Answer:
2
Explanation:
There are four quantum numbers:
Principal quantum number (n)
Azimuthal quantum number (l)
Magnetic quantum number (ml)
Principal quantum number (n)
It tell about the energy levels. It is designated by n.
For example,
If n =2
It means there are two energy level present.
Azimuthal quantum number (l)
The azimuthal quantum number describe the shape of orbitals. Its value for s, p, d, f... are 0, 1, 2, 3. For l=3
(n-1)
4-1 = 3
it means principle quantum is 4 and electron is present in f subshell.
Magnetic quantum number (ml)
It describe the orientation of orbitals. Its values are -l to +l. For l=3 the ml will be -3 -2 -1 0 +1 +2 +3.
Spin quantum number (ms)
The spin quantum number tells the spin of electron either its clock wise (+1/2) or anti clock wise (-1/2).
If the electron is added in full empty orbital its spin will be +1/2 because it occupy full empty. If electron is already present and another electron is added then its spin will be -1/2.
Or a gas with a volum of 475ml at a temperature of -25c is heated to 275c. What is the new volume of the gas if the pressure and number of moles are held constant
Answer:
The new volume of the gas is 1.04L
Explanation:
You have to apply Charles's Law to solve this:
In two different situations, when you have a gas with the same quantity and pressure, relation between volume and T° must be the same
Volume / Temperature = Constant
Temperature in K
So;
475 mL/248K = Vol₂ / 548K
(475 mL/248K) 548K = Vol₂
1049 mL = Vol₂
If 21.39 g acetylene is allowed to completely react with oxygen, how many grams of O2 are reacted? (The molar mass of acetylene is 26.04 g/mol)2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(g)acetyleneBe sure to use the p
Answer:
65.712 grams of oxygen has reacted.
Explanation:
[tex]2 C_2H_2(g) + 5 O_2(g)\rightarrow 4 CO_2(g) + 2 H_2O(g)[/tex]
Mass of acetylene = 21.39 g
Moles of acetylene = [tex]\frac{21.39 g}{26.04 g/mol}=0.8214 mol[/tex]
According to reaction , 2 moles of acetylene reacts with 5 moles of oxygen gas.
Then 0.8214 moles of oxygen gas will react with :
[tex]\frac{5}{2}\times 0.8214 mol=2.0535 mol[/tex] of oxygen gas.
Mass of 2.0535 moles of oxygen gas :
2.0535 mol × 32 g/mol = 65.712 g
65.712 grams of oxygen has reacted.
Shielding or screening occurs when one electron is blocked from the full effects of the nuclear charge so that the electron experiences only a part of the nuclear charge. Penetration occurs when an electron penetrates the electron cloud of the 1s orbital and now experiences the full effect of the nuclear charge. Penetration occurs when one electron is blocked from the full effects of the nuclear charge so that the electron experiences only a part of the nuclear charge. Shielding or screening occurs when an electron penetrates the electron cloud of the 1s orbital and now experiences the full effect of the nuclear charge. Shielding and penetration are essentially the same thing and occurs when an electron penetrates the electron cloud of the 1s orbital and now experiences the full effect of the nuclear charge. Shielding and penetration are essentially the same thing and occurs when one electron is blocked from the full effects of the nuclear charge so that the electron experiences only a part of the nuclear charge.
Answer:Shielding and penetration are essentially the same thing and occurs when one electron is blocked from the full effects of the nuclear charge so that the electron experiences only a part of the nuclear charge
Explanation:
Penetration is how well the outer electrons are shielded from the nucleus by the core electrons. The outer electrons therefore experience less of an attraction to the nucleus.
A sample of gas has a volume of 1.9L and a temperature of 21 degrees celsius. Heat is applied to the sample, leading to an increase in temperature to 27 degrees celsius. The sample is held at constant pressure. Solve for the final volume.
Answer:
1.94 L
Explanation:
21°C = 21 +273 = 294 K
27°C = 27 + 273 = 300 K
T1/V1 = T2/V2
294 K/1.9 L = 300 K/x L
x = (1.9*300)/294 ≈ 1.94 L
An experiment is performed in which different masses and shapes are dropped from different heights and the times it takes for each to reach the ground are recorded. Which of the following is the dependent variable? i.e. which variable is the RESULT of another variable?
a. time
b. mass
c. shape
d. height
Answer:
C
Explanation:
The answer is shape.
Basically, quantities can either be fundamental or derived. While fundamental are the basic quantities, derived are obtained from combining fundamentals.
To get a shape, we would need the combination of lengths. This makes the shape a derived quantity
Which procedure will most increase the energy output of a nuclear reactor? exposing the core to the outside world removing fissile material from the core decreasing the amount of steam produced raising the control rods out of the reactor
Raising the control rods out of the reactor
The reactor regulates the number of neutrons that are involved in the chain reaction. This is accomplished by the reactor absorbing some of the neutrons, produced in the splitting of the atoms, in its walls.
Explanation:
If the rods are taken out of the reactor, the rods would heat up very fast and most probably an explosion would occur. This is because most of the neutrons produced in the splitting of the radioactive atoms in the rod would go ahead and bombard other atoms in the rods hence spiking up the chain reaction rate. This would release a lot of energy at a go.
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Answer:
Raising the control rods out of the reactor
Explanation:
Raising the rods will allow the chain reaction to flow more freely, therefore increasing the energy output of the nuclear reactor
To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 2.2-L bulb, then filled it with the gas at 2.00 atm and 27.0 ∘C and weighed it again. The difference in mass was 5.1 g.
Identify the gas. Express your answer as a chemical formula.
Answer:
N2
Explanation:
We use the ideal gas equation to calculate the number of moles of the diatomic gas. Then from the number of moles we can get
Given:
P = 2atm
1atm = 101,325pa
2atm = 202,650pa
T = 27 degrees Celsius = 27 + 273.15 = 300.15K
V = 2.2L
R = molar gas constant = 8314.46 L.Pa/molK
PV = nRT
Rearranging n = PV/RT
Substituting these values will yield:
n = (202,650 * 2.2)/(8314.46* 300.15)
n = 0.18 moles
To get the molar mass, we simply divide the mass by the number of moles.
