ L. mmHg }mol^{-1}K^{-1}\times 408K\\\\n=\frac{780\times 0.4}{62.364\times 408}=0.01226mol[/tex]
According to mole concept:
1 mole of a compound contains [tex]6.022\times 10^{23}[/tex] number of molecules
So, 0.01226 moles of nitrogen gas will contain = [tex](0.012\times 6.022\times 10^{23})=7.38\times 10^{21}[/tex] number of molecules
Hence, the number of nitrogen molecules in the container are [tex]7.38\times 10^{21}[/tex]
Write a balanced chemical equation between magnesium chloride and sodium phosphate. Determine the grams of magnesium chloride that are needed to produce 1.33 x 10^{23} formula units of magnesium phosphate.
Answer:
The grams of MgCl₂ that are needed are 63.1 g
Explanation:
First of all, try to think the equation and ballance it. This is it:
3MgCl₂ + 2Na₃PO₄ → Mg₃(PO₄)₂ + 6NaCl
Let's convert our f.u in mol
1 mol = 6.02x10²³ formula units (Avogadro's number)
So f.u / Avogadro = mol
1.33x10²³ / 6.02x10²³ = 0.221 moles.
So 1 mol of phosphate sodium comes from 3 mol of magnesium chloride.
How many mol of magnesium chloride are necessary, for 0.221 mol of phosphate.?
0.221 moles .3 = 0.663 moles.
Molar mass of MgCl₂ is 95.2 g/m
0.663 moles . 95.2 g/m = 63.1 g
The balanced chemical equation between magnesium chloride and sodium phosphate is 3 MgCl2 + 2 Na3PO4 -> Mg3(PO4)2 + 6 NaCl. To determine the grams of magnesium chloride needed to produce a specific number of formula units of magnesium phosphate, stoichiometry calculations can be used.
Explanation:The balanced chemical equation between magnesium chloride (MgCl2) and sodium phosphate (Na3PO4) is:
3 MgCl2 + 2 Na3PO4 → Mg3(PO4)2 + 6 NaCl
To determine the grams of magnesium chloride needed to produce 1.33 x 10^23 formula units of magnesium phosphate (Mg3(PO4)2), we need to use stoichiometry.
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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Identify each of the following energy exchanges as primarily heat or work and determine whether the sign of is positive or negative for the system. a. A rolling billiard ball collides with another billiard ball. The first billiard ball (defined as the system) stops rolling after the collision. b. A book is dropped to the floor (the book is the system). c. A father pushes his daughter on a swing (the daughter and the swing are the system
Answer:
A) Work (positive)
B) Heat (negative) and work (positive)
C) Work (negative) and heat (negative)
Explanation:
A) This energy change is work given that the first ball is using this energy to move the other one. The system (fist ball) is doing the work so it has + sign.
B) The book loses part of its energy by friction (heat) and part by applying a force to the floor (work). The heat is taken from the system so it has - sign and the work is done by the system and it has + sign.
C) The daughter is receiving the work done by the father. This work done to the system has - sign. Also, you can say that the system (daughter and swing) loses energy by friction (heat) and because of that it slows down.
Note: the sign of work and heat is defined by convention.