Answer : The heat released per gram of the compound reacted with oxygen is, 71.915 kJ
Solution :
The balanced chemical reaction is,
[tex]2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)[/tex]
The expression for enthalpy change is,
[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]
[tex]\Delta H=[(n_{H_2O}\times \Delta H_{H_2O})+(n_{B_2O_3}\times \Delta H_{B_2O_3})]-[(n_{B_5H_9}\times \Delta H_{B_5H_9})+(n_{O_2}\times \Delta H_{O_2})][/tex]
where,
n = number of moles
Now put all the given values in this expression, we get
[tex]\Delta H=[(9\times -285.83)+(5\times -1271.94)]-[(2\times 73.2)+(12\times 0)]\\\\\Delta H=-9078.57kJ[/tex]
Now we have to calculate the heat released per gram of the compound reacted with oxygen.
As we know that,
1 mole of [tex]B_5H_9[/tex] has 63.12 grams of mass
So, 2 mole of [tex]B_5H_9[/tex] has [tex]2\times 63.12=126.24[/tex] grams of mass
As, 126.24 g of [tex]B_5H_9[/tex] release heat = 9078.57 kJ
So, 1 g of [tex]B_5H_9[/tex] release heat = [tex]\frac{9078.57}{126.24}=71.915kJ[/tex]
Therefore, the heat released per gram of the compound reacted with oxygen is, 71.915 kJ
The heat released per gram of the compound reacted with oxygen is -71.92 KJ/mol per gram of B5H9 reacted.
The equation goes as follows;
2B5H9(l) + 12O2(g) → 5B2O3(s) + 9H2O(l)
We have the following information;
ΔH°f B5H9(l) = 73.2 kJ/mol
ΔH°f B2O3(s) = −1271.94 kJ/mol
ΔH°f H2O(l) = −285.83 kJ/mol
Note that;
ΔHrxn = ∑ΔH°f (products) - ΔH°f (reactants)
ΔHrxn = ∑(5 × ( −1271.94 kJ/mol)) + (9 × ( −285.83 kJ/mol)) - ∑(2 × (73.2 kJ/mol) + (12 × 0)
ΔHrxn = -9078.57 kJ/mol
Since 1 mole of B5H9 = 63.12 g/mol
Two moles of B5H9 reacted so 2 moles × 63.12 g/mol = 126.24 g
Heat released per gram of B5H9 reacted = -9078.57 kJ/mol/126.24 g
= -71.92 KJ/mol per gram of B5H9 reacted.
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A mixture of gases A2 and B2 are introduced to a slender metal cylinder that has one end closed and the other fitted with a piston that makes a gas-tight seal so that the gases are a closed system. The cylinder is submerged in a large beaker of water whose temperature is 25∘C, and a spark is used to trigger a reaction in the cylinder. At the completion of the reaction, the piston has moved downward, and the temperature of the water bath has increased to 28∘C. If we define the system as the gases inside the cylinder, which of the following best describes the signs of q, w, and ΔE for this reaction?
(a) q < 0, w < 0, ΔE < 0
(b) q < 0, w > 0, ΔE < 0
(c) q < 0, w > 0, the sign of ΔE cannot be determined from the information given
(d) q > 0, w > 0, ΔE > 0
(e) q > 0, w < 0, the sign of ΔE cannot be determined from the information given
Answer:
q < 0, w > 0, the sign of ΔE cannot be determined from the information given
Explanation:
Determination of sign of q
Temperature of the water bath before the reaction = 25 °C
Temperature of the water bath after the completion of the reaction = 28 °C
After the completion of the reaction, temperature of the water bath is increased that means heat is released during the reaction and flows out of the system.
If heat is absorbed by the system, then q is indicated by positive sign and if heat is released by the system, then q is indicated by negative sign.
As in the given case, heat is released by the system, so sign of q is negative, or q < 0
Determination of sign of w
After the completion of the reaction, piston moved downward, that means volume of the system decreases or compression occur. During the compression, work is done on the system.
if work is done on the system, sign of w is positive.
If work is done by the system, sign of w is negative.
In the given case, work is done on the system, therefore sign of w is positive, or w > 0
Determination of sign of ΔE
Relationship between ΔE, q and w is given by first law of thermodynamics:
ΔE = q + w
In this case, q is positive and w is negative, so the sign of ΔE depends of magnitude of q and w. As magnitude of w and q cannot be determined in this case, thus, the given information is insufficient for the determination of sign of ΔE.
So, among the given option, option c is correct.
q < 0, w > 0, the sign of ΔE cannot be determined from the information given
The best option that describes the signs of q, w, and ΔE for this reaction is q < 0, w > 0, the sign of ΔE cannot be determined from the information given.
Option C is correct.
From the given information, if we examine the mixture of the gases A2 and B2 and take them to be the system, and the slender metal cylinder as well as the water bath to be the surroundings.
Then, we can infer that:
if there is a movement of heat flow(q) into the system, it signifies a positive signif there is an outflow of heat (q) out of the system, it signifies a negative signHence, when the reaction is said to be completed, we are being informed from the question that the temperature increased. This implies that there is an outflow and release of heat away from the system. As such, when heat flows away from the system, the sign (q) will be negative.
For the work done (w):
suppose work is done on a system show that the sign will be positive;and suppose the work is done by the system, then it will be negative.When the reaction goes into completion, the movement of the piston downward indicates that there is a decrease in the volume of the system.
As such, work is done on the system, hence, the work done (w) is positive.
For the change in the internal energy ΔE. If we look at the first law of thermodynamics, we know that:
ΔE = q + w
here,
we know q to be -ve and w to be +ve;However, the sign of ΔE largely depends on the magnitude of q and w. We can infer that there is little information given for the determination of ΔE
Therefore, we can conclude that q < 0 since it is negative, w > 0 since it is positive and the sign of ΔE cannot be determined from the information given.
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At a certain temperature, Kc = 0.0500 and ∆H = +39.0 kJ for the reaction below, 2 MgCl2(s) + O2(g) → 2MgO(s) + Cl2(g) Calculate Kc for the reaction, MgO(s) + ½ Cl2(g) → MgCl2(s) + ½ O2(g) and indicate whether the value of Kc will be larger or smaller at a lower temperature.
