Answer:
11.31g NaClO₂
Explanation:
Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45?
It is possible to answer this question using Henderson-Hasselbalch equation:
pH = pKa + log₁₀ [A⁻] / [HA]
Where pKa is -log Ka = 1.9547; [A⁻] is the concentration of the conjugate base (NaClO₂), [HA] the concentration of the weak acid
You can change the concentration of the substance if you write the moles of the substances:
[Moles HClO₂] = 250mL = 0.25L×(1.60mol /L) = 0.40 moles HClO₂
Replacing in H-H expression, as the pH you want is 1.45:
1.45 = 1.9547 + log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
-0.5047 = log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
0.3128 = [Moles NaClO₂] / [0.40 moles HClO₂]
0.1251 = Moles NaClO₂
As molar mass of NaClO₂ is 90.44g/mol, mass of 0.1251 moles of NaClO₂ is:
0.1251 moles NaClO₂ ₓ (90.44g / mol) =
11.31g NaClO₂Calculate the grams of solute needed to prepare 107 mL of a 2.75 M magnesium
chloride solution
Your Answer:
Answer
Answer: 28.02 g
Explanation:
The M stands for molarity. It is moles of solute/liters of solution. We can use the molarity to convert liters to mL, then make a proportion to find the grams.
[tex]\frac{2.75 mol}{L} *\frac{1L}{1000mL} =\frac{2.75 mol}{1000mL}[/tex]
Now that we have molarity in moles and mL, we can use the 107mL to get moles.
[tex]\frac{2.75moles}{1000mL} *107mL=0.29425mol[/tex]
We would multiply moles by molar mass to get grams. The molar mass of magnesium chloride is 95.211 g/mol.
[tex]0.29425mol*\frac{95.211g}{mol} =28.02g[/tex]
Potassium, a metal with one electron in the outermost shell, will react with how many chlorine atoms
Answer:
7 chlorine atoms
Explanation:
K=2.8.8.1
Cl=2.8.7
pottasium will give chlorine its I valence electron to form ions as follows
K=(2.8.8)+
Cl=(2.8.8)-
It will react with 1 chlorine atom.
Whilst one atom loses an electron to every other atom it results in the formation of?
An ionic bond is shaped by using the whole transfer of some electrons from one atom to every other. The atom losing one or more electrons becomes a cation—an undoubtedly charged ion. The atom gaining one or more electrons will become an anion—a negatively charged ion.
What number of bonds can chlorine form?In those compounds carbon, nitrogen, oxygen, and chlorine atoms have four, three, and one bonds, respectively. The hydrogen atom and the halogen atoms form the most effective covalent bond to different atoms in maximum stable neutral compounds.
Learn more about chlorine atom here: https://brainly.com/question/25310059
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If iron metal reacts with an aqueous solution of silver nitrate and zinc reacts with an aqueous solution of iron sulfate, rank these three metals Fe Zn Ag in order of decreasing reactivity.
Answer:
yes!you are right a cloudy formation will be formed when they will react.its because if nitrogen.
Paper is stable at room temperature in the presence of oxygen (O2) because it requires ____________ to start the reaction.
Answer:
Energy
Explanation:
Paper is oxidized in the presence of oxygen. This reaction however proceeds very slowly until energy is supplied to the system. This implies that the reaction is not spontaneous at room temperature. A spontaneous reaction takes place without any need for external supply of energy.
The need for supply of external energy must be as a result of the high activation energy required for the reaction to go to completion. If a chemical reaction has a high activation energy, it will require an external supply of energy in order for such reaction to occur.
Write empirical formula
Answer:
[tex]Pb(ClO_{3})_{4}\\Pb(MnO_{4})_{4}\\Fe(ClO_{3})_{3}\\\Fe(MnO_{4})_{3}\\[/tex]
Explanation:
[tex]Pb^{4+}(ClO_{3}^{-})_{4}--->Pb(ClO_{3})_{4}\\Pb^{4+}(MnO_{4}^{-})_{4}--->Pb(MnO_{4})_{4}\\Fe^{3+}(ClO_{3}^{-})_{3}--->Fe(ClO_{3})_{3}\\\Fe^{3+}(MnO_{4}^{-})_{3}--->Fe(MnO_{4})_{3}\\[/tex]
Organic chemistry too eazy but why 90% students avoid ?