5.1/0.18 = 28.5g/mol
This is the closest to the molar mass of diatomic nitrogen N2.
Hence, the gas is nitrogen gas
The diatomic gas could be identified using the ideal gas law and the given conditions. The calculated molar mass matched with the molar mass of Oxygen, so the diatomic gas is likely Oxygen (O2).
Explanation:By examining the given conditions and the difference in mass, we can identify the gas X2 using the ideal gas law. The ideal gas law is given by PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. Here, the pressure P = 2.00 atm, the volume V = 2.2 L, R = 0.0821 L.atm/mol.K and T = 27°C = 300.15 K. Inserting these values gives us the number of moles of gas. The molar mass of the gas can be calculated by dividing the mass of the gas by the number of moles. Using the molar mass and comparing it to the periodic table, the diatomic gas appears to be Oxygen (O2).
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Aqueous humor forms during capillary filtration in the __________?
Answer:
Ciliary body.
Explanation:
Ciliary body: It is the known for the part of the eye that includes the ciliary muscle, which helps in the control the ciliary epithelium and lens shape, which are helping in the production of aqueous humor.
Through active secretion mechanism helping in to produce eighty percent of aqueous humor, and through the plasma ultra-filtration mechanism twenty percent of aqueous humor is produced.
Ciliary body is the part of the layer which helps to deliver the nutrients, and oxygen to the eye tissues, and this layer is known as uvea.
The aqueous humor, a watery fluid in the anterior cavity of the eye, forms during capillary filtration in the ciliary body.
Explanation:The aqueous humor is a watery fluid that fills the anterior cavity of the eye, which includes the cornea, iris, ciliary body, and lens. It is produced during a process called capillary filtration.
Capillary filtration occurs when fluid moves from an area of high pressure to an area of lower pressure on the other side of the capillary wall. In the eye, this process takes place in the ciliary body, a part of the eye that has a rich capillary network, and results in the formation of aqueous humor.
The production of aqueous humor is essential for maintaining intraocular pressure and providing nutrients to the cornea and lens, which do not have their own blood supply.
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If a system has 2.00 × 10 2 kcal 2.00×102 kcal of work done to it, and releases 5.00 × 10 2 kJ 5.00×102 kJ of heat into its surroundings, what is the change in internal energy of the system?
Answer:
336.8 kilo Joules is the change in internal energy of the system.
Explanation:
The equation for first law of thermodynamics follows:
[tex]\Delta U=Q+W[/tex]
where,
Q = heat added to the system
ΔU = Change in internal energy
W = work done
We have :
Amount of heat given out by the system will be negatuive as heat relased by the system = Q
Q= [tex]-5.00\times 10^2 kJ[/tex]
Work done on the system will positive as work is done on the system:
w = [tex]2.00\times 10^2 kCal=836.8 kJ[/tex]
[tex]\Delta U=-5.00\times 10^2 kJ+836.8 kJ=336.8 kJ[/tex]
336.8 kilo Joules is the change in internal energy of the system.
The net change in internal energy of the system is calculated using the first law of thermodynamics and is found to be 3.368 × 105 Joules.
Explanation:The change in internal energy (ΔU) of a system can be calculated using the first law of thermodynamics, which states that the change in internal energy is equal to the heat (Q) added to the system minus the work (W) done by the system on its surroundings: ΔU = Q - W.
In your question, work is done on the system (2.00 × 102 kcal), which equates to 2.00 × 105 cal or 8.368 × 105 Joules (since 1 kcal = 4.184 kJ or 4184 Joules). Heat is released by the system (5.00 × 102 kJ), which is already in Joules. Since work is done on the system, it's positive, but heat released by the system is negative for the internal energy calculation. So, ΔU = Q - W = -5.00 × 102 kJ + 8.368 × 105 J.
Here's the calculation:
Q = -5.00 × 102 kJ = -5.00 × 105 J (since 1 kJ = 1000 J)W = +8.368 × 105 JΔU = -5.00 × 105 J + 8.368 × 105 JΔU = 3.368 × 105 JTherefore, the net change in internal energy of the system is 3.368 × 105 Joules.
A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out of oxalic acid , a diprotic acid that can be purchased inexpensively in high purity, and dissolves it in of distilled water. The student then titrates the oxalic acid solution with her sodium hydroxide solution. When the titration reaches the equivalence point, the student finds she has used of sodium hydroxide solution.Calculate the molarity of the student's sodium hydroxide solution. Be sure your answer has the correct number of significant digits.
Answer:
See explanation below to full answer
Explanation:
First of all, you are not providing the amounts of acid and hydroxide here, to do the calculations. However, in order to help you, I will use these values that are taken from a similar exercise. Then, replace your data with this procedure and you should get the correct answer.
For this part, I will say that the student weights about 210 mg of oxalic acid, (H2C2O4) and the volume of NaOH used to reach equivalent point was 150 mL in a beaker of 250 mL.
Now the equivalence point is the point where both moles of acid and hydroxide are the same. In other words:
nA = nB
The reaction that it's taking place is the following:
2NaOH + H2C2O4 ----------> Na2C2O4 + 2H2O
This means that 2 moles of NaOH reacts with 1 mole of H2C2O4, therefore the expression in (1) corrected is:
nB = 2 nA
So, we need to calculate first the moles of the acid. To do that we need the molar mass of the acid (the reported is 90.03 g/mol)
nA = 0.210 / 90.03 = 0.0023 moles
We have the moles of acid used, so the moles of the hydroxide is:
nB = 2 * 0.0023 = 0.0046 moles
We have the volume used of hydroxide, which is 150 mL, so finally the concentration is:
MB = 0.0046 / 0.150 = 0.031 M
Now, replace the actual values that you have in here, and you should get an accurate result.
Which of the following reactions is the least energetic? Question 18 options: ATP + H2O → ADP + Pi ATP + H2O → AMP + PPi AMP + H2O → Adenosine + Pi ADP + H2O → AMP + Pi All give off the same amount of energy.
Answer:
The correct answer is AMP+H2O→ Adenosine + pi
Explanation:
The above reaction is least energetic because there is no phosphoanhydride bond present with adenosine mono phosphate.Phospho anhydride bond is an energy rich bond.