Explanation:
Since, it is shown that the reaction has been reversed. Therefore, value of [tex]K_{c}[/tex] will become [tex]\frac{1}{K_{c}}[/tex].
Hence, new [tex]K_{c'} = \frac{1}{K_{c}}[/tex]
= [tex]\frac{1}{0.0500}[/tex]
= 20
Also, the number of moles of each reactant has been halved. So, [tex]K_{c''}[/tex] for the reaction [tex]MgO(s) + \frac{1}{2}Cl2(g) → MgCl_{2}(s) + \frac{1}{2} O2(g)[/tex] will also get halved.
Therefore, [tex]K_{c''}[/tex] = [tex]K_{c'}[/tex] = [tex](20)^{0.5}[/tex]
= 4.47
As the value of [tex]\Delta H[/tex] is given as +39.0 kJ. So, it means that the reaction is endothermic in nature. So, energy of reactants will be more than the products. Hence, according to Le Chatelier's principle reaction will move in the forward direction.
As a result, [tex]K_{c}[/tex] will also increase with increase in temperature.
The value of Kc for the reaction MgO(s) + ½ Cl2(g) → MgCl2(s) + ½ O2(g) is 1.00. The value of Kc will be smaller at a lower temperature for an exothermic reaction.
Explanation:To calculate Kc for the reaction MgO(s) + ½ Cl2(g) → MgCl2(s) + ½ O2(g), we need to use the equation Kc = K1*K2, where K1 is the equilibrium constant for the forward reaction and K2 is the equilibrium constant for the reverse reaction. In this case, K1 is equal to 0.0500, as given, and K2 can be found using the equation K2 = 1/K1. So, K2 = 1/0.0500 = 20. Therefore, Kc = 0.0500 * 20 = 1.00.
To determine whether the value of Kc will be larger or smaller at a lower temperature, we need to consider the effect of temperature on the equilibrium constant. In general, for exothermic reactions, the equilibrium constant decreases as the temperature decreases. Since the reaction given is exothermic (∆H = +39.0 kJ), the value of Kc will be smaller at a lower temperature.
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Identify the balanced chemical equation that represents a single displacement reaction. CF4 2Br2 ⟶ CBr4 2F2 3H2SO4 2Al ⟶ Al2(SO4)3 3H2 3H2SO4 2Al ⟶ Al2(SO4)3 H2 CF4 Br2 ⟶ ⟶ CBr4 F2
Answer:
3H₂SO₄ + 2Al₂(SO₄)₃ → Al₂(SO₄)₃ + 3H₂
Explanation:
3H₂SO₄ + 2Al₂(SO₄)₃ → Al₂(SO₄)₃ + 3H₂
In this type of reaction, one substance is replacing another:
A + BC → AC + B
In a single displacement reaction, atoms replace one another based on the activity series. Elements that are higher in the activity series. Also, if the element that is to replace the other in a compound is more reactive the reaction will occur. If it is less reactive, there will be no reation.
In the first equation, fluorine is more reactive than bromine. Therefore, bromine cannot replace bromine.
In the second equation, the displacement is between hydrogen and aluminium. Hydrogen is lower in the activity series, this implies that aluminum will replace it.
The balanced equation for a single displacement reaction is 3H2SO4 + 2Al
ightarrow Al2(SO4)3 + 3H2, where aluminum displaces hydrogen in sulfuric acid to form aluminum sulfate and hydrogen gas.
To identify the balanced chemical equation that represents a single displacement reaction, we can look at the provided options. A single displacement reaction is a type of chemical reaction where one element is replaced by another in a compound. We can consider the following valid general form for a single displacement reaction: A + BC
ightarrow B + AC, where A replaces B in the compound BC.
Now, the correct balanced equation among the provided examples that represents a single displacement reaction is:
3H2SO4 + 2AlThis equation follows the pattern of a single displacement reaction, where aluminum (Al) displaces the hydrogen (H) in sulfuric acid (H2SO4) to form aluminum sulfate (Al2(SO4)3) and hydrogen gas (H2). The equation is also balanced, with equal numbers of each type of atom on both sides of the equation.
Aluminum metal and chlorine gas react to from aluminum chloride 2Al(s) + 3Cl2(g) --> Al2Cl6(s) How many grams of Al2Cl6 can be produced from the reaction of 2.67 mol aluminum and 3.26 mol Cl2?
Answer:
290.1 grams.
Explanation:
2 Al + 3Cl2 ---> Al2Cl6
2 moles of Al react with 3 moles of Cl2 to form 1 mole of Al2Cl6
So 2.67 moles Al reacts with 3/2 * 2.67 moles Cl2 = 4 moles of chlorine. But we only have 3.26 moles of Cl2 so this will react with only part of the available Al.
3.26 reacts with 2/3 * 3.26 = 2.173 moles of aluminium.
So the number of moles of Al2Cl6 that can be produced = 2.173 * 1/2
= 1.087 moles.
= 1.087 * (2*26.98 + 6*35.45)
= 290.1 grams.
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The main component of smog is NO2(g). Smog is a product that is formed through the following series of intermediate chemical
reactions.
N2(g) + O2(g)—>2NO(g)
2NO(g) + O2(g)—>2NO2(g)
NON(g) light_» NO(g) + O(g)
O2(g)+O(g)—>Os(9)
NO(g) + O2(g)—>NO2(g) + O2(g)
What is the overall chemical equation for smog after the above intermediate reactions are combined?
N2(g)+302(9)+203(g)+20(g) NO(g)+ 4NO2(g) +203(9)
N2(g) + 302(g) + O2(g)+O() — 9NO(g)+3NO2(g) + Os(g)
16N2(g) + 302(g) → 3N02(0)
N2(g) +202(g) → 2NO2(g)
Answer:
N₂(g) + 2O₂(g) → 2NO₂(g).
Explanation;
Photo-chemical smog is considered to be a new type of air pollution that is basically a mixture of pollutants that are formed when volatile compounds of organic nature (Hydrocarbons) react and Nitrogen oxides (NOx) react with solar radiations of sun and lead to the formation of photochemical smog that looks like a brown haze in the skies of cities.
pounds (VOCs) react to sunlight, creating a brown haze above cities. It tends to occur more often in summer, because that is when we have the most sunlight.