Answer:
because it covers a large area as there are more then 1 lakh compounds of organic chemistry.
Calculate the pH and concentrations of H2A, HA−, and A2−, at equilibrium for a 0.236 M solution of Na2A. The acid dissociation constants for H2A are Ka1=7.68×10−5 and Ka2=6.19×10−9.
Answer:
[H₂A] = 5.0409x10⁻⁷M
[HA⁻] = 0.001951M
[A²⁻] = 0.234
11.29 = pH
Explanation:
When Na₂A is in equilibrium with water, the reactions that occurs are:
2Na⁺ + A²⁻(aq) + H₂O(l) ⇄ HA⁻(aq) + 2Na⁺(aq) + OH⁻(aq)
As sodium ion doesn't react:
A²⁻(aq) + H₂O(l) ⇄ HA⁻(aq) + OH⁻(aq)
Kb1 = KwₓKa2 = 1x10⁻¹⁴/ 6.19x10⁻⁹ = 1.6155x10⁻⁶ = [HA⁻] [OH⁻] / [A²⁻]
And HA⁻ will be in equilibrium:
HA⁻(aq) + H₂O(l) ⇄ H₂A(aq) + OH⁻(aq)
Kb2 = KwₓKa1 = 1x10⁻¹⁴/ 7.68x10⁻⁵ = 1.3021x10⁻¹⁰ = [H₂A] [OH⁻] / [HA⁻]
In the reaction, you have 2 equilibriums, for the first reaction, concentrations in equilibrium are:
[HA⁻] = X
[OH⁻] = X
[A²⁻] = 0.236M - X
Replacing in Kb1:
1.6155x10⁻⁶ = [HA⁻] [OH⁻] / [A²⁻]
1.6155x10⁻⁶ = [X] [X] / [0.236-X]
3.8126x10⁻⁶ - 1.6155x10⁻⁶X = X²
3.8126x10⁻⁶ - 1.6155x10⁻⁶X - X² = 0
Solving for X
X = -0.00195 → False solution. There is no negative concentrations
X = 0.001952.
Replacing, concentrations for the first equilibrium are:
[HA⁻] = 0.001952
[OH⁻] = 0.001952
[A²⁻] = 0.234
Now, in the second equilibrium:
[HA⁻] = 0.001952 - X
[OH⁻] = X
[H₂A] = X
Replacing in Kb1:
1.3021x10⁻¹⁰ = [H₂A] [OH⁻] / [HA⁻]
1.3021x10⁻¹⁰ = [X] [X] / [0.001952 - X]
2.5417x10⁻¹³ - 1.3021x10⁻¹⁰X = X²
2.5417x10⁻¹³ - 1.3021x10⁻¹⁰X - X² = 0
Solving for X
X = -5.04x10⁻⁷ → False solution. There is no negative concentrations
X = 5.0409x10⁻⁷
Replacing, concentrations for the second equilibrium are:
[HA⁻] = 0.001951M
[OH⁻] = 5.0409x10⁻⁷M
[H₂A] = 5.0409x10⁻⁷M
Thus, you have concentrations of H2A, HA−, and A2−
Now, for pH, the sum of both productions of [OH⁻] is:
[OH⁻] = 0.0019525
pOH = -log[OH⁻] = 2.709
As 14 = pH+ pOH
11.29 = pH
As per the question the pH and the cons of the H2A, the HA−, and the A2−, at equilibrium for a 0.236 M.
The pH needs to be in equilibrium from the mentioned elements and form a solution of Na2A. Thus the concentration of the ions is to be calculated with the dissociation of the constants for the H2A.Hence the [H₂A] = 5.0409x10⁻⁷M.[HA⁻] = 0.001951M A²⁻] = 0.234 will give 11.29 = pH.Learn more about the A2−, at equilibrium.
brainly.com/question/17086012.
➔ Which compound has both ionic and covalent bonds? A. Ammonium chloride B. Carbon dioxide C. Ethyl ethanoate D. Sodium chloride
Answer:
Choice A. Ammonium chloride.
Explanation:
Consider the bonds in each of the four compounds.
Ammonium chlorideAmmonium chloride [tex]\rm NH_4Cl[/tex] is an ionic compound. Each
The [tex]\rm {NH_4}^{+}[/tex] and [tex]\rm Cl^{-}[/tex] ions in [tex]\rm NH_4Cl[/tex] are connected with ionic bonds.