As a result hydrolysis of AMP generates very little amount of energy in comparison to the hydrolysis of ATP and ADP.
Classify the following as an endothermic or exothermic reaction:
a. making popcorn in a microwave oven.
b. a burning match.
c. boiling water.
d. burning rocket fuel.
e. the reaction inside a heat pack
Answer:
a. making popcorn in a microwave oven. Endothermic
b. a burning match. Exothermic
c. boiling water. Endothermic
d. burning rocket fuel. Exothermic
e. the reaction inside a heat pack Exothermic
Explanation:
In order to answer, we need to review the definitions of exothermic and endothermic reactions.
Exothermic reactions give out heat. They cause increase in the energy of the system.
Endothermic reactions absorb heat. They cause decrease in the energy of system.
By this definition,
a. making popcorn in a microwave oven. Endothermic as heat energy is provided to the corn which causes it to pop.
b. a burning match. Exothermic as heat energy is given out by a burning match.
c. boiling water. Endothermic as heat energy is provided to the water which causes it to boil.
d. burning rocket fuel. Exothermic as heat energy is given out by burning fuel.
e. the reaction inside a heat pack. Exothermic as reaction which takes place inside heat pack gives out heat. This heat provides comfort to painful joints and muscles.
Energy is absorbed in an endothermic reaction while energy is released in an exothermic reaction.
An exothermic process is a process in which energy is released. This implies that heat is evolved in the process. In an endothermic process, heat is absorbed in the process. We shall now classify the following process as endothermic or exothermic accordingly;
making popcorn in a microwave oven - Endothermic a burning match - Exothermicboiling water - Endothermicburning rocket fuel - Exothermic the reaction inside a heat pack - ExothermicLearn more: https://brainly.com/question/4612545
A flask with a mass of 321.9 g is filled with 15.7 mL of carbon tetrachloride. The mass of the flask and carbon tetrachloride is found to be 523.6 g. From this information, calculate the density of carbon tetrachloride. According to this problem, the density of CCl4 is Answer g/mL.
Answer:
Density of carbon tetrachloride = 12.8 g/mL
Explanation:
Given :
[tex]m_{flask}=321.9\ g[/tex]
[tex]m_{flask}+m_{CCl_4}=523.6\ g[/tex]
Mass of carbon tetrachloride: -
[tex]m_{flask}+m_{CCl_4}=523.6\ g[/tex]
[tex]m_{CCl_4}=523.6-m_{flask}\ g=523.6-321.9\ g=201.7\ g[/tex]
Mass of carbon tetrachloride = 201.7 g
Given, Volume = 15.7 mL
Considering the expression for density as:
[tex]Density=\frac {Mass}{Volume}[/tex]
So,
[tex]Density=\frac {201.7\ g}{15.7\ mL}[/tex]
Density of carbon tetrachloride = 12.8 g/mL
Look up the density of n-butyl chloride (1-chlorobutane). Assume that this alkyl halide was prepared instead of the bromide. Decide whether the alkyl chloride would appear as the upper or the lower phase at each stage of the separation procedure: after the reflux, after the addition of water, and after the addition of sodium bicarbonate.
Answer:
See explanation below
Explanation:
First, you need to know the density of each compound in order to know this.
The density of 1-chlorobutane is 0.88 g/mL,
The density of water is 1 g/mL
The density of sodium bicarbonate is 2.2 g/cm3.
therefore, the one that has a greater density will always go at the lower phase.
In this case, after the reflux, it will stay in the lower phase, basically because you don't have another solvent with a greater density than the butane.
After adding water, it will be in the upper phase, water has a greater density.
After adding bicarbonate, it will be in the upper phase too.
A large sport utility vehicle has a mass of 2700 kg.Calculate the mass of CO2 emitted into the atmosphere upon accelerating the SUV from 0.0 mph to 67.0 mph. Assume that the required energy comes from the combustion of octane with 30\% efficiency. (Hint: Use KE=1/2mv^2 to calculate the kinetic energy required for the acceleration.)
Answer:
[tex]m_{CO2}=260.7 g CO2[/tex]
Explanation:
First of all we need to calculate the energy required:
[tex]KE= 0.5*m*v^2[/tex]
where:
[tex]m=2700kg[/tex]
[tex]v=67 mph=29.95 m/s[/tex]
[tex]KE= 0.5*2700kg*(29.95)^2[/tex]
[tex]KE= 1210953 J=1210.953 kJ[/tex]
Octane's combustion enthalpy: [tex]\Delta H_{comb}=- 5450 kJ/mol[/tex]
The reaction:
[tex]C_8H_{18} + 25/2 O_2 longrightarrow 8 CO_2 +9 H_O[/tex]
Mass of CO2:
[tex]m_{CO2}=\frac{1210.953 kJ}{5450mol}*\frac{1}{0.3}*\frac{8 mol CO2}{1 mol}*\frac{44 g CO2}{mol CO2}[/tex]
[tex]m_{CO2}=260.7 g CO2[/tex]
An "empty" container is not really empty if it contains air. How may moles of nitrogen are in an "empty" two-liter cola bottle at atmospheric pressure and room temperature (25∘C)? Assume ideal behavior.What is the partial pressure of oxygen in air at atmospheric pressure (1 atm)? Assume ideal behavior
Answer:
1. 0.0637 moles of nitrogen.
2. The partial pressure of oxygen is 0.21 atm.
Explanation:
1. If we assume ideal behaviour, we can use the Law of ideal gases to find the moles of nitrogen, considering that air composition is mainly nitrogen (78%), oxygen (21%) and argon (1%):
[tex]V_{N_2}=V_{T}\times 0.78=2L \times 0.78 =1.56 L\\PV=nRT\\n_{N_2}=\frac{PV}{RT}=\frac{1 atm\times 1.56 L}{0.0821\frac{atmL}{molK}\times 298 K}\\n_{N_2}= 0.0637 mol[/tex]
2. Now, in order to find he partial pressure of oxygen we need to find the total moles of air, and then the moles of oxygen. Then, we use these results to determine the molar fraction of oxygen, to multiply it with total pressure and get the partial pressure of oxygen as follows:
[tex]n_{total}=\frac{1 atm \times 2L}{0.0821 \frac{atmL}{molK}298K}=0.0817 mol[/tex]
[tex]V_{O_2}=2L \times 0.21 = 0.42 L\\n_{O_2}=\frac {1atm \times 0.42 L}{0.0821 \frac{atm L}{mol K}298 K}=0.0172 mol\\X_{O_2}=\frac{n_{O_2}}{n_{total}}=\frac{0.0172 mol}{0.0817 mol}= 0.21 [/tex]
[tex]P_{O_2}=X_{O_2} \times P = 1 atm \times 0.21 = 0.21 atm[/tex]
As you see, the molar fraction and volume fraction are the same because of the assumption of ideal behaviour.