As mentioned in the question, the formation of smog involves a series of reactions:
N2(g) + O2(g)—>2NO(g)
2NO(g) + O2(g)—>2NO2(g)
NON(g) light_» NO(g) + O(g)
O2(g)+O(g)—>Os(9)
NO(g) + O2(g)—>NO2(g) + O2(g)
If we combine all these chemical equations, it gives an answer as:
N₂(g) + 2O₂(g) → 2NO₂(g).
This is because main components involve in the formation of smog is initial N2 and 2 molecules of O2.
Hope it helps!
Answer:
D
Explanation:
Edge 2020
Consider the combustion reaction for octane (C8H18), which is a primary component of gasoline. 2C8H18+25O2⟶16CO2+18H2O How many moles of CO2 are emitted into the atmosphere when 28.6 g C8H18 is burned?
To find out the number of moles of CO2 produced from 28.6 g of octane, convert the mass of octane to moles and then apply the stoichiometry of the balanced equation. This results in 2 moles of CO2 being emitted.
Explanation:The question asks how many moles of CO2 are produced when 28.6 g of C8H18 is burned according to the balanced chemical equation 2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g). The first step in this problem is to convert the given mass of octane to moles using its molar mass (114.23 g/mol for C8H18). With 28.6g of C8H18, we calculate the moles of octane:
moles of C8H18 = 28.6 g / 114.23 g/mol = 0.25 moles of C8H18
Next, we use the stoichiometric ratio from the balanced equation, which shows that 2 moles of C8H18 produce 16 moles of CO2. Therefore, for 0.25 moles of octane, the moles of CO2 produced are calculated as:
moles of CO2 = (0.25 moles C8H18/1) × (16 moles CO2/2 moles C8H18) = 2 moles CO2
So, 28.6 g of octane will produce 2 moles of CO2 when completely burned.
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Which of the following results from acid deposition? View Available Hint(s) Which of the following results from acid deposition? Chlorofluorocarbons are formed near the ground. Species diversity is increased. Objects made of stone, like marble headstones, are worn away. Food supply for birds is increased.
Answer:
Objects made of stone, like marble headstones, are worn away.
Explanation:
Acid deposition on earth is mostly as a result of acid rain. Acid rain forms which oxides of non-metals such as nitrogen and carbon dissolves in rain water to produce weak acid.
This weak acid that is produced easily wears away materials made up of marble and other sculptures. The most active of the acid is weak carbonic acid produced by dissolution of carbon dioxide in rain water.
Acid deposition leads to the deterioration of objects made of stone, such as marble headstones, because of the corrosion of metals and the wearing away of stone due to acidic precipitation.
Acid deposition results from the chemical reaction of air pollutants, like sulfur dioxide (SO2) and nitrogen oxides (NOx), with the atmosphere to form acids such as nitric and sulfuric acids. These acidic compounds can fall to Earth in various forms, either as wet precipitation like rain, sleet, or snow, or as dry particles. The impacts of acid deposition are numerous and widespread, including ecological and material damage.
One of the notable consequences of acid deposition is the deterioration of objects made of stone, such as marble headstones. Acid rain and acid particles contribute to the corrosion of metals and the deterioration of paint and stone, including culturally significant statues, monuments, and other stonework. This corrosive effect leads to the loss of structural integrity and the diminishment of aesthetic and historical value.
Therefore, among the listed options, the result of acid deposition is that objects made of stone, like marble headstones, are worn away.
Be sure to answer all parts. Nitric oxide (NO) reacts with molecular oxygen as follows: 2NO(g) + O2(g) → 2NO2(g) Initially NO and O2 are separated in two different chambers connected by a valve. When the valve is opened, the reaction quickly goes to completion. Determine what gases remain at the end and calculate their partial pressures. Assume that the temperature remains constant at 25°C. Initial conditions are as follows: NO: 3.90 L, 0.500 atm O2: 2.09 L, 1.00 atm
Answer:
The remain gases are [tex]o_{2}_{(g)}[/tex] and [tex]NO_{2}_{(g)}[/tex]
Pressure of [tex]O_{2}_{(g)}[/tex] [tex]1.09 atm O_{2}_{(g)}[/tex]
Pressure of [tex]NO_{2}_{(g)}[/tex] [tex]1.09 atm NO_{2}_{(g)}[/tex]
Explanation:
We have the following reaction
[tex]2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}[/tex]
Now we calculate the limit reagents, to know which of the two gases is completely depleted and which one is in excess.
Excess gas will remain in the tank when the reagent limits have run out and the reaction ends.
To calculate the limit reagent, we must calculate the mols of each substance. We use the ideal gas equation
[tex]PV= nRT[/tex]
We cleared the mols
[tex]n=\frac{PV}{RT}[/tex]
PV=nrT
replace the data for each gas
Constant of ideal gases
[tex]R= 0.082\frac{atm.l}{mol.K}[/tex]
Transform degrees celsius to kelvin
[tex]25+273=298K[/tex]
[tex]NO_{g}[/tex]
[tex]n=\frac{0.500atm.3.90l}{298k.0.082\frac{atm.l}{k.mol} } \\ \\ n=0.080mol NO_{(g)} \\[/tex]
[tex]O_{2}_{(g)[/tex]
[tex]n=\frac{1atm.2.09l}{298k.0.082\frac{atm.l}{k.mol} } \\ \\ n=0.086mol O_{2}_{(g)} \\[/tex]
Find the limit reagent by stoichiometry
[tex]2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}[/tex]
[tex]0.086mol O_{2}_(g).\frac{2mol NO_{(g)} }{1mol O_{2}_{(g)} } =0.17mol NO_{(g)}[/tex]
Using [tex]O_{2}_{(g)}[/tex]as the limit reagent produces more [tex]NO_{(g)}[/tex] than I have, so oxygen is my excess reagent and will remain when the reaction is over.
[tex]NO_{(g)}[/tex]
[tex]2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}\\ \\ 0.080mol NO_{(g)}.\frac{1mol O_{2}_ {(g)} }{2mol NO_{(g)} } =0.04mol O_{2}_{(g)}[/tex]
Using [tex]NO_{(g)}[/tex] as the limit reagent produces less [tex]O_{2}_{(g)}[/tex] than I have, so [tex]NO_{(g)}[/tex] is my excess reagent and will remain when the reaction is over.
