Answer:
Explanation:
This limit is a consequence of Heisenberg´s uncertainty principle:
Δp x Δx > = h
This state that the product of the uncertainty in momentum ( or velocity since p = mv ) times the uncertainty in position, Δx , must be greater or equal to Planck´s constant ( 6.626 x 10⁻³⁴ J·s ).
Later models refined this equation to:
Δp x Δx > = h/4π
This is the consequence of duality wave matter of the electron and Schrodinger´s equation, in which we can talk of probabilities of finding an electron and not confined to specific distances from the nucleus as in the Bohr atom.
Now think of think of this relation in terms of the uncertainty it describes. If we know the position of the electron with great exactitude, the velocity of the particle will be very high since the mass of hte electron is very small.
This a principle in nature and has nothing to do with the precision of our instruments for particles at the subatomic level.
The reason we do not observe this effect with everyday objects is that the obbects have masses so large compare to subatomic particles that the term mΔv becomes large enough, allowing us to know the position and velocity of macroscopic objects with small uncertainties:
Δp x Δx > = h/4π, Δp very large ( because the mass is very big ) then Δx is very small
The same does not have with small masses of the subatomic levels.
A 0.6-m3 rigid tank is filled with saturated liquid water at 135°C. A valve at the bottom of the tank is now opened, and one-half of the total mass is withdrawn from the tank in liquid form. Heat is transferred to water from a source of 210°C so that the temperature in the tank remains constant. Assume the surroundings to be at 25°C and 100 kPa.
The given question is incomplete. The complete question is as follows.
A 0.6-m3 rigid tank is filled with saturated liquid water at 135°C. A valve at the bottom of the tank is now opened, and one-half of the total mass is withdrawn from the tank in liquid form. Heat is transferred to water from a source of 210°C so that the temperature in the tank remains constant. Assume the surroundings to be at 25°C and 100 kPa. Determine the amount of heat transfer.
Explanation:
First, we will determine the initial mass from given volume and specific volume as follows.
[tex]m_{1} = \frac{V}{\alpha_{1}}[/tex]
= [tex]\frac{0.6}{0.001075}kg[/tex]
= 558.14 kg
Hence, the final mass and mass that has left the tank are as follows.
[tex]m_{2} = m_{out} = \frac{1}{2}m_{1}[/tex]
= [tex]\frac{1}{2} \times 558.14 kg[/tex]
= 279.07 kg
Now, the final specific volume is as follows.
[tex]\alpha_{2} = \frac{V}{m_{2}}[/tex]
= [tex]\frac{0.6}{279.07} m^{3}/kg[/tex]
= 0.00215 [tex]m^{3}/kg[/tex]
Final quality of the mixture is determined actually from the total final specific volume and the specific volumes of the constituents for the given temperature are as follows.
[tex]q_{2} = \frac{\alpha_{2} - \alpha_{liq135}}{(\alpha_{vap} - \alpha_{liq})_{135}}[/tex]
= [tex]\frac{0.00215 - 0.001075}{0.58179 - 0.001075}[/tex]
= [tex]1.85 \times 10^{-3}[/tex]
Hence, the final internal energy will be calculated as follows.
[tex]u_{2} = u_{liq135} + q_{2}u_{vap135}[/tex]
= [tex](567.41 + 1.85 \times 10^{-3} \times 1977.3) kJ/kg[/tex]
= 571.06 kJ/kg
Now, we will calculate the heat transfer as follows.
[tex]\Delta U = Q - m_{out}h_{out}[/tex]
[tex]m_{2}u_{2} - m_{1}u_{1} = Q - m_{out}h_{out}[/tex]
Q = [tex](279.07 \times 571.06 - 558.14 \times 567.41 + 279.07 \times 567.75) kJ[/tex]
= 1113.5 kJ
Thus, we can conclude that amount of heat transfer is 1113.5 kJ.
Final answer:
The question involves concepts of thermodynamics, specifically related to heat transfer and thermal equilibrium in a situation with a saturated liquid undergoing withdrawal and heating to maintain a constant temperature, demonstrating the application of energy conservation principles.
Explanation:
The question describes a thermodynamics scenario where heat transfer is involved to maintain the temperature of the system (a tank with water) constant. The water inside the rigid tank undergoes a process where half of its mass is withdrawn while the temperature is kept at 135°C through the addition of heat from a 210°C source. This is an example of applying the concepts of thermodynamics such as energy transfer, the properties of substances under varying conditions, and the concept of a saturated liquid.
When dealing with the thermal equilibrium between two bodies with different initial temperatures, the overall heat lost by the hotter body (pan) is equal to the heat gained by the colder body (water). The final temperature at equilibrium can be calculated using the principle of conservation of energy. Similarly, the expansion of a fluid within a radiator and thermal contraction of coffee in a glass can be described using thermodynamic principles and the relevant coefficients of thermal expansion and specific heat capacities.
Hydrogen chloride gas and oxygen gas react to form water and chlorine gas. A reaction mixture initially contains 53.2 g of hydrogen chloride and 26.5 g of oxygen gas. Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant remains? Enter to 1 decimal place.
In a reaction between hydrogen chloride and oxygen to form water and chlorine gas, using the provided masses and the balanced chemical equation, hydrogen chloride is found to be the limiting reactant, leaving 14.8 g of excess oxygen gas after the reaction.
To solve this problem, we first need to write the balanced chemical equation for the reaction between hydrogen chloride (HCl) and oxygen (O₂) to form water (H₂O) and chlorine gas (Cl₂). The reaction is as follows:
4 HCl(g) + O₂(g) ⇒ 2 H₂O(g) + 2 Cl₂(g)
Using the given masses of reactants, we calculate the molar amounts of HCl and O₂. The molar mass of HCl is approximately 36.5 g/mol and that of O₂ is 32.0 g/mol. Thus:
53.2 g HCl x (1 mol HCl / 36.5 g) = 1.46 mol HCl
26.5 g O₂ x (1 mol O₂ / 32.0 g) = 0.828 mol O₂
According to the balanced equation, it takes 4 moles of HCl to react with 1 mole of O₂. So, we divide the molar amounts by their respective coefficients to find the limiting reactant:
1.46 mol HCl / 4 = 0.365 mol
0.828 mol O₂ / 1 = 0.828 mol
Since 0.365 mol < 0.828 mol, HCl is the limiting reactant. The reaction will consume all of the HCl, leaving some O₂ in excess. To find the amount of excess O₂, we calculate how much O₂ is needed to react with the available HCl:
1.46 mol HCl x (1 mol O₂ / 4 mol HCl) = 0.365 mol O₂ needed
We subtract the O₂ needed from the initial amount to find the excess:
0.828 mol O₂ - 0.365 mol O₂ = 0.463 mol excess O₂
0.463 mol O₂ x 32.0 g/mol = 14.8 g excess O₂
Therefore, the mass of the excess reactant O2 remaining is 14.8 g.