Final answer:
Using the ideal gas law, we can determine the moles of nitrogen in an 'empty' container and calculate the partial pressure of oxygen in air at atmospheric pressure. The moles of nitrogen is 78% of the total moles of air in the container. The partial pressure of oxygen is 21% of the atmospheric pressure.
Explanation:
To calculate the number of moles of nitrogen in a two-liter container at atmospheric pressure and room temperature, we can use the ideal gas law PV = nRT. Given room temperature 25°C (which is 298.15 K), a volume of 2.00 liters (2.00 x 10-3 m3), and atmospheric pressure (1 atm or 101.325 kPa), we can solve for n, the number of moles of nitrogen gas (N2).
The air is approximately 78% nitrogen by moles. Therefore, to find the moles of nitrogen, we first calculate the moles of air using the ideal gas law and then multiply this by 0.78.
For the partial pressure of oxygen, we acknowledge that air is around 21% oxygen by moles. Thus, the partial pressure of oxygen would be 0.21 times the total atmospheric pressure, which results in a partial pressure of 0.21 atm.
Which substance is the limiting reactant when 2.0 g of sulfur reacts with 3.0 g of oxygen and 4.0 g of sodium hydroxide according to the following chemical equation: [tex]2S(s) + 3O_2(g) + 4NaOH(aq) \rightarrow 2Na_2SO_4(aq) + 2H_2O(l)[/tex]a. O₂(g)b. NaOH(aq)c. S(s)d. None of these substances is the limiting reactant.
Answer:
The limiting reactant is NaOH (option B)
Explanation:
2S(s) + 3O₂(g) + 4NaOH(aq) → 2Na₂SO₄(aq) + 2H₂O(l)
The reaction is ballanced. OK
We need to know how many moles do we have from each compound.
Mass / Molar weight = Mol
Molar weight S = 32 g/m
Molar weight O₂ = 32 g/m
Molar weight NaOH = 40 g/m
Mol S: 2g/ 32g/m = 0.0625 mol
Mol O₂: 3g / 32 g/m = 0.09375 mol
Mol NaOH: 4g/ 40g/m = 0.1 mol
Now, we can play with the reactants. The base is: 2 moles of S, react with 3 mol of O₂ and 4 moles of hydroxide to make 2 moles of sulfate and 2 moles of water. Pay attention to the rules of three.
2 moles of S __ react with __ 3 moles of O₂ __ and __ 4 moles of NaOH
0.0625 moles S __________ 0.09375 moles O₂ ___ 0.125 moles NaOH
The limiting reactant is the NaOH. I need to use 0.125 moles and I only have 0.1 moles.
Let's do the same with O₂
3 moles of O₂ __ react with __ 2 moles of S __ and __ 4 moles of NaOH
0.09375 moles of O₂ _______ 0.0625 mol of S _____ 0.125 moles NaOH
The heat of vaporization of water at 100°c is 40.66 kj/mol. Calculate the quantity of heat that is absorbed/released when 5.00 g of steam condenses to liquid water at 100°c.
Answer : The quantity of heat released is -11.30 kJ
Explanation :
First we have to calculate the number of moles of water.
[tex]\text{Moles of water}=\frac{\text{Mass of water}}{\text{Molar mass of water}}[/tex]
Molar mass of water = 18 g/mole
[tex]\text{Moles of water}=\frac{5.00g}{18g/mole}=0.278mole[/tex]
Now we have to calculate the amount of heat released.
[tex]\Delta H=-\frac{q}{n}[/tex]
where,
[tex]\Delta H[/tex] = heat of vaporization = 40.66 kJ/mol
q = heat released = ?
n = number of moles of water = 0.278 mole
[tex]40.66kJ/mol=-\frac{q}{0.278mol}[/tex]
[tex]q=-11.30kJ[/tex]
In vaporization process, the amount of heat is absorbed but in the process of condensation the amount of heat is released.
Therefore, the quantity of heat released is -11.30 kJ
To calculate the quantity of heat absorbed/released when 5.00 g of steam condenses to liquid water at 100°C, use the equation Q = mL. Convert the mass of steam to moles, calculate the heat absorbed or released. Substitute the values and calculate the heat absorbed/released.
Explanation:To calculate the quantity of heat that is absorbed/released when 5.00 g of steam condenses to liquid water at 100°C, we can use the equation Q = mL, where Q is the heat absorbed or released, m is the mass of the substance, and L is the latent heat of vaporization. First, we need to convert the mass of steam to moles using the molar mass of water. Then, we can calculate the heat absorbed or released using the given latent heat of vaporization.
First, calculate the moles of water vapor:
moles = (mass of water vapor) / (molar mass of water)
Next, calculate the heat absorbed or released using the formula:
Q = (moles of water vapor) * (latent heat of vaporization)
Plugging in the given values, we get:
Q = (5.00 g / (molar mass of water)) * (latent heat of vaporization)
Finally, calculate the molar mass of water using the atomic masses of hydrogen and oxygen:
molar mass of water = (2 * atomic mass of hydrogen) + atomic mass of oxygen
Substitute the molar mass of water and the given latent heat of vaporization into the equation, and calculate the value of Q to find the quantity of heat absorbed/released.
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What type of compound involves the transfer of electrons?
Answer: Electrovalent or Ionic Compounds
Explanation:
Electrovalent Compounds Form bonds that are characterised by transfer of electrons from metallic atoms to non-metal licenses atoms during a chemical reaction.
The metallic atom after donating their valence electrons, become positively charged, while the non-metal license atoms becomes negatively charged after acquiring extra electrons.