Calculate the moles that are formed of [tex]NO_{2}_{(g)}[/tex]
[tex]2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}\\ \\ 0.080mol NO_{(g)}.\frac{2mol NO_{2}_ {(g)} }{2mol NO_{(g)} } =0.080mol NO_{2}_{(g)}[/tex]
We know that for all [tex]NO_{(g)}[/tex] to react, 0.04 mol [tex]O_{2}_{(g)}[/tex] is consumed.
we subtract the initial amount of [tex]O_{2}_{(g)}[/tex] less than necessary to complete the reaction. And that gives us the amount of mols that do not react.
[tex]0.086-0.04= 0.046[/tex]
The remain gases are[tex]O_{2}_{(g)}[/tex] and [tex]NO_{2}_{(g)}[/tex]
calculate the volume that gases occupy
[tex]0.080 mol NO_{2}_{(g)} .\frac{22.4l NO_{2}_{(g)} }{1molNO_{2}_{(g)} }} =1.79 lNO_{2}_{(g)}[/tex]
[tex]0.046 mol O_{2}_{(g)} .\frac{22.4 l O_{2}_{(g)} }{1molO_{2}_{(g)} }} =1.03 l O_{2}_{(g)}[/tex]
Calculate partial pressures with the ideal gas equation
[tex]PV= nRT[/tex]
[tex]P=\frac{nRT}{V}[/tex]
Pressure of [tex]O_{2}_{(g)}[/tex]
[tex]P=\frac{0.046mol.0.082\frac{atm.l}{K.mol} 298K}{1.03l}= 1.09 atmO_{2}_{(g)}[/tex]
Pressure of [tex]NO_{2}_{(g)}[/tex]
[tex]P=\frac{0.080mol.0.082\frac{atm.l}{K.mol} 298K}{1.79l}= 1.09 atmNO_{2}_{(g)}[/tex]
Answer:
It will remain O₂ and NO₂, with partial pressures: pO₂ = 0.186 atm, and pNO₂ = 0.325 atm.
Explanation:
First, let's identify the initial amount of each reactant using the ideal gas law:
PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas equation (0.082 atm.L/mol.K), and T is the temperature (25°C + 273 = 298 K).
n = PV/RT
NO:
n = (0.500*3.90)/(0.082*298)
n = 0.0798 mol
O₂:
n = (1.00*2.09)/(0.082*298)
n = 0.0855 mol
By the stoichiometry of the reaction, we must found which reactant is limiting and which is in excess. The limiting reactant will be totally consumed. Thus, let's suppose that NO is the limiting reactant:
2 moles of NO ------------------ 1 mol of O₂
0.0798 mol ------------------- x
By a simple direct three rule:
2x = 0.0798
x = 0.0399 mol of O₂
The number of moles of oxygen needed is lower than the number of moles in the reaction, so O₂ is the limiting reactant, and NO will be totally consumed. The number of moles of NO₂ formed will be:
2 moles of NO --------------- 2 moles of NO₂
0.0798 mol ---------------- x
By a simple direct three rule:
x = 0.0798 mol of NO₂
And the number of moles of O₂ that remains is the initial less the total that reacts:
n = 0.0855 - 0.0399
n = 0.0456 mol of O₂
The final volume will be the total volume of the containers, V = 3.90 + 2.09 = 5.99 L, so by the ideal gas law:
PV = nRT
P = nRT/V
O₂:
P = (0.0456*0.082*298)/5.99
P = 0.186 atm
NO₂:
P = (0.0798*0.082*298)/5.99
P = 0.325 atm
The recommended procedure for preparing a very dilute solution is not to weigh out a very small mass or measure a very small volume of stock solution. Instead, it is done by a series of dilutions. A sample of 0.6597 g of KMnO4 was dissolved in water and made up to the volume in a 500.0−mL volumetric flask. A 2.000−mL sample of this solution was transferred to a 1000−mL volumetric flask and diluted to the mark with water. Next, 10.00 mL of the diluted solution was transferred to a 250.0−mL flask and diluted to the mark with water. (a) Calculate the concentration (in molarity) of the final solution. Enter your answer in scientific notation.
Answer: [tex]6.4\times 10^{-7}M[/tex].
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]
where,
n= moles of solute
[tex]Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{0.6597g}{158g/mol}=0.004mole[/tex]
[tex]V_s[/tex] = volume of solution in ml = 500 ml
[tex]Molarity=\frac{0.004\times 1000}{500}=0.008M[/tex]
According to the dilution law:
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of stock solution = 0.008 M
[tex]V_1[/tex] = volume of stock solution = 2 ml
[tex]M_2[/tex] = molarity of resulting solution = ?
[tex]V_2[/tex] = volume of resulting solution = 1000 ml
[tex]0.008\times 2 ml=M_2\times 1000ml[/tex]
[tex]M_2=0.000016M[/tex]
According to the dilution law:
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of stock solution = 0.000016 M
[tex]V_1[/tex] = volume of stock solution = 10 ml
[tex]M_2[/tex] = molarity of resulting solution = ?
[tex]V_2[/tex] = volume of resulting solution = 250 ml
[tex]0.000016\times 10 ml=M_2\times 250ml[/tex]
[tex]M_2=0.00000064M=6.4\times 10^{-7}M[/tex]
Scientific notation is defined as the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.
Therefore, the concentration of of the final solution is [tex]6.4\times 10^{-7}M[/tex].
A solution is prepared by dissolving 0.6597 g of KMnO₄ in a 500.0 mL volumetric flask and then is subjected to 2 successive dilutions. The final molarity of the solution is 6.680 × 10⁻⁷ M.
A solution is prepared by dissolving 0.6597 g of KMnO₄ in a 500.0 mL volumetric flask and then is subjected to 2 successive dilutions.
The initial molarity of the solution is:
[tex]M = \frac{mass\ solute }{molar\ mass\ solute \times liters\ solution} = \frac{0.6597g}{158.03g/mol \times 0.5000L } = 8.349 \times 10^{-3} M[/tex]
First dilutionA 2.000−mL (V₁) sample of this solution was transferred to a 1000−mL (V₂) volumetric flask and diluted to the mark with water. We can calculate the final concentration using the dilution rule.
