The standard Gibbs free energy change (ΔG°) for the reaction Pb2+(aq) + Zn(s) → Pb(s) + Zn2+(aq) is -121.9 kJ/mol and for the reaction Br2(l) + 2Cl-(aq) → 2Br-(aq) + Cl2(g) is -55.9 kJ/mol.
(a) Pb2+(aq) + Zn(s) → Pb(s) + Zn2+(aq)
The reduction half-reaction: Pb2+(aq) + 2e- → Pb(s) E° = -0.13 V
The oxidation half-reaction: Zn(s) → Zn2+(aq) + 2e- E° = -0.76 V
To calculate the standard cell potential (E°cell) for the reaction, we subtract the reduction potential from the oxidation potential:
E°cell = E°red + E°ox = (-0.13 V) - (-0.76 V) = 0.63 V
Using the equation ΔG° = -nFE°, where n is the number of electrons transferred and F is the Faraday constant, we can calculate the standard Gibbs free energy change (ΔG°) for the reaction:
ΔG° = -nFE° = -(2 mol)(96485 C/mol)(0.63 V) = -121871.8 J/mol
Converting the value to kilojoules and rounding to three significant figures:
ΔG° = -121.9 kJ/mol
(b) Br2(l) + 2Cl-(aq) → 2Br-(aq) + Cl2(g)
The reduction half-reaction: 2e- + 2Cl-(aq) → Cl2(g) E° = 1.36 V
The oxidation half-reaction: Br2(l) → 2Br-(aq) + 2e- E° = 1.07 V
E°cell = E°red + E°ox = (1.36 V) - (1.07 V) = 0.29 V
ΔG° = -nFE° = -(2 mol)(96485 C/mol)(0.29 V) = -55871.4 J/mol
Converting the value to kilojoules and rounding to two significant figures:
ΔG° = -55.9 kJ/mol
(c) MnO2(s) + 4H+(aq) + Cu(s) → Mn2+(aq) + 2H2O(l) + Cu2+(aq)
The reduction half-reaction: 2H+(aq) + 2e- → H2(g) E° = 0.00 V
The oxidation half-reaction: MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l) E° = 1.23 V
E°cell = E°red + E°ox = (0.00 V) - (1.23 V) = -1.23 V
ΔG° = -nFE° = -(2 mol)(96485 C/mol)(-1.23 V) = 238188.9 J/mol
Converting the value to kilojoules and rounding to three significant figures:
ΔG° = 238.2 kJ/mol
(a) The standard Gibbs free energy change (ΔG°) for the reaction Pb2+(aq) + Zn(s) → Pb(s) + Zn2+(aq) is -121.9 kJ/mol.
(b) The standard Gibbs free energy change (ΔG°) for the reaction Br2(l) + 2Cl-(aq) → 2Br-(aq) + Cl2(g) is -55.9 kJ/mol.
(c) The standard Gibbs free energy change (ΔG°) for the reaction MnO2(s) + 4H+(aq) + Cu(s) → Mn2+(aq) + 2H2O(l) + Cu2+(aq) is 238.2 kJ/mol.
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Write a balanced equation for the reaction between chromium metal and hydrochloric acid to form chromium(III) chloride
and a reactive gas.
Answer: Cr(s) + HCl(aq) ⇒ CrCl3 (aq) + H2(g)
Explanation:
If you have 2 moles of iron chloride (FeCl3) and dissolve it in water to a final volume of 1L of water, what is the molarity and osmolarity of the solution?
A. Molarity of 0.5 mol/L and osmolarity of 2 osmol/L
B. Molarity of 0.5 mol/L and osmolarity of 8 osmol/L
C. Molarity of 2 mol/L and osmolarity of 2 osmol/L
D. Molarity of 2 mol/L and osmolarity of 8 osmol/L
The molarity of the solution is 2 mol/L, and the osmolarity of the solution is 8 osmol/L (D).
Molarity is defined as the number of moles of solute per liter of solution. In this case, we have 2 moles of iron chloride (FeCl3) dissolved in a final volume of 1 liter of water. Therefore, the molarity of the solution is:
Molarity = moles of solute / volume of solution
Molarity = 2 mol / 1 L
Molarity = 2 mol/L
Osmolarity, on the other hand, takes into account the number of particles in a solution that can contribute to osmotic pressure. Iron chloride dissociates into ions in water, forming Fe3+ and 3 Cl- ions. So, each mole of FeCl3 dissociates into 4 particles (1 Fe3+ ion and 3 Cl- ions).