A typical example of electrovalent compounds can be found between the association of Group 1(Alkali Metals) elements and the Group 7(Halogen Family) elements.
Answer: The type of compound involves the transfer of electrons is called the ionic compounds.
Explanation: ionic compounds are compounds in which one atom or molecule completely transfers an electron to another.
Ions that have gained an electron are negatively charged and they are called anions while ions that have lost an electron are positively charged and they are called cations.
Given the partial equation ? MnO4−+ ? SO32− → ? Mn2++ ? SO42− what must the coefficients be so that the electrons are balanced? Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4, where 1 indicates the absence of a coefficient).
Answer:
The coefficients should be: 2, 5, 2, 5
Explanation:
Given redox reaction: MnO₄⁻ + SO₃²⁻ → Mn²⁺+ SO₄²⁻
To balance the given redox reaction in acidic medium, the oxidation and the reduction half-reactions should be balanced first.
Reduction half-reaction: MnO₄⁻ → Mn²⁺
Oxidation state of Mn in MnO₄⁻ is +7 and the oxidation state of Mn in Mn²⁺ is +2. Therefore, Mn accepts 5e⁻ to get reduced from +7 to +2 oxidation state.
⇒ MnO₄⁻ + 5e⁻ → Mn²⁺
Now the total charge on reactant side is (-6) and the total charge on product side is +2. Therefore, to balance the total charge, 8H⁺ must be added to the reactant side.
⇒ MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺
To balance the number of hydrogen and oxygen atoms, 4H₂O must be added to the product side.
⇒ MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O .....equation 1
Oxidation half-reaction: SO₃²⁻ → SO₄²⁻
Oxidation state of S in SO₃²⁻ is +4 and the oxidation state of S in SO₄²⁻ is +6. Therefore, S loses 2e⁻ to get oxidized from +4 to +6 oxidation state.
⇒ SO₃²⁻ → SO₄²⁻ + 2e⁻
Now the total charge on reactant side is (-2) and the total charge on product side is (-4). Therefore, to balance the total charge, 2H⁺ must be added to the product side.
⇒ SO₃²⁻ → SO₄²⁻ + 2e⁻ + 2H⁺
To balance the number of hydrogen and oxygen atoms, 1 H₂O must be added to the reactant side.
⇒ SO₃²⁻ + H₂O → SO₄²⁻ + 2e⁻ + 2H⁺ .....equation 2
Now, to cancel the electrons transferred, equation (1) is multiplied by 2 and equation (2) is multiplied by 5.
Balanced Reduction half-reaction:
MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O ] × 2
⇒ 2MnO₄⁻ + 10e⁻ + 16H⁺ → 2Mn²⁺ + 8H₂O .....equation 3
Balanced Oxidation half-reaction:
SO₃²⁻ + H₂O → SO₄²⁻ + 2e⁻ + 2H⁺ ] × 5
⇒ 5SO₃²⁻ + 5H₂O → 5SO₄²⁻ + 10e⁻ + 10H⁺ .....equation 4
Now adding equation 3 and 4, to obtain the overall balanced redox reaction:
2MnO₄⁻ + 5SO₃²⁻ + 6H⁺ → 2Mn²⁺ + 5SO₄²⁻ + 3H₂O
Therefore, the coefficients should be: 2, 5, 2, 5
Using relative enthalpy and entropy values, determine how the process is affected after each of the following temperature or pressure changes. Consider that a more effective reaction produces more product or more product in a shorter amount of time.
Reaction: SO₂ (g) + 2H₂S (g) ↔ 3S(s) + 2H₂O (g)Substance | ΔG kJ/mol | ΔH kJ/molH₂O(g) | -228.6 | -241.8H₂O(l) | -237.1 | -285.8SO₂(g) | -300.4 | -296.9SO₃(g) | -370.4 | -395.2H₂S(g) | -33.01 | -20.17S(s) | 0 | 0Categorize into: "More Effective" ~ "Less Effective" ~ "Equally Effective"a. Temp. decreases while maintaining container sizeb. Temp. increases while maintaining container sizec. Pressure decreases while maintaining container sized. Pressure increases while maintaining container size
Changes in pressure and temperature can affect a reaction's effectiveness. A drop in temperature or a rise in pressure makes the given reaction more effective, while a rise in temperature or a drop in pressure makes it less effective.
Explanation:In the given reaction, the relative enthalpy and the entropy can provide insights on how the temperature and pressure changes can affect the reaction.
a. Temperature decreases: When temperature decreases, reactions that are exothermic (ΔH < 0) become more effective because the system attempts to increase its temperature by favoring the exothermic reaction. Therefore, in this case, considering ΔHSO₂(g) and ΔHH₂S(g) are both negative, the reaction is 'More Effective' at lower temperatures.b. Temperature increases: The increase in temperature generally favors the endothermic processes (ΔH > 0) and any process with positive ΔS (entropy). But in this case, the reaction is exothermic and creating more order (ΔS is negative because gases are forming a solid), so an increase in temperature would make the reaction 'Less Effective.'c. Pressure decreases: When pressure decreases, it favors the side of the equation with more gaseous moles. Here, there are 3 gas moles on the reactant side and 2 on the product side. Thus, reducing pressure would favor the reactant side, making the reaction 'Less Effective.'d. Pressure increases: Increasing pressure favors the side with fewer moles of gas, which is the product side in this case, so the reaction becomes 'More Effective.'Learn more about Chemical reaction effectiveness here:https://brainly.com/question/34566703
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More effective reaction conditions for the specified equilibrium include decreasing temperature and increasing pressure, leading to more product formation. Increasing temperature and decreasing pressure are less effective. Effectiveness is based on shifting equilibrium towards the product side.