[tex]C_2 = \frac{C_1 \times V_1 }{V_2} = \frac{(8.349 \times 10^{-3}M )\times 2.000mL }{1000mL} = 1.670 \times 10^{-5}M[/tex]
Second dilution10.00 mL (V₁) of the diluted solution was transferred to a 250.0−mL (V₂) flask and diluted to the mark with water. We can calculate the final concentration using the dilution rule.
[tex]C_2 = \frac{C_1 \times V_1 }{V_2} = \frac{(1.670 \times 10^{-5}M )\times 10.00mL }{250.0mL} = 6.680 \times 10^{-7}M[/tex]
A solution is prepared by dissolving 0.6597 g of KMnO₄ in a 500.0 mL volumetric flask and then is subjected to 2 successive dilutions. The final molarity of the solution is 6.680 × 10⁻⁷ M.
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A balloon with a volume of 38.2 L at 91 psi is allowed to expand into a larger can with a total volume of 77.4 L. What is the pressure in psi of the gas in the larger container?
Answer:
45 psiExplanation:
1) Data:
a) V₁ = 38.2 liter
b) P₁ = 91 psi
c) V₂ = 77.4 liter
d) P₂ = ?
2) Formula:
According to Boyle's law, at constant temperature, the pressure and volume of a fixed amount of gas are inversely related:
PV = constant ⇒P₁V₁ = P₂V₂3) Solution:
Solve for the unknown: P₂ = P₁V₁ /v₂Substitute the values: V₂ = 91 psi × 38.2 liter / 77.4 liter = 44.9 9si ≈ 45 psi.Hydrogen fuel cells burn with _______ and produce _______.
Hydrogen fuel cells burn with Oxygen and produces Water.
It is commonly used to produce electricity.
Answer:
with oxygen to form water
Explanation:
In a flame of pure hydrogen gas, burning in air, the hydrogen (H2) reacts with oxygen (O2) to form water (H2O) and releases heat. If carried out in atmospheric air instead of pure oxygen (as is usually the case), hydrogen combustion may yield small amounts of nitrogen oxides, along with the water vapor.
Horizontal rows of the periodic table are called
Horizontal rows of the periodic table are called periods.
Periods are the horizontal rows of the periodic table that correspond to successive layers of electron shells surrounding the atomic nucleus. The atomic number rises as one moves over a period from left to right, adding more protons and electrons in the process.
The elements' characteristics gradually change as a result of this process, changing in reactivity and with an increase in atomic radius. While all elements in the same period have the same number of electron shells, the amounts of valence electrons in each element cause the elements' chemical characteristics to vary.
Which one of the following substances has three unpaired d electrons? Which one of the following substances has three unpaired d electrons? [Cu(NH3)4]2+ [V(H2O)6]4+ [Ag(NH3)2]+ [Zn(NH3)4]2+ [Cr(CN)6]3-
Answer:
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Explanation:
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how would you use a grignard reaction of an aldehyde or ketone to synthesize 2-pentanol? draw the grignard reagent and the aldehyde or ketone you would use below.
You start from butanal which will react with methyl magnesium bromide (Gringard reagent) forming a compound with a new carbon-carbon bond. After that using an acid media the magnesium bromide salt is eliminated with formation of the 2-pentanol.
Calculate the total mass of the protons and electrons in 19 9F. Use 1.007825 amu as the mass of 11H (mass of a proton and an electron). Express your answer in atomic mass units to 6 significant figures.
Explanation:
Atomic mass means the sum of total number of protons and neutrons present in an atom.
When an atom is neutral then number of protons equal the number of electrons.
Mass of a proton equals to 1.007276 u, mass of neutron equals to 1.008664 u.
For example, in a [tex]^{19}_{9}F[/tex] atom there are total 9 protons and number of neutrons is 19 - 9 = 10.
Since, it is a neutral atom so number of electrons will also be 9.
So, total mass will be calculated as follows.
Total mass = 9 (mass of electron + mass of proton)
= 9 (1.007825 u)
= 9.07043 u
Therefore, we can conclude that total mass of protons and electrons in [tex]^{19}_{9}F[/tex] is 9.07043 u.
The total mass of the protons and electrons in 19 9F is approximately 9.07065 amu. This calculation is made by multiplying the number of each particle by their respective masses and summing the results.
Explanation:In order to calculate the total mass of the protons and electrons in 19 9F, you first need to understand that the number '9' represents the atomic number, which is the number of protons in the atom, and hence, in a neutral atom, the number of electrons as well. The mass of a proton is roughly 1.0073 amu and the mass of an electron is approximately 0.00055 amu. With 9 protons and 9 electrons, the total mass would be (9 protons * 1.0073 amu/proton)+(9 electrons * 0.00055 amu/electron).
So, the mass of the protons and electrons is equal to 9.0657 amu for the protons and 0.00495 amu for the electrons. Adding these together gives a total mass of approx 9.07065 amu.
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You are performing an experiment in a lab to attempt a new method of producing pure elements from compounds. The only problem is that you do not know what element will form. By your previous calculations you know that you will have 3 moles of product. When it is complete, you weigh it and determine you have 209.1 grams. What element have you produced?
You need to calculate the atomic weight of that element.
number of moles = mass / atomic weight
atomic weight = mass / number of moles
atomic weight = 209.1 / 3 = 69.7 g/mol
The element produced will be gallium, Ga.
Answer:
The element produced is gallium, Ga
Explanation:
Given:
Moles of the product = 3 moles
Weight of the product = 209.1 g
To determine:
The identity of the product formed i.e. the unknown element
Calculation:
The identity of element can be deduce from its atomic weight and comparing the calculated weight to that of the elements in the periodic table.
[tex]Number\ of\ moles = \frac{mass\ of\ the\ element}{atomic\ weight}[/tex]
[tex]Atomic\ weight = \frac{mass\ of\ the\ element}{number\ of\ moles}[/tex]
In this case:
[tex]Atomic\ weight = \frac{209.1\ g}{3\ moles} = 69.7\ g/mol[/tex]
From the periodic table, the element with an atomic mass = 69.7 g/mol is gallium, Ga
A mineral is observed to become soft and crumbly and develop a reddish color as it undergoes weathering, what process has most likely taken place?