Since we have 2 moles of FeCl3, the total number of particles in the solution is 2 mol x 4 particles/mol = 8 particles. Osmolarity is calculated by dividing the total number of particles by the volume of the solution:
Osmolarity = total particles / volume of solution
Osmolarity = 8 particles / 1 L
Osmolarity = 8 osmol/L
Therefore, the correct answer is option D: Molarity of 2 mol/L and osmolarity of 8 osmol/L.
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When we celebrate a new year we are really celebrating what?
Which of these would be found in a wetland?
Mrs. Cornell and vultures
Cattails and salamanders
Maple trees and cats
Oak trees and squirrels
Answer:
jwjsjdjdhdvjeiejjrjrje
calculate the solubility of mgco3 in water at 25∘c, given the ksp is 4.0 × 10−5.
The solubility of MgCO3 in water at 25°C, given the Ksp is 4.0 × 10−5 is 2 × 10^-3 M.
The chemical equation of magnesium carbonate is MgCO3.
The solubility of MgCO3 is x.
The Ksp value of MgCO3 is 4.0 × 10^-5.
We have to calculate the solubility of MgCO3 in water at 25°C.
The solubility product of MgCO3 is given as:
Ksp = [Mg2+][CO32-]Ksp = [Mg2+][CO32-] = (x)(x) = x2Ksp = 4.0 × 10^-5=x^2x = √Ksp = √4.0 × 10^-5=2 × 10^-3 M
Therefore, the solubility of MgCO3 in water at 25°C, given the Ksp is 4.0 × 10−5 is 2 × 10^-3 M.
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if 4.10 grams of chromium is heated with 9.30 grams of chlorine what mass of chromium(III) chloride is produced?
Answer:
4.1g of Cr will make (317/104) x 4.1 = 12.5g of CrCl3
a 900 w carbon-dioxide laser emits light with a wavelength of 10μm into a 3.0-mm-diameter laser beam.
The 900 W carbon-dioxide laser emits light with a wavelength of 10 μm into a 3.0 mm diameter laser beam. This means that the laser is producing a high-power output in the infrared region of the electromagnetic spectrum.
The wavelength of 10 μm corresponds to the mid-infrared range, which is commonly used in various applications such as industrial cutting and welding, medical procedures, and scientific research.
The 3.0 mm diameter laser beam represents the spatial profile of the laser output. It indicates that the laser beam has a relatively small cross-sectional area, allowing for focused and concentrated energy delivery.
The combination of a high-power output and a specific wavelength allows the carbon-dioxide laser to efficiently perform tasks that require precision and controlled energy deposition, making it suitable for a wide range of industrial and scientific applications.
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In the following molecules, the primary intermolecular attractive force is
Answer:
Dipole-dipole
Explanation:
2 questions. will give brainliest if i can figure out how
What happens to the electrons of a reducing agent during a reaction?
a.Electrons are gained
b.Electrons are lost
c.Electrons are held
d.Electrons are doubled
When is a substance reduced during an oxidation reduction reaction?
a.After one substance is oxidized
b.Before one substance is oxidized
c.Same time as one substance is being oxidized
d.Only oxidation occurs
1= b. electrons are lost
2= a. after one substance is oxidized
compute the equilibrium constant for the spontaneous reaction between sn2 (aq) and fe(s) . express your answer using two significant figures.
The equilibrium constant for the given reaction is 0.93 (to two significant figures). Hence, the answer is 0.93.