The reaction in question is SO₂ (g) + 2H₂S (g) ↔ 3S(s) + 2H₂O (g). To determine the effectiveness of the reaction under different temperature and pressure changes, one must consider how each change affects the equilibrium. Here’s the analysis:
Temp. decreases while maintaining container size: As the reaction is exothermic (ΔH for H₂S and SO₂ are negative), decreasing the temperature shifts the equilibrium to the right, producing more products. Thus, this condition is More Effective.Temp. increases while maintaining container size: With an exothermic reaction, increasing the temperature shifts the equilibrium to the left, producing fewer products. Thus, this condition is Less Effective.Pressure decreases while maintaining container size: Since the number of gas molecules decreases from reactants to products (3 moles to 2 moles), decreasing pressure shifts the equilibrium to the left. Therefore, this condition is Less Effective.Pressure increases while maintaining container size: Increasing pressure favors the side with fewer gas molecules. Therefore, the equilibrium shifts to the right, making this condition More Effective.The internal energy of a system ________. refers only to the energies of the nuclei of the atoms of the component molecules is the sum of the kinetic energy of all of its components is the sum of the potential and kinetic energies of the components is the sum of the rotational, vibrational, and translational energies of all of its components none of the above
Answer:
The most appropriate answer here would be :
Internal energy of a system is the sum of the rotational, vibrational, and translational energies of all of its components
Explanation:
Internal energy of a system is the total energy the system possess. It is represented by U (I'll be referring to internal energy as U now). This option is particularly true for ideal gases. In ideal monoatomic gases, U is the sum of translational kinetic energies only. In di and polyatomic gases, U is the sum of translational and rotaional kinetic energies. Also, vibrational kinetic energies come into play as we increase the temperature and this also adds to U. But, in real substances such as real gases, solids, liquids, there is also interatomic forces and these accounts for intermolecular potential energies. Intermolecular potential energies also add to U in these type of systems. But even for real gases, under many circumstances, the intermolecular potential energy can be neglected.
So, the most appropriate answer here is: Internal energy of a system is the sum of the rotational, vibrational, and translational energies of all of its components
The internal energy of a system is the total of both kinetic and potential energies of its atoms and molecules. It includes various energy forms such as translational, vibrational, and rotational energy. Thus, option C is correct.
The internal energy of a system is the sum of all the kinetic and potential energies of its component atoms and molecules. This encompasses various forms of energy, including translational, vibrational, and rotational kinetic energy, as well as potential energy from molecular interactions and chemical bonds.
The correct choice is C. It is the sum of the potential and kinetic energies of the components.
Complete Question: -
The internal energy of a system ________.
A. refers only to the energies of the nuclei of the atoms of the component molecules
B. is the sum of the kinetic energy of all of its components
C. is the sum of the potential and kinetic energies of the components
D. is the sum of the rotational, vibrational, and translational energies of all of its components none of the above
What sublevels are filling across the transition elements
The transition elements fill d sublevels, coming after the s sublevel of the same principal energy level. Lanthanides begin filling the 4f sublevel after the 6s, positioned two principal energy levels behind. Many transition element compounds display bright colors from d electron transitions.
Explanation:The sublevels that are filling across the transition elements are primarily the d sublevels. The electron configurations of these elements have their outermost s sublevel either completely filled or missing one electron. However, the defining characteristic of transition elements is the filling of the inner d sublevel, which typically occurs after the s sublevel of the same principal energy level has been filled. When discussing the f-block elements, specifically the lanthanides, these elements begin filling their 4f sublevels after the 6s sublevel. This is due to the behaviour of electron filling, where the f sublevels are two principal energy levels behind the current one being filled.
Many transition element compounds are known for their brightly colored appearances as a result of inner-level d electron transitions. Unlike the transition elements, the lanthanides are not grouped together in the periodic table and instead are inserted in a separate row, reflecting their unique electron configurations and properties.
If 50.00 mL of 1.05 M sodium hydroxide is added to 25.00 mL of 1.88 M hydrochloric acid, with both solutions originally at 24.66°C, what will be the final solution temperature? (Assume that no heat is lost to the surrounding air and that the solution produced in the neutralization reaction has a density of 1.02 g/mL and a specific heat of 3.98 Jg⁻¹°C⁻¹.)
Final answer:
To find the final solution temperature, we need to calculate the heat exchanged by the reaction and the heat exchanged by the solution and set them equal to each other. By plugging in the given values and solving the equation, we find that the final solution temperature will be 24.77 °C.
Explanation:
To find the final solution temperature, we can use the principle that the heat given off by the reaction is equal to that taken in by the solution. We need to calculate the heat exchanged by the reaction and the heat exchanged by the solution and set them equal to each other.
First, we calculate the heat exchanged by the reaction using the equation:
q_reaction = C_reaction * ΔT_reaction
where C_reaction is the heat capacity of the reaction solution and ΔT_reaction is the change in temperature of the reaction.
Next, we calculate the heat exchanged by the solution using the equation:
q_solution = m_solution * C_solution * ΔT_solution
where m_solution is the mass of the solution, C_solution is the specific heat of the solution, and ΔT_solution is the change in temperature of the solution.
Now we can set the two heat exchanges equal to each other and solve for the final solution temperature:
q_reaction = q_solution
C_reaction * ΔT_reaction = m_solution * C_solution * ΔT_solution
Plugging in the given values:
C_reaction = C_solution = 3.98 Jg⁻¹°C⁻¹
m_solution = (50.00 mL of NaOH * 1.02 g/mL) + (25.00 mL of HCl * 1.02 g/mL) = 76.50 g
ΔT_reaction = (28.9 °C - 24.66 °C) = 4.24 °C
ΔT_solution = ?
Now we can solve for ΔT_solution:
3.98 Jg⁻¹°C⁻¹ * 4.24 °C = 76.50 g * 3.98 Jg⁻¹°C⁻¹ * ΔT_solution
ΔT_solution = (3.98 Jg⁻¹°C⁻¹ * 4.24 °C) / (76.50 g * 3.98 Jg⁻¹°C⁻¹) = 0.1107 °C
Finally, we calculate the final solution temperature:
Final Temperature = 24.66 °C + 0.1107 °C = 24.77 °C
The final temperature of the solution after the reaction is approximately 33.51°C.
To find the final temperature of the solution after the neutralization reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl), we can follow these steps:
The reaction between NaOH and HCl can be written as:
[tex]\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \][/tex]
Calculate the moles of NaOH and HCl:
[tex]\[ \text{Moles of NaOH} = 1.05 \, \text{M} \times 0.05000 \, \text{L} = 0.0525 \, \text{moles} \] \[ \text{Moles of HCl} = 1.88 \, \text{M} \times 0.02500 \, \text{L} = 0.0470 \, \text{moles} \][/tex]
Since HCl is the limiting reagent (0.0470 moles compared to 0.0525 moles of NaOH), the reaction will produce 0.0470 moles of water.