Answer:
Corrosion
Explanation:
The corrosion is a natural process that occurs because of the influence of the water and wind. It is part of the processes of erosion. It manages to soften the minerals, make them less stable and more crumbly. It can easily be noticed as it gives the mineral reddish, or rather dark orange color. This process occurs where the humidity is high, as the water plays a crucial role for this process to occur. Apart from occurring in nature, this process is also a big problem with the refined metals, as it gradually destroys them, and taking in consideration that the metals are included in all sorts of objects, it can make a lot of damage.
Compared with its corresponding unsaturated fatty acid, a saturated fatty acid has _____.
more hydrogen
less hydrogen
more oxygen
less oxygen
Answer: more hydrogen
Explanation:
Saturated fatty acids are defined as the fatty acids in which a single bond is present between carbon and carbon atoms and hydrocarbon chain is attached to carboxylic group. Example: [tex]C_{18}H_{34}O_2[/tex]
Unsaturated fatty acids are defined as the fatty acids which have double or triple covalent C-C bonds and hydrocarbon chain is attached to carboxylic group. Example: [tex]C_{18}H_{32}O_2[/tex]
Thus a saturated fatty acid has more hydrogens than unsaturated fatty acid.
Rank the following elements that make up the majority of Earth’s crust according to increasing ionization energy Mg Ca Si O
The order of increasing ionization energy of the elements are; Ca < Mg < Si < O.
Ionization energy is a periodic trend that decreases down the group but increases across the period.
Ionization energy decreases down the group due to addition of more shells and increased screening of outermost electrons by the inner electrons.
Ionization energy increases across the period due to increase in size of the nuclear charge.
Hence, the order of increasing ionization energy of the elements are; Ca < Mg < Si < O.
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A solution has a pH of 4.20. Using the relationship between pH and pOH, what is the concentration of OH?? A. 9.9 × 10-1 M B. 6.2 × 10-1 M C. 6.3 × 10-5 M D. 6.7 × 10-6 M E. 1.6 × 10-10M
The correct answer is option E.
Explanation:
The pH of the solution is defined as negative logarithm of hydrogen ions concentration in a solution. Mathematically written as:
[tex]pH=-\log[H^+][/tex]
The pOH of the solution is defined as negative logarithm of hydroxide ions concentration in a solution. Mathematically written as:
[tex]pOH=-\log[OH^-][/tex]
The sum of ph and pOH is equal to 14.
pH + pOH = 14
A solution has a pH of 4.20.
[tex]pOH=14-pH=14-4.20=9.8[/tex]
[tex]pOH=9.8=-\log[OH^-][/tex]
[tex][OH^-]=1.58\times 10^{-10} M\approx 6.0\times 10^{-10} M[/tex]
Hence, the correct answer is option E.
Answer:
E. 1.6 x 10 -10 M is the correct answer
The standard enthalpy change ΔH o rxn for the thermal decomposition of silver nitrate according to the following equation is 78.67 kJ: AgNO3(s) → AgNO2(s) + 1 2 O2(g) The standard enthalpy of formation of AgNO3(s) is −123.02 kJ/mol. Calculate the standard enthalpy of formation of AgNO2(s).
Answer: The enthalpy of the formation of [tex]AgNO_2(s)[/tex] is coming out to be -44.35 kJ/mol.
Explanation:
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H^o[/tex]
The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)][/tex]
For the given chemical reaction:
[tex]AgNO_3(s)\rightarrow AgNO_2(s)+\frac{1}{2}O_2(g)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(AgNO_2)})+(\frac{1}{2}\times \Delta H^o_f_{(O_2)})]-[(1\times \Delta H^o_f_{(AgNO_3)})][/tex]
We are given:
[tex]\Delta H^o_f_{(AgNO_3)}=-123.02kJ/mol\\\Delta H^o_f_{(O_2)}=0kJ/mol\\\Delta H^o_{rxn}=78.67kJ[/tex]
Putting values in above equation, we get:
[tex]78.67=[(1\times \Delta H^o_f_{(AgNO_2)})+(\frac{1}{2}\times 0)]-[1\times (-123.0))]\\\\\Delta H^o_f_{(AgNO_2)}=-44.35kJ/mol[/tex]
Hence, the enthalpy of the formation of [tex]AgNO_2(s)[/tex] is coming out to be -44.35 kJ/mol.
The standard enthalpy of formation AgNO2(s) is calculated from the standard enthalpy change for the thermal decomposition of silver nitrate and the standard enthalpy of formation of AgNO3(s). It is found to be -44.35 kJ/mol.
Explanation:This question is based on the calculation of the standard enthalpy of formation for AgNO2(s). The standard enthalpy change ΔH for the thermal decomposition of silver nitrate is given as 78.67 kJ according to the following equation: AgNO3(s) → AgNO2(s) + 1 2 O2(g). Additionally, we know that the standard enthalpy of formation of AgNO3(s) is −123.02 kJ/mol.
The standard enthalpy of formation for AgNO2 can be calculated using the formula ΔHf [AgNO2] = ΔH [reaction] + ΔHf [AgNO3]. Substituting the given values, we get ΔHf [AgNO2] = 78.67 kJ/mol + (-123.02 kJ/mol) = -44.35 kJ/mol. Therefore, the standard enthalpy of formation of AgNO2(s) is -44.35 kJ/mol.
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One disadvantage of vacuum distillation is it is inherently less efficient than fractional distillation at atmospheric pressure.
a. True
b. False
Answer:
One disadvantage of vacuum distillation is it is inherently less efficient than fractional distillation at atmospheric pressure. :True -a.
How many moles of KBr will be produced from 10.51 moles of BaBr2?
BaBr2 + K2SO4 → KBr + BaSO4
Select one:
a. 21.0
b. 5.26
c. 17.9
d. 7.88
To find the number of moles of KBr produced from 10.51 moles of BaBr₂, we use the balanced equation to see that 1 mole of BaBr₂ gives 2 moles of KBr. Multiplying 10.51 by 2 yields 21.02 moles of KBr. The correct answer is thus 21.0 moles of KBr.