Problem Page Question It takes to break a carbon-carbon single bond. Calculate the maximum wavelength of light for which a carbon-carbon single bond could be broken by absorbing a single photon.
This is a incomplete question. The complete question is:
It takes 348 kJ/mol to break a carbon-carbon single bond. Calculate the maximum wavelength of light for which a carbon-carbon single bond could be broken by absorbing a single photon. Round your answer to correct number of significant digits
Answer: 344 nm
Explanation:
[tex]E=\frac{Nhc}{\lambda}[/tex]
E= energy = 348kJ= 348000 J (1kJ=1000J)
N = avogadro's number = [tex]6.023\times 10^{23}[/tex]
h = Planck's constant = [tex]6.626\times 10^{-34}Js [/tex]
c = speed of light = [tex]3\times 10^8ms^{-1}[/tex]
[tex]348000=\frac{6.023\times 10^{23}\times 6.626\times 10^{-34}\times 3\times 10^8}{\lambda}[/tex]
[tex]\lambda=\frac{6.023\times 10^{23}\times 6.626\times 10^{-34}\times 3\times 10^8}{348000}[/tex]
[tex]\lambda=3.44\times 10^{-7}m=344nm[/tex] [tex]1nm=10^{-9}m[/tex]
Thus the maximum wavelength of light for which a carbon-carbon single bond could be broken by absorbing a single photon is 344 nm
Consider the above unbalanced equation. What volume of CO2 is produced at 270. mm Hg and 38.5°C when 0.820 g of C4H8 reacts with excess O2? Use molar masses with at least as many significant figures.
Answer:
The volume of CO2 is 4.20 L
Explanation:
Step 1: Data given
Pressure = 270 mm Hg = 260 /760 = 0.355263 atm
Temperature : 38.5 °C = 311.65 K
Mass of C4H8 = 0.820 grams
Step 2: The balanced equation
C4H8 + 6O2 → 4CO2 + 4H2O
Step 3: Calculate moles C4H8
Moles C4H8 = mass C4H8 / molar mass C4H8
Moles C4H8 = 0.820 grams / 56.11 g/mol
Moles C4H8 = 0.0146 moles
Step 4: Caalculate moles CO2
For 1 mol C4H8 we need 6 moles O2 to produce 4 moles CO2 and 4 moles H2O
For 0.0146 moles we'll have 4*0.0146 = 0.0584 moles CO2
Step 6: Calculate Volume CO2
p*V = n*R*T
V = (n*R*T) /p
⇒ with V = the volume of CO2 = TO BE DETERMINED
⇒ with n = the moles of CO2 = 0.0584 moles
⇒ with R = the gas constant = 0.08206 L*atm / mol*K
⇒ with T = The temperature = 311.65 K
⇒ with p = the pressure = 0.355263 atm
V = (0.0584 * 0.08206 * 311.65) / 0.355263
V = 4.20 L
The volume of CO2 is 4.20 L
How many inner, outer, and valence electrons are present in an atom of each of the following elements?
(a) Br (b) Cs (c) Cr (d) Sr (e) F
Answer:
For a: The number of inner electrons, outer electrons and valence electrons are 28, 7 and 7 respectively.
For b: The number of inner electrons, outer electrons and valence electrons are 54, 1 and 1 respectively.
For c: The number of inner electrons, outer electrons and valence electrons are 23, 1 and 1 respectively.
For d: The number of inner electrons, outer electrons and valence electrons are 36, 2 and 2 respectively.
For e: The number of inner electrons, outer electrons and valence electrons are 2, 7 and 7 respectively.
Explanation:
Outer shell electrons are the electrons which are not tightly held by the electrons. They are called as valence electrons. The electrons present in the highest principle quantum number are known as valence electrons.
Inner shell electrons are the electrons which are tightly held by the electrons. They are called as core electrons.
Inner electrons = Total number of electrons - Valence electrons
Total number of electrons in an atom is equal to the atomic number of the element.
For the given options:
Option a: Bromine (Br)Bromine is the 35th element of the periodic table having electronic configuration of [tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^5[/tex]
Highest principle quantum number is 4
Number of valence electrons = 7
Number of outer electrons = 7
Number of inner electrons = 35 - 7 = 28
Option b: Cesium (Cs)Cesium is the 55th element of the periodic table having electronic configuration of [tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^1[/tex]
Highest principle quantum number is 6
Number of valence electrons = 1
Number of outer electrons = 1
Number of inner electrons = 55 - 1 = 54
Option c: Chromium (Cr)Chromium is the 24th element of the periodic table having electronic configuration of [tex]1s^22s^22p^63s^23p^64s^13d^{5}[/tex]
Highest principle quantum number is 4
Number of valence electrons = 1
Number of outer electrons = 1
Number of inner electrons = 24 - 1 = 23
Option d: Strontium (Sr)Strontium is the 38th element of the periodic table having electronic configuration of [tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^2[/tex]
Highest principle quantum number is 5
Number of valence electrons = 2
Number of outer electrons = 2
Number of inner electrons = 38 - 2 = 36
Option e: Fluorine (F)Fluorine is the 9th element of the periodic table having electronic configuration of [tex]1s^22s^22p^5[/tex]
Highest principle quantum number is 2
Number of valence electrons = 7
Number of outer electrons = 7
Number of inner electrons = 9 - 7 = 2
The number of inner, outer, and valence electrons for each element is as follows: Br has 2 inner electrons, 28 outer electrons, and 7 valence electrons; Cs has 2 inner electrons, 118 outer electrons, and 1 valence electron; Cr has 2 inner electrons, 24 outer electrons, and 6 valence electrons; Sr has 2 inner electrons, 36 outer electrons, and 2 valence electrons; F has 2 inner electrons, 7 outer electrons, and 7 valence electrons.