2.Calculate the heat released during the reaction:
The enthalpy change for the neutralization of strong acid and base (like HCl and NaOH) is typically [tex]\(-57.3 \, \text{kJ/mol}\).[/tex]
The total heat released q can be calculated as:
[tex]\[ q = \text{moles of HCl} \times \Delta H_{\text{neutralization}} \] \[ q = 0.0470 \, \text{moles} \times -57.3 \, \text{kJ/mol} = -2.6931 \, \text{kJ} = -2693.1 \, \text{J} \][/tex]
(The negative sign indicates that the heat is released, but we will use the magnitude for temperature calculation.)
3. Determine the total mass of the solution:
The total volume of the solution is:
[tex]\[ \text{Volume} = 50.00 \, \text{mL} + 25.00 \, \text{mL} = 75.00 \, \text{mL} \][/tex]
Given the density of the solution is 1.02 g/mL, the total mass (\(m\)) is:
[tex]\[ m = 75.00 \, \text{mL} \times 1.02 \, \text{g/mL} = 76.50 \, \text{g} \][/tex]
4.Calculate the temperature change:
[tex]\[ \Delta T = \frac{q}{mc} \] \[ \Delta T = \frac{2693.1 \, \text{J}}{76.50 \, \text{g} \times 3.98 \, {J/gC}} = \frac{2693.1}{304.47} \approx 8.85 \°C} \][/tex]
5.Calculate the final temperature:
The initial temperature of both solutions is 24.66°C. Thus, the final temperature [tex](\(T_f\))[/tex] is:
[tex]\[ T_f = 24.66 \, \°C} + 8.85 \, \°C} = 33.51 \, \°C} \][/tex]
So, the final temperature of the solution after the reaction is approximately 33.51°C.
Impure samples have melting point ranges that are both Blank 1. Fill in the blank, read surrounding text. and Blank 2. Fill in the blank, read surrounding text. compared to a pure sample.
Answer:
1 Lower
2 wider
Explanation:
It is lower and wider in range because impurities affects the crystalline lattice of sample structure theory causing a deviation from real melting point of pure sample.
Final answer:
Impure samples display a wider and lower melting point range compared to a pure sample due to melting point depression caused by impurities.
Explanation:
Impure samples have melting point ranges that are both wider and lower compared to a pure sample. This is due to the presence of impurities which cause a phenomenon known as melting point depression. When assessing the purity of a substance, melting point determination is crucial as a pure sample typically has a very narrow melting point range of 1 - 2 0C. In contrast, an impure sample will start melting at a lower temperature and continue to melt over a broader range, with the extent of this range depending on the amount and type of impurity present.
For example, if we examine the melting points of samples of benzoic acid contaminated with acetanilide, as the quantity of impurity increases, the onset of melting begins at a progressively lower temperature, and the breadth of the melting range expands. This makes the melting point range a valuable tool for a rough assessment of a sample's purity.
In cellular chemical pathways, the product(s) of any particular reaction are often quickly consumed by the next reaction in the pathway. This would tend to keep the product concentration ___ and drive the reaction ____.
Answer:
Tend to keep the product concetration low and therefore drive the reaction righward
Explanation:
The fact the products of a reaction are quickly consumed by the next one would tend to keep the product concetration low and therefore drive the reaction righward (to the products).
This happens because the system will not achive equilibrium between the reactants and the product, and will keep producing it util the system achives equilibrium or the reactants dry out.
A 20.0–milliliter sample of 0.200–molar K2CO3 solution is added to 30.0 milliliters of 0.400–molar Ba(NO3)2 solution. Barium carbonate precipitates. The concentration of barium ion, Ba2+, in solution after reaction is_________.
Answer:
[Ba^2+] = 0.160 M
Explanation:
First, let's calculate the moles of each reactant with the following expression:
n = M * V
moles of K2CO3 = 0.02 x 0.200 = 0.004 moles
moles of Ba(NO3)2 = 0.03 x 0.400 = 0.012 moles
Now, let's write the equation that it's taking place. If it's neccesary, we will balance that.
Ba(NO3)2 + K2CO3 --> BaCO3 + 2KNO3
As you can see, 0.04 moles of K2CO3 will react with only 0.004 moles of Ba(NO3) because is the limiting reactant. Therefore, you'll have a remanent of
0.012 - 0.004 = 0.008 moles of Ba(NO3)2
These moles are in total volume of 50 mL (30 + 20 = 50)
So finally, the concentration of Ba in solution will be:
[Ba] = 0.008 / 0.050 = 0.160 M
A certain substance X condenses at a temperature of 123.3 degree C . But if a 650. g sample of X is prepared with 24.6 g of urea ((NH2)2 CO) dissolved in it, the sample is found to have a condensation point of 124.3 degree C instead. Calculate the molal boiling point elevation constant Kb of X.Round your answer to 2 significant digits.
Answer:
The molal boiling point elevation constant is 1.59 ≈ 1.6 [tex]Kkgmol^{-1}[/tex]
Explanation:
To solve this question , we will make use of the equation ,
Δ[tex]T_{b} = i*K_{b} *m[/tex]
where ,
Δ[tex]T_{b}[/tex] is the change in boiling point of the substance [tex]X[/tex] ( °[tex]C[/tex] or [tex]K[/tex])[tex]i[/tex] is the Vant Hoff Factor which = 1 in this case ( no unit )[tex]K_{b}[/tex] is the mola boiling point elevation constant of X ( [tex]Kkgmol^{-1}[/tex])[tex]m[/tex] is the molality of the solution which has [tex](NH_{2})_{2} CO[/tex] as the solute and [tex]X[/tex] as the solution ([tex]molkg^{-1}[/tex])Δ[tex]T_{b}[/tex] = [tex]124.3 -123.3 = 1[/tex] °[tex]C[/tex] or [tex]K[/tex];[tex]i[/tex]=1;[tex]m[/tex]= [tex]\frac{moles of solute}{weight of solvent(kg)}[/tex][tex]molkg^{-1}[/tex]∴ [tex]m = \frac{\frac{24.6}{60} }{\frac{650}{1000} }[/tex]
as the weight of [tex](NH_{2})_{2} CO[/tex] is [tex]60g[/tex] and thus number of moles = [tex]\frac{24.6}{60}[/tex]and the weight of solvent in [tex]kg[/tex] is [tex]\frac{650}{1000}[/tex]4. [tex]K_{b}[/tex] ⇒ ?