Given, spontaneous reaction between Sn2(aq) and Fe(s) is:Sn2(aq) + Fe(s) → Sn(s) + Fe2+(aq)The standard reduction potentials for Sn2+(aq) and Fe2+(aq) are given as:Sn2+(aq) + 2e- → Sn(s); E°red = -0.14 VFe2+(aq) + 2e- → Fe(s); E°red = -0.44 VThe cell potential (E°cell) for the spontaneous reaction can be calculated using the formula: E°cell = E°reduction at cathode - E°reduction at anodeE°cell = E°red of Fe2+ - E°red of Sn2+E°cell = (-0.44 V) - (-0.14 V)E°cell = -0.3 VThe standard equilibrium constant (K°) can be calculated using the relation: ΔG° = -RTlnK°Here, ΔG° is the change in Gibbs free energy for the reaction, R is the gas constant, and T is the temperature in Kelvin.Assuming T = 298 K, we can use R = 8.314 J/mol*K.ΔG° = -nFE°cellHere, n is the number of electrons transferred in the reaction and F is the Faraday constant (96485 C/mol).n = 2 for the given reaction.ΔG° = -2 * 96485 C/mol * (-0.3 V)ΔG° = 58.32 kJ/molSubstituting the values of R, T, and ΔG° in the equation ΔG° = -RTlnK°, we get:lnK° = -ΔG°/RTlnK° = -(58.32 kJ/mol) / (8.314 J/mol*K * 298 K)lnK° = -0.0737K° = e-lnK°K° = e0.0737K° = 0.93The equilibrium constant for the given reaction is 0.93 (to two significant figures). Hence, the answer is 0.93.
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What is FALSE about nitrogen gas and carbondioxide?
1- Both are compounds
2- Both are gases
3- Elements in both substances are non-metals
4- Both are made up of atoms
Answer: Both are compounds
Explanation:
[tex]\text{N}_{2}[/tex] is an element, not a compound.
Considering the limiting reactant concept, how many moles of C are produced from the reaction of 1.00 mol A and 1.00 mol B?
A(g) + 2 B(g) → 3 C(g)
Please show all work, thank you.
1.50 mol of C are produced from the reaction of 1.00 mol A and 1.00 mol B.
The balanced chemical equation for the given reaction is A(g) + 2 B(g) → 3 C(g). This equation can be used to determine the number of moles of C produced from the reaction of 1.00 mol A and 1.00 mol B. The limiting reactant concept must be considered to solve this problem.Limiting reactant:The limiting reactant is the reactant that is completely consumed in a chemical reaction. In other words, the reactant that runs out first limits the amount of product that can be formed.To determine the limiting reactant, the stoichiometry of the chemical reaction must be used. This involves using the mole ratios from the balanced chemical equation to determine how much product can be formed from each reactant.Let's calculate the number of moles of C that can be produced from 1.00 mol A:A(g) + 2 B(g) → 3 C(g)Molar ratio of A to C = 1 : 3From the balanced equation, 1 mol of A produces 3 mol of C. Therefore, 1.00 mol A will produce 3.00 mol C.Let's calculate the number of moles of C that can be produced from 1.00 mol B:A(g) + 2 B(g) → 3 C(g)Molar ratio of B to C = 2 : 3From the balanced equation, 2 mol of B produces 3 mol of C. Therefore, 1.00 mol B will produce 1.50 mol C.The actual amount of product that can be formed is limited by the smaller of these two values. In this case, 1.50 mol C is the smaller value, so 1.50 mol C is produced from the reaction of 1.00 mol A and 1.00 mol B.Answer:1.50 mol of C are produced from the reaction of 1.00 mol A and 1.00 mol B.
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what were the advantages and disadvantages of each measuring device?
To assess the advantages and disadvantages of different measuring devices, specific devices need to be mentioned. Without specific devices, a comprehensive analysis of their advantages and disadvantages is not possible.
The advantages and disadvantages of measuring devices vary depending on the specific device and its application. Different devices have unique features and limitations that affect their accuracy, precision, ease of use, and suitability for different measurements. For example, consider two commonly used measuring devices: a ruler and a digital caliper. A ruler, while simple and inexpensive, may have limitations in terms of precision and accuracy due to human error in reading the markings. On the other hand, a digital caliper offers higher precision and accuracy, as it provides digital readouts and can measure small distances with greater precision. However, digital calipers may be more expensive and require batteries for operation.
Other factors to consider when evaluating measuring devices include durability, range of measurements, calibration requirements, ease of calibration, portability, and the specific needs of the measurement task at hand. To provide a more detailed analysis, it would be helpful to specify the measuring devices being compared.
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What is the proper disposal of all chemicals in this reaction? Briefly explain any hazards associated with barium nitrate and silver nitrate
Barium nitrate is toxic if ingested or inhaled, while silver nitrate is corrosive and can cause severe skin and eye irritation.