Explanation:(a) Br has 2 inner electrons, 28 outer electrons, and 7 valence electrons.
(b) Cs has 2 inner electrons, 118 outer electrons, and 1 valence electron.
(c) Cr has 2 inner electrons, 24 outer electrons, and 6 valence electrons.
(d) Sr has 2 inner electrons, 36 outer electrons, and 2 valence electrons.
(e) F has 2 inner electrons, 7 outer electrons, and 7 valence electrons.
Learn more about Inner, outer, valence electrons here:https://brainly.com/question/29276667
#SPJ3
Draw an orbital diagram showing valence electrons, and write the condensed ground-state electron configuration for each:
(a) Ga (b) Zn (c) Sc
Answer : The condensed ground-state electron configuration for each is:
(a) [tex][Ar]4s^24p^{1}[/tex]
(b) [tex][Ar]4s^23d^{10}[/tex]
(c) [tex][Ar]4s^23d^{1}[/tex]
Explanation :
Electronic configuration : It is defined as the representation of electrons around the nucleus of an atom.
Number of electrons in an atom are determined by the electronic configuration.
Noble-Gas notation : It is defined as the representation of electron configuration of an element by using the noble gas directly before the element on the periodic table.
(a) The given element is, Ga (Gallium)
As we know that the gallium element belongs to group 13 and the atomic number is, 31
The ground-state electron configuration of Ga is:
[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^1[/tex]
So, the condensed ground-state electron configuration of Ga in noble gas notation will be:
[tex][Ar]4s^24p^{1}[/tex]
(b) The given element is, Zn (Zinc)
As we know that the zinc element belongs to group 12 and the atomic number is, 30
The ground-state electron configuration of Zn is:
[tex]1s^22s^22p^63s^23p^64s^23d^{10}[/tex]
So, the condensed ground-state electron configuration of Zn in noble gas notation will be:
[tex][Ar]4s^23d^{10}[/tex]
(c) The given element is, Sc (Scandium)
As we know that the scandium element belongs to group 3 and the atomic number is, 21
The ground-state electron configuration of Sc is:
[tex]1s^22s^22p^63s^23p^64s^23d^{1}[/tex]
So, the condensed ground-state electron configuration of Sc in noble gas notation will be:
[tex][Ar]4s^23d^{1}[/tex]
Phosgene, a poisonous gas, when heated will decompose into carbon monoxide and chlorine in a reversible reaction: COCl2 (g) <-----> CO (g) + Cl2 When 2.00 mol of phosgene is put into an empty 1.00 L flask and 395 ˚C and allowed to come to equilibrium, the final mixture contains 0.0398 mol of chlorine. Find Keq. Group of answer choices
Answer:
8.08 × 10⁻⁴
Explanation:
Let's consider the following reaction.
COCl₂(g) ⇄ CO (g) + Cl₂(g)
The initial concentration of phosgene is:
M = 2.00 mol / 1.00 L = 2.00 M
We can find the final concentrations using an ICE chart.
COCl₂(g) ⇄ CO (g) + Cl₂(g)
I 2.00 0 0
C -x +x +x
E 2.00 -x x x
The equilibrium concentration of Cl₂, x, is 0.0398 mol / 1.00 L = 0.0398 M.
The concentrations at equilibrium are:
[COCl₂] = 2.00 -x = 1.96 M
[CO] = [Cl₂] = 0.0398 M
The equilibrium constant (Keq) is:
Keq = [CO].[Cl₂]/[COCl₂]
Keq = (0.0398)²/1.96
Keq = 8.08 × 10⁻⁴
A 7.41 mass % aqueous solution of sodium chloride has a density of 1.14 g/mL. Calculate the molarity of the solution. Give your answer to 2 decimal places.
Answer:
Molarity for solution is 1.44 M
Explanation:
Molarity = Mol of solute / 1L of solution
We would need the volume of solution (To be calculated with density)
We would need the moles of solute (To be calculated with mass and molar mass of solute)
7.41 % by mass means 7.41 g of solute in 100 g of solution
So, moles of solute → 7.41 g / 58.45 g/mol = 0.127 mol
Let's determine the volume by density
Density = Mass / volume
1.14 g/mL = 100 g / Volume
Volume = 100 g / 1.14 g/mL → 87.7 mL
To reach molarity we must have the volume in L
87.7 mL . 1L / 1000 mL = 0.0877 L
Molarity → mol /L = 0.127 mol / 0.0877L → 1.44 M
How much heat energy, in kilojoules, is required to convert 72.0 gg of ice at −−18.0 ∘C∘C to water at 25.0 ∘C∘C ? Express your answer to three significant figures and include the appropriate units. View Available Hint(s)
Answer: The enthalpy change is 34.3 kJ
Explanation:
The conversions involved in this process are :
[tex](1):H_2O(s)(-18^0C)\rightarrow H_2O(s)(0^0C)\\\\(2):H_2O(s)(0^0C)\rightarrow H_2O(l)(0^0C)\\\\(3):H_2O(l)(0^0C)\rightarrow H_2O(l)(25^0C)[/tex]
Now we have to calculate the enthalpy change.
[tex]\Delta H=[m\times c_{s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{l}\times (T_{final}-T_{initial})][/tex]
where,
[tex]\Delta H[/tex] = enthalpy change = ?
m = mass of water = 72.0 g
[tex]c_{s}[/tex] = specific heat of ice = [tex]2.09J/g^0C[/tex]
[tex]c_{l}[/tex] = specific heat of liquid water = [tex]4.184J/g^0C[/tex]
n = number of moles of water = [tex]\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{72.0g}{18g/mole}=4.00moles[/tex]
[tex]\Delta H_{fusion}[/tex] = enthalpy change for fusion = 6010 J/mole
Now put all the given values in the above expression, we get
[tex]\Delta H=[72.0g\times 2.09J/g^0C\times (0-(-18)^0C]+4.00mole\times 6010J/mole+[72.0g\times 4.184J/g^)C\times (25-0)^0C][/tex][tex]\Delta H=34279.8J=34.3kJ[/tex] (1 KJ = 1000 J)
Therefore, the enthalpy change is 34.3 kJ
Final answer:
The total amount of heat energy required to convert 72.0 g of ice at − 18.0 °C to water at 25.0 °C is 34.2 kJ. This includes heating the ice to 0 °C, melting it, and then heating the water to 25.0 °C.