∴
[tex]1=1*K_{b} *\frac{\frac{24.6}{60} }{\frac{650}{1000} }[/tex]
⇒ [tex]K_{b}[/tex] = [tex]1.59[/tex] ≈ 1.6 [tex]Kkgmol^{-1}[/tex]
The molal boiling point elevation constant (Kb) of substance X is 4.1.
Explanation:The molal boiling point elevation constant (Kb) can be calculated using the formula: ΔT = Kb × m
Where ΔT is the change in boiling point, Kb is the molal boiling point elevation constant, and m is the molality of the solution.
In this case, the change in boiling point (ΔT) is 1 degree C (124.3 - 123.3), the molality (m) can be calculated by dividing the molal mass of urea by the mass of the solvent water, which gives a value of 0.0246 kg urea / 0.100 kg water = 0.246 mol/kg, and the formula becomes: 1 = Kb × 0.246
Now, rearrange the equation to solve for Kb: Kb = 1 / 0.246 = 4.07
Rounding to 2 significant digits, the molal boiling point elevation constant Kb of substance X is 4.1.
What is the density of a block of marble that occupies 310 cm3 and has a mass of 853 g?1. Diamond has a density of 3.26 g/cm3 .2.What is the mass of a diamond that has a volume of 0.350 cm3 ?3. What is the volume of a sample of liquid mercury that has a mass of 76.2 g, given that the density of mercury is 13.6 g/mL?4. What is the density of a sample of ore that has a mass of 74.0 g and occupies 20.3 cm3 ?
Answer:53gm
Explanation:
To calculate the density of a substance, divide its mass by its volume. For the mass of a diamond or volume of mercury, multiply or divide, respectively, the given quantity by the substance's density. Densities are significant as they indicate how much matter is contained within a space.
The density of a substance is defined as its mass per unit volume. The formula for density (d) is d = mass (m) / volume (v), where the mass is measured in grams (g) and the volume in cubic centimeters (cm3) for solids and liquids, or in milliliters (mL) as 1 mL equals 1 cm3.
To find the density of a block of marble, we use the formula with the given values: d = 853 g / 310 cm3.
To find the mass of a diamond with a known density, multiply the volume by the density: mass = 0.350 cm3
x 3.26 g/cm3.
The volume of liquid mercury given its mass and density can be calculated by rearranging the formula: volume = 76.2 g / 13.6 g/mL.
Lastly, to find the density of a sample of ore, apply the formula: d = 74.0 g / 20.3 cm3.
Remember that densities can vary greatly among different materials and are particularly high for substances such as gold and mercury.
A voltaic cell is constructed with an Ag/Ag+ half-cell and a Pb/Pb2+ half-cell. Measurement shows that the silver electrode is positive.
(a) Write balanced half-reactions and the overall spontaneous reaction. (Type your answer using the format [NH4]+ for NH4+. Use the lowest possible coefficients.)
reduction:
__Ag+(aq) + __e- -> ___(s)
oxidation:
__(s) ->__Pb2+(aq) +__e-
overall reaction:
__Ag+(aq) +__(s) ->__(s) + __ Pb2+(aq)
(b) The cation flow is towards which electrode?
lead OR silver
What process occurs at the lead electrode?
oxidation OR reduction
In which direction do the electrons flow?
from the lead electrode to the silver electrode OR from the silver electrode to the lead electrode
Answer:
a) Reduction:
Ag⁺(aq) + e⁻ → Ag(s)
Oxidation:
Pb(s) → Pb⁺²(aq) + 2e⁻
Overall reaction:
2Ag⁺(aq) + Pb(s) → 2Ag(s) + Pb²⁺
b) Silver; oxidation; from the lead electrode to the silver electrode.
Explanation:
a) Ag⁺ had lost 1 electron, so need to gain 1 electron to become Ag(s). Pb needs to lose 2 electrons to become Pb⁺².
Reduction:
Ag⁺(aq) + e⁻ → Ag(s)
Oxidation:
Pb(s) → Pb⁺²(aq) + 2e⁻
Overall reaction:
2Ag⁺(aq) + Pb(s) → 2Ag(s) + Pb²⁺ (it will need 2Ag⁺ to gaind the 2 electrons released by Pb)
b) The cation formed in the redox reaction is Pb²⁺, so, to equilibrate the charges, it will flow towards the silver (Ag) electrode.
The lead (Pb) is being oxidized, so oxidation is happening at it.
The electrons flow from the oxidation (anode) to the reduction (cathode), so they flow from the lead electrode to the silver electrode.
In the voltaic cell, the Ag+ is reduced to Ag in the silver half-cell, while Pb is oxidized to Pb2+ in the lead half-cell. The cations flow towards the silver electrode and the electrons flow from the lead to the silver electrode. Hence, the overall reaction is 2Ag+(aq) + Pb(s) -> 2Ag(s) + Pb2+(aq).
Explanation:In a voltaic cell with an Ag/Ag+ half-cell and a Pb/Pb2+ half-cell, the silver half-cell acts as the cathode or reduction half-cell which gains electrons, while the lead half-cell acts as the anode or oxidation half-cell and loses electrons. Therefore, the balanced half-reactions and overall spontaneous reactions are:
(a) Balanced Half-Reactions and Overall Reaction:
Reduction: Ag+(aq) + 1e- -> Ag(s)
Oxidation: Pb(s) -> Pb2+(aq) + 2e-
Overall Reaction: 2Ag+(aq) + Pb(s) -> 2Ag(s) + Pb2+(aq)
(b) The Cation Flow and Electrons Flow:
The cation flow is towards the silver electrode and the electron flow is from the lead electrode to the silver electrode. In the voltaic cell, the process that occurs at the lead electrode is oxidation.
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