Barium nitrate is toxic if ingested or inhaled, posing risks to the respiratory, cardiovascular, and nervous systems. It can also harm aquatic life if released into water bodies. Silver nitrate is corrosive and can cause severe irritation and burns to the skin and eyes. To properly dispose of these chemicals, consult local waste management authorities for specific guidelines. Follow any recommended neutralization procedures, use suitable containers for storage, and label them clearly with the contents and hazards. Engage a licensed waste disposal company or hazardous waste facility to ensure proper collection and disposal. Adhering to local regulations and best practices is crucial to minimize the environmental and health risks associated with these chemicals.
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how might using water somethimes require changing its original state?
How many atoms are in 0.650 mole of zinc?
NEED TO KNOW ASAP PLEASE
20 POINTS: Why it's important for science fair participants to not only create an informative and attractive presentation, but also to detail their information, ideas, and research in a science report.
Answer:
To show what they know!
Explanation:
If you're not showing how you did your experiment thoroughly, then nobody will understand what you did! For instance, if you spent 2 weeks studying really hard about how music affects different kinds of animals, and only put down that music affects animals and no showing your work. Then no one will know how much effort and work you put into it! If you put all the minor details into the presentation, then more than likely everyone will know where your coming from!
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calculate the ph when 9.0 ml of 0.150 m koh is mixed with 20.0 ml of 0.300 m hbro (ka = 2.5 × 10⁻⁹)
To calculate the pH when 9.0 mL of 0.150 M KOH is mixed with 20.0 mL of 0.300 M HBrO (Ka = 2.5 × 10⁻⁹), we need to consider the reaction between KOH and HBrO.
The reaction between KOH (potassium hydroxide) and HBrO (hypobromous acid) can be written as follows:
KOH + HBrO → KBrO + H₂O
Initially, we have 9.0 mL of 0.150 M KOH and 20.0 mL of 0.300 M HBrO. To calculate the pH, we need to determine the final concentrations of the species involved in the equilibrium. Let's assume that the reaction goes to completion, meaning all the KOH and HBrO react completely.
First, we need to find the moles of KOH and HBrO.
For KOH:
Moles of KOH = volume (L) × concentration (mol/L)
Moles of KOH = 9.0 mL × (1 L / 1000 mL) × 0.150 mol/L
For HBrO:
Moles of HBrO = volume (L) × concentration (mol/L)
Moles of HBrO = 20.0 mL × (1 L / 1000 mL) × 0.300 mol/L
Next, we need to consider the stoichiometry of the reaction. Since KOH and HBrO react in a 1:1 ratio, the moles of KOH and HBrO will be equal in the balanced equation. Let's assume the moles of KOH and HBrO are both "x".
Now, we can set up an equilibrium expression for the reaction between HBrO and water:
Ka = [KBrO][H₂O] / [HBrO]
Using the given Ka value (2.5 × 10⁻⁹), we can substitute the concentrations:
2.5 × 10⁻⁹ = x² / (0.150 - x)
Since x is much smaller compared to 0.150, we can simplify the expression by assuming that (0.150 - x) ≈ 0.150:
2.5 × 10⁻⁹ ≈ x² / 0.150
Rearranging the equation, we find:
x² ≈ 2.5 × 10⁻⁹ × 0.150
x² ≈ 3.75 × 10⁻¹⁰
Taking the square root of both sides, we get:
x ≈ √(3.75 × 10⁻¹⁰)
x ≈ 6.12 × 10⁻⁶
This value represents the concentration of H₃O⁺ ions. Now, we can calculate the pH using the equation:
pH = -log[H₃O⁺]
Plugging in the value for [H₃O⁺], we find:
pH = -log(6.12 × 10⁻⁶)
pH ≈ 5.21
Therefore, when 9.0 mL of 0.150 M KOH is mixed with 20.0 mL of 0.300 M HBrO, the pH of the resulting solution is approximately 5.21
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How is oxide different from a neutral oxygen atom
Answer: The main difference between oxide and oxygen is that oxide is a chemical compound with at least one oxygen atom while oxygen is an element whose atomic number is 8.
Explanation: let me know if it was right or wrong
Consider the following balanced equation:
2KCIO3(s) → 2KCl(s) + 302(g)
How many moles of O2 will be obtained by decomposing
3.50 moles of KCIO3?