Explanation:
To calculate the amount of heat energy required in kilojoules to convert 72.0 g of ice at − 18.0 °C to water at 25.0 °C, we must consider three stages of the process: heating the ice to 0 °C, melting the ice, and then heating the resulting water to 25.0 °C.
First, we use the specific heat capacity of ice to heat it from − 18.0 °C to 0 °C:The total energy Qtotal is the sum of Q1, Q2, and Q3, which is 2664 J + 23976 J + 7566 J = 34206 J.
To convert this value to kilojoules, we divide by 1,000: 34206 J / 1000 = 34.206 kJ. Therefore, the answer to three significant figures is 34.2 kJ.
If one wished to obtain 0.050 moles of isopentyl alcohol, how many milliliters should one obtain? Enter only the number to two significant figures.
Answer:
1100 millimeters
Explanation:
1 mole of isopentyl alcohol = 22.4L = 22.4×1000 mL = 22,400L
0.050 moles of isopentyl alcohol = 0.050 × 22,400mL = 1120mL = 1100mL (to two significant figures)
A parallel-plate capacitor is charged and then is disconnected from the battery. By what factor does the stored energy change when the plate separation is then doubled? It becomes four times larger. It becomes one-half as large. It stays the same. It becomes two times larger.
When the plate separation of a charged parallel-plate capacitor is doubled, the stored energy becomes one-half as large due to the inverse relationship between capacitance and the distance between plates.
When a parallel-plate capacitor is disconnected from a battery and the plate separation is doubled, the stored energy changes in the following manner: The stored energy becomes one-half as large. This happens because the energy (U) stored in a capacitor is given by U = (1/2)CV2, where C is the capacitance and V is the voltage across the plates. When the capacitor is disconnected from the battery, the charge Q and voltage V across the plates remain constant.
However, the capacitance C of a parallel-plate capacitor is directly proportional to the area of the plates (A) and inversely proportional to the distance (d) between them, as given by C = ε0A/d, where ε0 is the vacuum permittivity. When distance d is doubled, capacitance C is halved, therefore, the energy stored, which is proportional to C, also halves because it is dependent on the capacitance.
While mercury is very useful in barometers, mercury vapor is toxic. Given that mercury has a ΔHvap of 59.11 kJ/mol and its normal boiling point is 356.7°C, calculate the vapor pressure in mm Hg at room temperature, 25°C.
Answer:
P = 2.65 E-3 mm Hg
Explanation:
liquid-vapor equilibrium:
⇒ ec. Clausius-Clapeyron:
Ln (P2/P1) = - ΔHv/R [ (1/T2) - (1/T1) ]∴ R = 8.314 E-3 KJ/K.mol
∴ ΔHv = 59.11 KJ/mol
∴ T2 = 25°C ≅ 298 K
∴ T1 = 356.7°C = 629.7 K
normal boiling point:
∴ P = 1 atm = P1 = 101.325 KPa
vapor pressure (P2):
⇒ Ln P2 - Ln (101.325) = - (59.11 KJ/mol)/(8.314 E-3 KJ/K.mol)*[ (1/298) - (1/629.7) ]
⇒ Ln P2 - 4.618 = - (7109.694 K)*[ 1.7676 E-3 K-1 ]
⇒ Ln P2 = 4.618 - 12.567
⇒ P2 = e∧(-7.9494)
⇒ P2 = (3.5313 E-4 KPa)×(7.50062mm Hg/KPa) = 2.65 E-3 mm Hg
"A student prepares a solution by dissolving 1.66 g of solid KOH in enough water to make 500.0 mL of solution. Calculate the molarity of K+ ions in this solution.
A 35.00 mL sample of this KOH solution is added to a 1000 mL volumetric flask, and water is added to the mark. What is the new molarity of K+ ions in this solution?"
Answer:
For 1: The molarity of [tex]K^+\text{ ions}[/tex] in this solution is 0.0592 M
For 2: The new molarity of [tex]K^+\text{ ions}[/tex] in this solution is [tex]2.07\times 10^{-3}M[/tex]
Explanation:
For 1:To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]
Given mass of KOH = 1.66 g
Molar mass of KOH = 56.1 g/mol
Volume of solution = 500.0 mL
Putting values in above equation, we get:
[tex]\text{Molarity of solution}=\frac{1.66\times 1000}{56.1g/mol\times 500.0}\\\\\text{Molarity of solution}=0.0592M[/tex]
1 mole of KOH produces 1 mole of potassium ions and 1 mole of hydroxide ions
So, molarity of [tex]K^+\text{ ions}=0.0592M[/tex]
Hence, the molarity of [tex]K^+\text{ ions}[/tex] in this solution is 0.0592 M
For 2:To calculate the molarity of the diluted solution, we use the equation:
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of the concentrated KOH solution having [tex]K^+\text{ ions}[/tex]
[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of diluted KOH solution having [tex]K^+\text{ ions}[/tex]
We are given:
[tex]M_1=0.0592M\\V_1=35.00mL\\M_2=?M\\V_2=1000mL[/tex]
Putting values in above equation, we get:
[tex]0.0592\times 35.00=M_2\times 1000\\\\M_2=\frac{0.0592\times 35.0}{1000}=2.07\times 10^{-3}M[/tex]
Hence, the new molarity of [tex]K^+\text{ ions}[/tex] in this solution is [tex]2.07\times 10^{-3}M[/tex]
Write the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by each:
(a) P (b) Mg (c) Se
Answer:
Part A:
Charge is [tex]P^{3-}[/tex]
Configuration is [tex]1s^2 2s^22p^63s^23p^6[/tex]
Part B:
Charge is [tex]Mg^{2+}[/tex]
Configuration is [tex]1s^2 2s^22p^6[/tex]
Part C:
Charge is [tex]Se^{2-}[/tex]
Configuration is [tex]1s^2 2s^22p^63s^23p^64s^23d^{10}4p^6[/tex]
Explanation:
Monatomic ions:
These ions consist of only one atom. If they have more than one atom then they are poly atomic ions.