The question asks how many moles of KBr will be produced from 10.51 moles of BaBr₂ when it reacts with K₂SO₄ to form KBr and BaSO₄. The balanced chemical equation for this reaction is:
BaBr₂ + K₂SO₄ → 2 KBr + BaSO₄
From the balanced equation, we can see that 1 mole of BaBr₂ produces 2 moles of KBr. Therefore, to find the number of moles of KBr produced from 10.51 moles of BaBr2, we simply multiply the number of moles of BaBr₂ by 2.
10.51 moles of BaBr₂ x 2 moles of KBr / 1 mole of BaBr₂ = 21.02 moles of KBr
The closest answer from the given options that matches our calculation is 21.0 moles of KBr, which is option (a).
Write balanced molecular equation for the reaction between nitric acid and calcium hydroxide.
Answer:
2HNO3 + Ca(OH)2 → 2H2O + Ca(NO3)2
Explanation:
HNO3 + Ca(OH)2 → ?
This is an acid-base reaction. It is also a double displacement reaction, in which the positive ions change partners.
Thus, H pairs with OH and Ca pairs with NO3.
HNO3 + Ca(OH)2 → HOH + Ca(NO3)2
HOH is our old friend, H2O (water) so, after balancing atoms, the equation becomes:
2HNO3 + Ca(OH)2 → 2H2O + Ca(NO3)2
How many grams of aluminum will be required to produce 4.5 g of copper metal?
Answer:
Mass of aluminum required is 1.277 grams
Explanation:
the reaction between copper and aluminium is a redox reaction and this can be written as:
[tex]3Cu^{+2}+2Al(s) ---> 2Al^{+3} +3Cu(s)[/tex]
Thus we need two moles of aluminum to reduce or to produce three moles of copper metal.
The moles of copper need to produce
= [tex]\frac{mass}{atomicmass}=\frac{4.5}{63.5}=0.071mol[/tex]
These will produced from = [tex]\frac{2X0.071}{3}=0.0473mol[/tex] of aluminum
The mass of aluminum required = moles X atomic mass = 0.0473X27=1.277g
To produce 4.5 g of copper metal, approximately 1.28 g of aluminum will be required, based on stoichiometry calculations using a typical chemical reaction.
Explanation:To determine the amount of aluminum needed to produce 4.5 g of copper metal, we need to use stoichiometry. Initially, we'll need to know the balanced chemical equation for the reaction in which aluminum is used to produce copper.
Typically, this equation might be something like: 2 Al(s) + 3 CuSO4(aq) -> Al2(SO4)3(aq) + 3 Cu(s). From this equation, we see that 2 moles of aluminum will produce 3 moles of copper. In terms of mass, 54 g (2 moles) of aluminum will produce about 190 g (3 moles) of copper (calculated using atomic masses of Al = 27 g/mol and Cu = 63.5 g/mol).
Therefore, to find the amount of aluminum required to produce 4.5 g of copper, we can use a simple ratio: (54 g Al / 190 g Cu = x g Al / 4.5 g Cu). Solving for 'x' gives us approximately 1.28 g of aluminum.
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When 10.0 grams of CH4 reacts completely with 40.0 grams of O2 such that there are no reactants left over, 27.5 grams of carbon dioxide are formed. How many grams of water are formed? CH4+ 2O2 → CO2 + 2H2O
Answer:
[tex]\boxed{\text{27.4 g CO$_{2}$; 22.5 g H$_{2}$O}}}[/tex]
Explanation:
We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 16.04 32.00 44.01 18.02
CH₄ + 2O₂ → CO₂ + 2H₂O
m/g: 10.0 40.0
1. Moles of CH₄
[tex]\text{Moles of CH}_{4} = \text{10.0 g CH}_{4} \times \dfrac{\text{1 mol CH}_{4}}{\text{16.04 g CH}_{4}} = \text{0.6234 mol CH}_{4}[/tex]
2. Mass of CO₂
(i) Calculate the moles of CO₂
The molar ratio is (1 mol CO₂ /1 mol CH₄)
[tex]\text{Moles of CO$_{2}$} = \text{0.6234 mol CH$_4$} \times \dfrac{\text{1 mol CO$_{2}$}} {\text{1 mol CH$_{4}$}} = \text{0.6234 mol CO$_{2}$}[/tex]
(ii) Calculate the mass of CO₂
[tex]\text{Mass of CO$_{2}$} = \text{0.6234 mol CO$_{2}$} \times \dfrac{\text{44.01 g CO$_{2}$}}{\text{1 mol CO$_{2}$}} = \textbf{27.4 g CO$_{2}$}\\\\\text{The mass of carbon dioxide formed is } \boxed{\textbf{27.4 g CO$_{2}$}}[/tex]
3. Mass of H₂O
(i) Calculate the moles of H₂O
The molar ratio is (2 mol H₂O /1 mol CH₄)
[tex]\text{Moles of H$_{2}$O}= \text{0.6234 mol CH}_{4} \times \dfrac{\text{2 mol H$_{2}$O}}{\text{1 mol CH$_{4}$}} = \text{1.247 mol H$_{2}$O}[/tex]
(ii) Calculate the mass of H₂O
[tex]\text{Mass of H$_{2}$O} = \text{1.247 mol H$_{2}$O } \times \dfrac{\text{18.02 g H$_{2}$O}}{\text{1 mol H$_{2}$O}} = \textbf{22.5 g H$_{2}$O}\\\\\text{The mass of water formed is } \boxed{\textbf{22.5 g H$_{2}$O}}[/tex]
How is the enthalpy of reaction related to the enthalpies of formation for the products and reactants?
Answer:
ΔH rxn = ∑ enthalpy of formation for the products - ∑enthalpy of formation for the reactantsExplanation:
The enthalpy of reaction is the change of enthalpy of the chemical reaction. It is equal to the heat released or absorbed during the chemical reaction.
The enthalpy of reaction is equal to the difference of the enthalpy at the end of the reaction and the enthalpy at the begining of the reaction.
Since the substances at the end are the products, and the substances at the begining are the reactants, this is equivalent to say that the enthalpy of reaction is equal to the enthalpy of formation for the products less the enthalpy of formation for the reactants.