0.530 mole
O 3.00 moles
O 2.30 moles
5.25 moles
5.25 moles
General Formulas and Concepts:Chemistry
Atomic Structure
CompoundsMolesAqueous Solutions
States of matterStoichiometry
Analyzing reactions RxNUsing Dimensional AnalysisExplanation:Step 1: Define
[RxN - Balanced] 2KClO₃ (s) → 2KCl (s) + 3O₂ (g)
[Given] 3.50 mol KClO₃
[Solve] mol O₂
Step 2: Identify Conversions
[RxN] 2 mol KClO₃ (s) → 3 mol O₂ (g)
Step 3: Stoichiometry
[DA] Set up conversion: [tex]\displaystyle 3.50 \ mol \ KClO_3(\frac{3 \ mol \ O_2}{2 \ mol \ KClO_3})[/tex][DA} Multiply [Cancel out units]: [tex]\displaystyle 5.25 \ mol \ O_2[/tex]WHAT IT IZ BRAINLY anyways I got a question for y’all do you think someone can forget who they rlly loved ? Can you ever forget who you really loved ? I’m tryna prove sum
Answer:
Its a 50% chance of it happeing
Explanation:
HELP ME! is iodine solution polar, non-polar, or ionic?
Please don't attach links, i will report you
Answer:
Since the iodine-iodine bond is a pure covalent bond, iodine is a non-polar molecule. The electronegativity difference between the two Iodine atoms is zero.
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A 0.330 gram sample of an unknown, monoprotic acid (the molar ratio between acid and base is 1-to-1) was dissolved in 50.0 mL of water and titrated to equivalence point with 22.00 mL of 0.150 M NaOH. Determine the molar mass of the the unknown aci
The molar mass of the unknown monoprotic acid will be approximately 100 g/mol.
To determine the molar mass of unknown monoprotic acid, we will use the concept of stoichiometry.
First, let's calculate the number of moles of NaOH used in the titration:
Moles of NaOH = concentration of NaOH × volume of NaOH used
= 0.150 mol/L × 0.02200 L
= 0.00330 mol
Since the acid and base have a 1-to-1 molar ratio, the number of moles of the unknown acid is also 0.00330 mol.
Next, let's calculate the molar mass of the unknown acid:
Molar mass of the unknown acid = mass of the unknown acid / moles of the unknown acid
We are given the mass of the unknown acid as 0.330 grams and the moles of the unknown acid as 0.00330 mol.
Molar mass = 0.330 g / 0.00330 mol
= 100 g/mol
Therefore, the molar mass is 100 g/mol.
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.Rank these systems in order of decreasing entropy.
Rank from highest to lowest entropy. To rank items as equivalent, overlap them
1 mol of methane gas at 273 K and 40 L 1 mol of helium gas at 273 K and 20 L 1/2 mol of liquid helium at 100K 1 mol of helium gas at 273 K and 40 L 1 mol of hydrogen gas at 273 K and 40 L 1/2 mol of helium gas at 100 K and 20 L 1/2 mol of helium gas at 273 K and 20 L Greatest entropy Least entropy
Entropy is a measure of the disorder of a system. It is the measure of the number of possible arrangements (or microstates) of the system, multiplied by Boltzmann's constant.
A higher number of arrangements results in greater entropy. The system with the greatest entropy is that which has the greatest number of arrangements.The system with the greatest entropy is 1 mol of methane gas at 273 K and 40 L. At this state, the gas is present in a larger volume and hence has more microstates, making it more disordered than any other state. Therefore, the greatest entropy is obtained from this system. Next in the rank is 1 mol of helium gas at 273 K and 40 L. Hydrogen gas, 1 mol at 273 K and 40 L, follows helium gas at 273 K and 40 L in rank. 1 mol of helium gas at 273 K and 20 L ranks lower in the list than hydrogen gas. The next state with the lowest entropy is 1/2 mol of helium gas at 273 K and 20 L, followed by 1/2 mol of liquid helium at 100K. The state with the lowest entropy is 1/2 mol of helium gas at 100 K and 20 L. Therefore, the systems in order of decreasing entropy are:
1. 1 mol of methane gas at 273 K and 40 L
2. 1 mol of helium gas at 273 K and 40 L
3. 1 mol of hydrogen gas at 273 K and 40 L
4. 1 mol of helium gas at 273 K and 20 L
5. 1/2 mol of helium gas at 273 K and 20 L
6. 1/2 mol of liquid helium at 100 K
7. 1/2 mol of helium gas at 100 K and 20 L
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Which of these objects is made of cells?