Examples of Mono Atomic ions: [tex]Na^+, Cl^-, Ca^2^+[/tex]
Part A:
For P:
Phosphorous (P) has 15 electrons so it require 3 more electrons to stabilize itself.
Charge is [tex]P^{3-}[/tex]
Full ground-state electron configuration of the mono atomic ion:
[tex]1s^2 2s^22p^63s^23p^6[/tex]
Part B:
For Mg:
Magnesium (Mg) has 12 electrons so it requires 2 electrons to lose to achieve stable configuration.
Charge is [tex]Mg^{2+}[/tex]
Full ground-state electron configuration of the mono atomic ion:
[tex]1s^2 2s^22p^6[/tex]
Part C:
For Se:
Selenium (Se) has 34 electrons and requires two electrons to be stable.
Charge is [tex]Se^{2-}[/tex]
Full ground-state electron configuration of the mono atomic ion:
[tex]1s^2 2s^22p^63s^23p^64s^23d^{10}4p^6[/tex]
The electron configuration of atoms or ions depends on the number of electrons present.
The electron configuration refers to the arrangement of electrons in an atom or ions. Electrons are arranged in energy levels and each energy level is composed of orbitals.
The monoatomic ion most likely formed by P is P^3-. The electron configuration of this ion is 1s2 2s2 2p6 3s2 3p6. This is because P is in group 15 and attains a stable octet by gaining three electrons.
The monoatomic ion most likely formed by Mg is Mg^2+. Mg is a group 2 element and attains a stable octet by loss of two electrons. The electron configuration of this ion is 1s² 2s² 2p^6.
Se is a group 16 element, the monoatomic ion formed by Se is Se^2-. Se attains a stable octet by gain of two electrons. The electron configuration of this ion is; 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6.
Learn more: https://brainly.com/question/5624100
An aqueous solution is saturated with both a solid and a gas at 5 ∘C∘C. What is likely to happen if the solution is heated to 85 ∘C∘C ? View Available Hint(s)
Here is the complete question
An aqueous solution is saturated with both a solid and a gas at 5 °C. What is likely to happen if the solution is heated to 85 °C ?
View Available Hint(s)
a.) Some gas will bubble out of solution and more solid will dissolve.
b.) Some gas will bubble out of the solution and some solid will precipitate out of the solution.
c.) Some solid will precipitate out of solution.
d.) More gas will dissolve and more of the solid will dissolve.
Answer:
a.) Some gas will bubble out of solution and more solid will dissolve.
Explanation:
Temperature increase usually increases the dissolution of solids in liquids. From the question; Some gas will be bubble out of the solution as the temperature is being increased to 85 °C because that aqueous solution is saturated(i.e equal amount of solute and solvent in the solution) with both solid and gas at 5 °C, but when the solution is heated to 85 °C, the solution becomes supersaturated( i.e the solute is now more at the given temperature than the solvent).
Answer:
As the temperature increases, the solubility of the solid increases and the solubility of the gas decreases. When the solution that is saturated between a solid and a gas at 5 ° C and heated to 85 ° C, the gas comes out of the solution first.
Explanation:
A saturated solution is one that has the maximum amount of solute that is dissolved. An unsaturated solution is one that has a low amount of solute compared to the saturated solution. According to Henry's law, the solubility of a gas at a specific temperature is directly proportional to its partial pressure, that is
C ∝ p
C = kp
Where
p is the partial pressure
k ia a proportionality constant
C is the concentration of the gas
Draw the Lewis structure (including all lone pair electrons and any formal charges) for one of the four possible isomers of C3H9N.
Answer:
As shown in the attachment
Explanation:
The four possible isomers are as shown in the attachment.
a. Lithium and sodium are the most similar because they are both________ elements located in the same__________ , and therefore have similar properties.b. Nitrogen and oxygen are not the most similar because although they are both______ elements, are each located in a different________
Answer:
Lithium and sodium are the most similar because they are both alkali elements located in the same group, and therefore have similar properties.
Nitrogen and oxygen are not the most similar because although they are both non metals elements, are each located in a different group
Explanation:
Li and Na are both alkali elements from group 1 that shares some similities. The both can be obtained by the water hydrolysis. These are common reactions:
Metal from group 1 + H₂O → Base + H₂
Metal from group 1 + O₂ → oxides
Metal from group 1 + group 17 → ionic halides
Both form cations with 1+ charge, they can release only 1 e-
N is an element from group 15 and O, from group 16. They are both non metal.
Nitrogen can make a variety of oxides.
They react in water to produce nitric acid:
N₂O₃ + H₂O → 2HNO₃
N₂O₅ + H₂O → 2HNO₃
It has an anion with -3, as oxidation state. (Nitride)
The N with H, makes a well known hidride → ammonia
N₂ + 3H₂ → 2NH₃
The Oxygen also makes a well known hidride → water
2H₂ + O₂ → 2H₂O
Both are covalent hidrides.
N can have many oxidation's states. O always acts with -2 except for the peroxydes, with -1. O can have a great power of oxidation, that N does not have.
O₂ always acts as a reactant, at combustion reactions.
Lithium and sodium are similar as they are both alkali elements, found in the same group of the periodic table, leading to similar properties. Nitrogen and oxygen, while both nonmetals, are in separate groups, yielding different properties.
Explanation:a. Lithium and sodium are the most similar because they are both alkali elements located in the same group, and therefore have similar properties. Lithium and sodium both belong to the alkali metals group in the periodic table. They share similar chemical behaviors, as they have only one electron in a valence s subshell outside a filled set of inner shells. This fact influences their reactivity and the compounds they can form.
b. Nitrogen and oxygen are not the most similar because although they are both nonmetal elements, they are each located in a different group. Even though nitrogen and oxygen are close to each other in the periodic table, they do not share the same group, hence they have different properties and different numbers of valence electrons.
Learn more about Chemical Elements Similarities here:https://brainly.com/question/35865624
#SPJ3
How accurately can an umpire know the position of a baseball (mass = 0.142 kg) moving at 100.0 mi/h ± 1.00% (44.7 m/s ± 1.00%)?
Final answer:
The accuracy of an umpire knowing the position of a baseball depends on the speed of the baseball and the percentage uncertainty in that speed. The formula Δx = v × Δt can be used to calculate the uncertainty in the position of the baseball. For example, if the umpire measures the position of the baseball over a time interval of 1 second, the uncertainty in the position would be 44.7 meters.