In the form of equation that is:
ΔH rxn = ∑ enthalpy of formation for the products - ∑enthalpy of formation for the reactants.The enthalpy of a reaction (ΔH_rxn) is calculated by subtracting the sum of the enthalpies of formation of the reactants from the sum of the enthalpies of formation of the products. This calculation uses Hess's Law and accounts for the stoichiometric coefficients in the balanced equation. As such, the enthalpy change can indicate whether a reaction is exothermic or endothermic.
The enthalpy of a reaction (ΔH_rxn) is determined using the enthalpies of formation (ΔH_f) for the reactants and products. According to Hess's Law, the enthalpy change of any chemical reaction is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants. Mathematically, this is expressed as:
ΔH_rxn = ∑ΔH_f(products) - ∑ΔH_f(reactants)
The enthalpies of formation are multiplied by the respective number of moles of each substance (as given by their coefficients in the balanced chemical equation) before summing up. For example, in the combustion of methane:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)The reaction releases 890.5 kJ at 25°C because the energy needed to break the chemical bonds in the reactants differs from the energy released when forming the bonds in the products, reflecting an exothermic process.
Please answer these questions fast! Im on a time crunch right now
Answer:
1st image, 17.79 g Mg(NO3)2
2nd image 1.47 mol NH3
3rd image Molar Mass Al(OH)3 = 78 g/mol
Explanation:
1st image
Molar Mass Mg(NO3)2 = 148. 31 g/mol
0,12 mol Mg(NO3)2 x (148, 31 g Mg(NO3)2 / 1 mol Mg(NO3)2) = 17.79 gMg(NO3)2
2nd image
Molar Mass NH3 = 17g/mol
25 g NH3 x ( 1 mol NH3/ 17 g NH3) = 1.47 mol NH3
3rd image
Molar Mass Al(OH)3 = 1 Al *(27) + 3 O * (16) + 3 H * (1) = 78 g/mol
The amount of gas that occupies 60.82 l at 31.0 °c and 367 mm hg is ________ mol.
Answer:
1.17mole
Explanation:
Given parameters:
Volume of gas = 60.82L or 60.82dm³
Temperature of the gas = 31.0°C or 304K
Pressure on the gas = 367mmHg
To make calculation easier we convert the pressure from mmHg to atm.
1 atm = 760mmHg
367mmHg gives [tex]\frac{367}{760}[/tex]atm = 0.48atm
Solution
Assuming ideality, we can find the number of moles of the gas used. The ideal gas law combines Boyle's law, Charles's Law and Avogadro's law. It is expressed below:
PV = nRT
P is the pressure
V is the volume
n is the number of moles
R is the gas constant given as 0.082atmdm³mol⁻¹K⁻¹
T is the temperature in kelvin
The unknown is n, which is the number of moles.
Making n the subject of the formula, we have ;
n = [tex]\frac{PV}{RT}[/tex]
n = [tex]\frac{0.48 x 60.82}{0.082x304}[/tex]= 1.17mole
The question is asking to identify the number of moles in a gas given certain conditions using the Ideal Gas Law (PV=nRT). We do this by first converting all units to the necessary format and then substituting them into the rearranged Ideal Gas Law equation to solve for n (moles).
Explanation:The student's question is basically asking about the amount of moles in a gas under specified conditions. We can decipher this answer through the Ideal Gas Law equation which states that PV=nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. To apply the equation, we first need to convert the given conditions to the correct units compatible with R (Ideal Gas Constant) which is typically 0.0821 L·atm/K·mol.
First, volume (V) is already given in liters (60.82 L)Second, we convert temperature (T) from Celsius to Kelvin by adding 273.15 to the stated degree Celsius, so 31.0 °c becomes 304.15 K Third, we need to convert pressure (P) from mmHg to atm. The conversion factor is approximately 1 atm = 760 mmHg, so pressure becomes approximately 0.483 atmOnce we have these conditions correctly converted, we substitute them into the Ideal Gas Law. Rearranging the equation to solve for n (moles), the equation becomes n = PV/RT. Thus, substituting our numbers in gives n = (0.483 atm*60.82 L)/ (0.0821 L·atm/K·mol * 304.15 K).
Simplifying this will give the answer for the amount of moles.
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Match each vocabulary word with its definition. match the items in the left column to the items in the right column. 1 . periodicity unlike charges 2 . shell like charges 3 . nonmetals charge on a neutron 4 . noble gases location of an electron 5 . neutral high ionization energies 6 . repel repeating nature of atomic structure 7 . attract least reactive family
Answer:
Explanation:
1 . periodicity
repeating nature of atomic structure
Periodicity is the repeating nature or trend of an atom on the periodic table. Periodicity can be inform of ionization energy, atomic radius, nuclear charge etc.
2 . shell
location of an electron
Electrons are located in shells. It is where electrons are found in an atom and the probability of finding electrons there are high
3 . non-metals
high ionization energies
Non-metals have high ionization energies which is the energy required to remove a loosely bonded electron in an atom. Metals have low ionization energies.
4 . noble gases
least reactive family
Noble gases are called inert gases. They have complete electronic configuration and this makes them unreactive.
5 . neutral
charge on a neutron
Neutrons have no charge on them. They are subatomic particles found in the nucleus alongside the protons
6 . repel
like charges
Like charges repel one another
7 . attract
unlike charges
Unlike charges attracts. Positve attracts negative charges.
Answer:
sorry in need question for points
Explanation:
1 . periodicity
repeating nature of atomic structure
Periodicity is the repeating nature or trend of an atom on the periodic table. Periodicity can be inform of ionization energy, atomic radius, nuclear charge etc.
2 . shell
location of an electron
Electrons are located in shells. It is where electrons are found in an atom and the probability of finding electrons there are high
3 . non-metals
high ionization energies
Non-metals have high ionization energies which is the energy required to remove a loosely bonded electron in an atom. Metals have low ionization energies.
4 . noble gases
least reactive family
Noble gases are called inert gases. They have complete electronic configuration and this makes them unreactive.
5 . neutral
charge on a neutron
Neutrons have no charge on them. They are subatomic particles found in the nucleus alongside the protons
6 . repel
like charges
Like charges repel one another
7 . attract
unlike charges
Unlike charges attracts. Positve attracts negative charges.