A. rocks B. snow C. trees D. water
Answer:
I think it's trees
Explanation:
I'm not 100% sure tho
Hydrogen-3 has a half-life of 12.3 years. How many years will it take for 539.3 mg 3^H to decay to 2.1 mg 3^H? time to decay:________________________________________________ years Sodium-24 has a half-life of 14.8 hours. How much of a 563.3 mg sodium-24 sample will remain after 3.7 days? mass remaining:_____________________________________________ mg.
Approximately 4.41 mg of the sodium-24 sample will remain after 3.7 days.
Part A - Hydrogen-3 decay. The half-life of Hydrogen-3 is 12.3 years, and we have to find the time that would take for 539.3 mg 3^H to decay to 2.1 mg 3^H. Let's use the formula given below and find out how many years it will take. Amount = Initial amount * (1/2)^(Time/half life). Initial amount = 539.3 mg. The amount left = 2.1 mg. Time = ?Half-life = 12.3 years. Amount left/Initial amount = (1/2)^(Time/half-life)2.1/539.3 = (1/2)^(Time/12.3)log(2.1/539.3) = log(1/2)^(Time/12.3)(Time/12.3) = log(539.3/2.1) / log2. Time = 12.3 * log(539.3/2.1) / log2Time = 12.3 * 11.3949 / 0.30103. Time = 466.95 years. Therefore, it would take approximately 466.95 years for 539.3 mg 3^H to decay to 2.1 mg 3^H.
Part B - Sodium-24 decay. The half-life of Sodium-24 is 14.8 hours. Let's find out how much of a 563.3 mg sodium-24 sample will remain after 3.7 days. Initial amount = 563.3 mg. Half-life = 14.8 hours. We need to find the amount left after 3.7 days (88.8 hours). Amount = Initial amount * (1/2)^(Time/half-life). Amount = 563.3 * (1/2)^(88.8/14.8)Amount = 563.3 * (1/2)^6Amount = 4.41 mg. Therefore, approximately 4.41 mg of the sodium-24 sample will remain after 3.7 days.
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the formulas of three important acids and three important bases and describe their uses.
Answer:
18929eu9sisjwiejwjxjjejxme
What does the first ionization energy represent?
when we drop a feather and a stone at the same time from a height the stone reach the faster than a feather
Answer:
See explanation
Explanation:
When an a heavy object and a light object are thrown down from the same height, the two objects are expected to hit the ground at the same time since the two objects are both accelerated to the same extent (approximately 10m/s^2) under gravity.
However, due to air resistance, the heavier object reaches the ground first before the lighter object.
Hence, when we drop a feather and a stone at the same time from a height the stone will hit the ground faster than the feather.
Identify which properties are common to each of the following chemical families
(a) alkali metals
(b) alkaline earth metals
(c) halogens
(d) noble gases
The noble gases have a full outer shell of valence electrons, making them stable and unreactive. They are colorless, odorless gases at room temperature and have very low boiling points. Their lack of reactivity makes them useful in a variety of applications, including lighting and welding.
The properties that are common to each of the following chemical families include:
(a) Alkali metals The alkali metals have a single valence electron in their outermost shell, which is easily lost to form an ion with a charge of +1. They are the most reactive metals, reacting with water and air to produce hydrogen gas and an oxide layer, respectively. They are silvery-white and have a soft texture.
(b) Alkaline earth metals The alkaline earth metals have two valence electrons in their outermost shell, which they readily lose to form ions with a charge of +2. They are less reactive than the alkali metals, but they still react with oxygen to form an oxide layer on their surface. They are also silvery-white in color and have a harder texture than the alkali metals.
(c) Halogens The halogens have seven valence electrons in their outermost shell, making them highly reactive nonmetals. They readily form ions with a charge of -1 by gaining an electron. They are diatomic molecules at room temperature and can be found in a variety of colors and states of matter.
(d) Noble gases The noble gases have a full outer shell of valence electrons, making them stable and unreactive. They are colorless, odorless gases at room temperature and have very low boiling points. Their lack of reactivity makes them useful in a variety of applications, including lighting and welding. These properties are common to each of the following chemical families.
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