Explanation:
The accuracy with which an umpire can know the position of a baseball depends on several factors, including the speed of the baseball and the percentage uncertainty in that speed.
Given that the baseball is moving at 100.0 mi/h ± 1.00% (44.7 m/s ± 1.00%), and assuming the umpire can accurately measure the speed, the uncertainty in the position of the baseball can be calculated using the formula:
Δx = v × Δt
where Δx is the uncertainty in the position, v is the velocity of the baseball, and Δt is the time interval over which the position is being measured.
For example, if the umpire measures the position of the baseball over a time interval of 1 second, the uncertainty in the position would be 44.7 m/s × 1 s = 44.7 meters.
Refractometry is an analytical method based on the difference between the speed of light as it travels through a substance (v) and its speed in a vacuum (c). In the procedure, light of known wavelength passes through a fixed thickness of the substance at a known temperature. The index of refraction equals c/v. Using yellow light (λ = 589 nm) at 20°C, for example, the index of refraction of water is 1.33 and that of diamond is 2.42. Calculate the speed of light in (a) water and (b) diamond.
Answer:
a). [tex]2.25 \times 10^{8}[/tex] m/s
b). 1.23 m/s
Explanation:
Given :
Index of refraction for water, [tex]n_{w}[/tex] = 1.33
Index of refraction of diamond, [tex]n_{d}[/tex] = 2.42
We know that,
Refractive index of any material is given by,
[tex]n = \frac{c}{v}[/tex]
where c = sped of light in air
v = speed of light in any medium
a). Speed of light in water
[tex]v = \frac{c}{n_{w}}[/tex]
[tex]v = \frac{3\times 10^{8}}{1.33}[/tex]
[tex]v = 2.25\times 10^{8}[/tex] m/s
b). Speed of light in diamond
[tex]v =\frac{c}{n_{d}}[/tex]
[tex]v =\frac{3\times 10^{8}}{2.42}[/tex]
= 1.23 m/s
Calculate the mass of glucose C6H12O6 that contains a million ×1.0106 carbon atoms. Be sure your answer has a unit symbol if necessary, and round it to 2 significant digits.
Answer:
The mass of glucose that contains a million [tex]1.0\times 10^6[/tex] carbon atoms is [tex]4.98\times 10^{-17} g[/tex].
Explanation:
Number of carbon atoms = [tex]1.0\times 10^6 [/tex]
1 molecule of glucose has 6 carbon atoms, then [tex]1.01\times 10^6 [/tex] will be in N molecules of glucose:
[tex]N=\frac{1.0\times 10^6}{6}[/tex]molecules of glucose
1 mole = [tex]N_A=6.022\times 10^{23} molecules[/tex]
Moles of glucose = n
[tex]N=n\times N_A[/tex]
[tex]n=\frac{N}{N_A}=\frac{\frac{1.0\times 10^6}{6}}{6.022\times 10^{23}}[/tex]
[tex]n=2.768\times 10^{-19} moles[/tex]
Mass of [tex]2.768\times 10^{-19} [/tex] moles of glucose;
[tex]2.768\times 10^{-19} mol\times 180=4.9817\times 10^{-17} g\approx 4.98\times 10^{-17} g[/tex]
Calculate the mass of glucose C₆H₁₂O₆ that contains a million (1.0 × 10⁶) carbon atoms. Be sure your answer has a unit symbol if necessary, and round it to 2 significant digits.
The mass of glucose that contains a million carbon atoms is 5.04 × 10⁻⁷ g.
1 molecule of glucose contains 6 carbon atoms. The number of molecules of glucose that contain 1.0 × 10⁶ carbon atoms are:
[tex]1.0 \times 10^{6}\ C\ atoms \times \frac{1\ Glucose\ molecule}{6\ C\ atoms} = 1.7 \times 10^{5} \ Glucose\ molecule[/tex]
We will convert molecules into moles using Avogadro's number: there are 6.02 × 10²³ molecules in 1 mole of molecules.
[tex]1.7 \times 10^{5} \ molecule \times \frac{1mol}{6.02 \times 10^{23} \ molecule} = 2.8 \times 10^{-19} \ mol[/tex]
We will convert moles to mass using the molar mass of glucose (180.16 g/mol).
[tex]2.8 \times 10^{-19} \ mol \times \frac{180.16g}{1mol} =5.04 \times 10^{-7} g[/tex]
The mass of glucose that contains a million carbon atoms is 5.04 × 10⁻⁷ g.
You can learn more about Avogadro's number here: https://brainly.com/question/13302703
In-Lab Question 5a. Comparing your actual mass to the calculated target mass, would you predict the absorbance of your solution made from solid to be greater than or less than that of the unknown solution? greater than less than the same cannot tell
Answer:
less than
Explanation:
The absorbance of a solution is a function of the concentration (amount) of a substance in the solution. Mathematically, if the concentration is increased, the absorbance of the solution will also increase and if the concentration is decreased, the absorbance will decrease. There will be a decrease in the value of the absorbance because the calculated and predicted masses are not the same.
Final answer:
The absorbance of a solution made from a solid may be greater or less than that of an unknown solution, depending on their respective concentrations and properties.
Explanation:
When comparing the absorbance of a solution made from a solid to that of an unknown solution, we need to consider the concentration or the size of the container. A solution made from a solid may have a higher absorbance because it contains a higher concentration of molecules that can interact with light. On the other hand, the absorbance of the unknown solution will depend on its specific properties. Therefore, whether the absorbance of the solution made from a solid is greater, less than, or the same as the absorbance of the unknown solution cannot be determined without further information.
Ketones undergo a reduction when treated with sodium borohydride, NaBH4. The product of the above reaction has the following spectroscopic properties; propose a structure. MS: M+ = 86 IR: 3400 cm-1 1H NMR: 1.56 δ (4H, triplet); 1.78 δ (4H, multiplet); 3.24 δ (1H, quintet); 3.58 δ (1H, singlet) 13C NMR: 24.2, 35.5, 73.3 δ
Answer:
The product is cyclohexanol
Explanation:
Firstly,
A ketone undergo a borohydride reduction reaction to form an alcohol as below,
R-CO-R' ⇒ R-CO(OH)-R'
IR Spectrum confirms that alcohol group is existed with the peak at 3400 cm⁻¹From 1H-NMR, the product has 10 hydrogen atoms, the MS suggest that the formula is C₅H₁₀O (M = 86). With this formula, the alcohol is monosaturated. Since, the substance already underwent reduction reaction, the only way to suggest a monosaturated compound is a cyclic alcohol. So the compound is cyclopentanol.Check with other spectroscopic properties,3 signals of 13C NMR confirms the structure is symmetrical, δ 24.2, (-CH₂-CH₂-CH(CH₂-)-OH), δ 35.5 (-CH₂-CH₂-CH(CH₂-)-OH), δ 73.3 (-CH₂-CH₂-CH(CH₂-)-OH).1H NMR confirms,1.56 δ (4H, triplet) - (-CH₂-CH₂-CH-OH) ; triplet as coupling with 2 H,
1.78 δ (4H, multiplet) - (-CH₂-CH₂-CH-OH); multiplet as coupling with 2H of CH₂, 1 H of CH
3.24 δ (1H, quintet); - (-CH₂-CH₂-CH(CH₂-)-OH), coupling with4 H of 2 group of CH₂
3.58 δ (1H, singlet); - (-CH₂-CH₂-CH(CH₂-)-OH), hydrogen of alcohol group, not tend to coupling with other hydrogen
How is n1 in the Rydberg equation related to the quantum number n in the Bohr model of the atom?
Explanation:
The n in Bohr model of the atom is principle quantum number.
The Rydberg n integer stats represent electron orbits at various integral distances from the atom in Bohr's conceptualization of the atom. Subsequent models discovered that the values for n1 and n2 match the two orbitals ' principle quantum numbers.
Final answer:
The Rydberg equation and the Bohr model are related through the principal quantum number n. In the Rydberg equation, n1 and n2 represent initial and final energy levels during an electron transition, similar to the energy levels denoted by n in the Bohr model of the atom. This relationship explains the emission spectra of atoms such as hydrogen.
Explanation:
The Rydberg equation relates to the Bohr model through the quantum number n. In the Rydberg equation, n1 corresponds to the lower energy level (or orbit) from which an electron transitions to a higher energy level designated by n2 where n2 > n1. The principle quantum number n in the Bohr model signifies distinct energy levels or orbits in which electrons can reside around the nucleus, with the lowest energy state starting at n=1 and increasing integers signifying higher energy states.
When an electron jumps between energy levels, light is emitted, and the wavelength of this light can be calculated using the Rydberg formula, which includes the Rydberg constant and the initial and final principal quantum numbers. For the hydrogen atom, transitions to the ground state (n = 1) produce the Lyman series in the ultraviolet band, while transitions to the first excited state (nf = 2) result in the Balmer series in the visible band, and so on for higher energy levels.
Write the formula and name of the compound formed from the following ionic interactions:
(a) The 2+ ion and the 1- ion are both isoelectronic with the atoms of a chemically unreactive Period 4 element.
(b) The 2+ ion and the 2- ion are both isoelectronic with the Period 3 noble gas.
(c) The 2+ ion is the smallest with a filled d subshell; the anion forms from the smallest halogen.
(d) The ions form from the largest and smallest ionizable atoms in Period 2.
Answer:
Explanation:
a ) period 4 noble gas is krypton
isoelectronic with it are Sr ⁺² and Br⁻
compound is SrBr₂
b ) Period 3 noble gas is Argon
isoelectronic with it are Mg⁺² and O⁻²
compound is MgO
c) 2+ ion is the smallest with a filled d subshell is Zn⁺² , smallest halogen
is F⁻
compound is ZnF₂
d ) ions from the largest and smallest ionizable atoms in Period 2
Li⁺ and F⁻
compound is LiF
The compounds resulting from the described ionic interactions are Strontium Sulfide (SrS), Magnesium Oxide (MgO), Iron(II) Fluoride (FeF2), and Beryllium Fluoride (BeF2).
Explanation:The compound formed by ionic interactions can be determined by considering the properties of the ions given.
The chemically unreactive Period 4 element that these ions are isoelectronic with is Kr (Krypton). The 2+ cation may be Sr (Strontium) and the 1- anion is S (Sulfur). Thus, the formula for this compound would be SrS and the name would be Strontium Sulfide. The Period 3 noble gas is Ar (Argon). One possibility is that the 2+ ion is Mg (Magnesium) and the 2- ion is O (Oxygen). Thus, the formula for this compound is MgO, and the name is Magnesium Oxide. The smallest 2+ ion with a filled d subshell is Fe (Iron) and the anion of the smallest halogen is F (Fluoride). Thus, the formula is FeF2 and the name is Iron(II) Fluoride. The largest atom in Period 2 that can form an ion is Be (Beryllium) and the smallest is F (Fluorine). Thus, the formula is BeF2 and the name is Beryllium Fluoride. Learn more about Ionic Compounds here:https://brainly.com/question/3222171
#SPJ12
A biochemist carefully measures the molarity of glycerol in 913 mL of photobacterium cell growth medium to be 81.3 μM.
Unfortunately, a careless graduate student forgets to cover the container of growth medium and a substantial amount of the solvent evaporates. The volume of the cell growth medium falls to 11.1 mL.
a. Calculate the new molarity of glycerol in the photobacterium cell growth medium. Round each of your answers to 3 significant digits.
Answer : The new molarity of glycerol in the photobacterium cell growth medium is, [tex]6.69\times 10^3\mu M[/tex]
Explanation :
Formula used :
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the initial molarity and volume of glycerol.
[tex]M_2\text{ and }V_2[/tex] are the new molarity and volume of glycerol .
We are given:
[tex]M_1=81.3\mu M\\V_1=913mL\\M_2=?\\V_2=11.1mL[/tex]
Putting values in above equation, we get:
[tex]81.3\mu M\times 913mL=M_2\times 11.1mL\\\\M_2=6.69\times 10^3\mu M[/tex]
Hence, the new molarity of glycerol in the photobacterium cell growth medium is, [tex]6.69\times 10^3\mu M[/tex]
Final answer:
The new molarity of glycerol after the evaporation of the solvent from 913 mL to 11.1 mL is 6.69 × 10^6 µM, calculated using the conservation of moles and by adjusting the concentration for the reduced volume.
Explanation:
The molarity of a solution is defined as the number of moles of solute per liter of solution. When the solvent evaporates from a solution and the volume decreases, the molarity of the solute increases because the same amount of solute is now present in a smaller volume of solvent. The original volume of the photobacterium cell growth medium is 913 mL with a glycerol concentration of 81.3 µM (micro molar). The new volume of the cell growth medium after evaporation is 11.1 mL. To find the new molarity, we can use the concept of conservation of moles of solute, which states that the number of moles in the solution before and after evaporation are the same.
First, we convert the original volume from milliliters to liters:
913 mL = 0.913 L
Then we calculate the moles of glycerol originally in the solution:
moles of glycerol = (81.3 µM) × (0.913 L) = 0.0000813 moles/L × 0.913 L = 7.42449 × [tex]10^{-5}[/tex] moles
Since the amount of glycerol remains the same, and only the volume has changed, we can find the new molarity by dividing the moles of glycerol by the new volume in liters:
New volume in liters: 11.1 mL = 0.0111 L
New molarity = moles of glycerol / new volume
= 7.42449 × [tex]10^{-5}[/tex] moles / 0.0111 L
Performing the calculation gives us:
New molarity = 6.68873 M (rounded to three significant digits). However, this result is not in micro molarity. So to convert this to µM,
1 M = 1,000,000 µM
New molarity in µM = 6.68873 × 1,000,000 µM = 6.69 × 106 µM (or 6,690,000 µM, rounded to three significant digits)
Calculate the number of moles of carbon dioxide formed if 31.6 mL of sodium bicarbonate reacts with excess hydrochloric acid.
Answer:
0.827 mol
Explanation:
Sodium bicarbonate (NaHCO₃) reacts with hydrochloric acid (HCl) by the equation:
NaHCO₃ + HCl → CO₂ + NaCl + H₂O
Thus, the stoichiometry between sodium bicarbonate and CO₂ is
1 mol : 1 mol.
The molar mass of sodium bicarbonate is 84.007 g/mol, thus the mass in 1 mol is 84.007 g, and the density of it is 2.2 g/mL, thus the volume in 1 mol is:
V = 84.007/2.2 = 38.185 mL
According to Proust's law, the ratio reaction remains constant so:
38.185 mL of NaHCO₃/1 mol of CO₂ = 31.6 mL of NaHCO₃/n
38.185n = 31.6
n = 31.6/38.185
n = 0.827 mol of CO₂
Identify each element below, and give the symbols of the other elements in its group:
(a) [Ar] 4s²3d¹⁰4p²
(b) [Ar] 4s²3d⁷
(c) [Kr] 5s²4d⁵
Answer:
Answer in explanation
Explanation:
Argon has 18 electrons. So to get the element in question, we only need to add 18 to the number of the filled electrons.
a. Germanium, atomic number 32
Other group members:
Silicon Si , Carbon C , Tin Sn , Lead Pb and Flerovium Fl
b. Cobalt , atomic number 27
Other group members:
Rhodium Rh , Iridium Ir and Meitnerium Mt
c. Technetium , atomic number 43
Krypton is element 36
Other group members are :
Manganese Mn , Rhenium Re and Bohrium Bh
As an instructor is preparing for an experiment, he requires 225 g phosphoric acid. The only container readily available is a 150-mL Erlenmeyer flask. Is it large enough to contain the acid, the density of which is 1.83 g/mL?
Answer:
Yes, the 150-mL Erlenmeyer will be large enough to contain the acid.
Explanation:
As an instructor is preparing for an experiment, he requires 225 g phosphoric acid. Considering that the density of the phosphoric acid is 1.83 g/mL, we can find the volume occupied by the acid using the following expression.
density = mass / volume
volume = mass / density
volume = 225 g / (1.83 g/mL)
volume = 123 mL
The phosphoric acid occupies 123 mL so the 150-mL Erlenmeyer will be large enough to contain it.
For which blocks of elements are outer electrons the same as valence electrons? For which are d electrons often included among valence electrons?
Answer:
1. Group 1 — 3
2. Transition metals
Explanation:
d-block elements
d-Block Elements:These elements are also known as transition elements as their positioning and transition of properties lies between s and p block elements.
d-block elements have number of valence electrons equal to their group number, which is equal to the number of electrons in the "valence shell".
For example, Consider a transition metal or d block element Scandium. It's atomic number is 21.
Electronic configuration of Scandium(Sc)- [Ar] 3d¹ 4s², it has three electrons in its outermost shell and has a valency of three.
The electron configuration of scandium indicates that the ultimate shell(orbit) of scandium has a complete of electrons. But the electron configuration of scandium within side the Aufbau approach indicates that its ultimate electron([tex]3d^1\\[/tex]) has entered the d-orbital. Thus we can say that scandium has 3 valence electrons.
Therefore, we can say that d block elements have same number of outer electrons and valence electrons.
Learn more:
https://brainly.com/question/16089134
Before Mendeleev published his periodic table, Döbereiner grouped elements with similar properties into "triads," in which the unknown properties of one member could be predicted by averaging known values of the properties of the others. To test this idea, predict the values of the following quantities:
(a) The atomic mass of K from the atomic masses of Na and Rb
(b) The melting point of Br₂ from the melting points of Cl₂ (-101.0°C) and I₂ (113.6°C) (actual value = - 7.2°C)
Answer:
a) The atomic mass of the potassium is 54.23 amu.
b) The melting point of bromine gas is 6.3°C.
Explanation:
a) Atomic mass of Na =22.99 amu
Atomic mass of Rb = 85.47 amu
Döbereiner triad = Na , K ,Rb
Taking average of atomic masses of Na and Rb
Atomic mass of the K = [tex]\frac{22.99 amu+85.47 amu}{2}=54.23 amu[/tex]
The atomic mass of the potassium is 54.23 amu.
b) Melting point of chlorine gas =-101.0°C
Melting point of iodine gas =113.6°C
Döbereiner triad = Cl, Br , I
Melting point of bromine gas :
=[tex]\frac{-101.0^oC +113.6^oC}{2}=6.3^oC[/tex]
The melting point of chlorine gas is 6.3°C.