What make [tex]\rm NH_4Cl[/tex] special is that its cation [tex]\rm {NH_4}^{+}[/tex] is polyatomic. In other words, each [tex]\rm {NH_4}^{+}[/tex] ion contains more than one atoms. These atoms (one [tex]\rm N[/tex] atom and four [tex]\rm H[/tex] atoms) are connected with covalent bonds. Therefore, [tex]\rm NH_4Cl[/tex] has both ionic and covalent bonds.
Carbon dioxideCarbon dioxide [tex]\rm CO_2[/tex] is a covalent compound. Each [tex]\rm CO_2[/tex] molecule contains two [tex]\rm C=O[/tex] double bonds in total. [tex]\rm CO_2[/tex] molecules have no ionic bond.
Ethyl ethanoateThe name "ethyl ethanoate" might sound like the name of a salt (think about sodium ethanoate.) However, in reality, ethyl ethanoate [tex]\rm CH_3COOCH_2CH3[/tex] is an ester. The "ethyl" here refers to the [tex]\rm -OCH_2CH3[/tex] part, originating from ethanol. On the other hand, "ethanoate" refers to the [tex]\rm CH_3C(O)-[/tex] part, which can be obtained from ethanoic acid.
These two parts are connected with a covalent [tex]\rm C-O[/tex] single bond. (The [tex]\rm C[/tex] in ethanoic acid is connected to the [tex]\rm O[/tex] in ethanol.) As a result, there's no ionic bond in ethyl ethanoate, either.
Sodium chlorideSodium chloride [tex]\rm NaCl[/tex] is an ionic compound. Both the [tex]\rm Na^{+}[/tex] ion and the [tex]\rm Cl^{-}[/tex] are monoatomic. While the [tex]\rm Na^{+}[/tex] and [tex]\rm Cl^{-}[/tex] in sodium chloride are connected with ionic bonds, neither [tex]\rm Na^{+}[/tex] nor [tex]\rm Cl^{-}[/tex] contains covalent bond.
Question 39 (1 point)
What is the volume (in L) of 4.20 grams of CO2 gas at STP? (
molar masses are,
Carbon = 12.01 g/mol, and oxygen - 16.00 g/mol)
Your Answer:
Answer:
volume = 2128L
Explanation:
volume at stp = 22.4dm³ =22400L
CO2 = 12 + 2(16) = 44
4.20/44 = volume/22400L
0.095 = volume/22400L
volume = 0.095 x 22400
volume = 2128L
Assuming the same temperature and pressure for each gas, how many milliliters of carbon dioxide are produced from 16.0 mL of CO?
2 CO(g) + O2(g)
2 CO2(g)
Express your answer with the appropriate units.
Answer:
[tex]V_{CO_2}=16.0mL[/tex]
Explanation:
Hello,
In this case, given that the same temperature and pressure is given for all the gases, we can notice that 16.0 mL are related with two moles of carbon monoxide by means of the Avogadro's law which allows us to understand the volume-moles relationship as a directly proportional relationship. In such a way, since in the chemical reaction:
[tex]2CO(g)+O_2(g)\rightarrow 2CO_2(g)[/tex]
We notice two moles of carbon monoxide yield two moles of carbon dioxide, therefore we have the relationship:
[tex]n_{CO}V_{CO}=n_{CO_2}V_{CO_2}[/tex]
Thus, solving for the yielded volume of carbon dioxide we obtain:
[tex]V_{CO_2}=\frac{n_{CO}V_{CO}}{n_{CO_2}} =\frac{2mol*16.0mL}{2mol}\\ \\V_{CO_2}=16.0mL[/tex]
Best regards.
Sort the resources into the correct categories.
are replaced by natural processes
Renewable Resources
Nonrenewable Resources
cannot be replaced in a short time
are used more quickly than replaced
have fixed amounts
are considered unlimited
are replaced faster than used
Intro
✓ Done
Answer:
Renewable Resources: are considered unlimited, are replaced faster than used.
Nonrenewable Resources: are used more quickly than replaced, have fixed amounts, cannot be replaced in a short time.
Explanation:
Renewable resources are natural resources that are able to naturally regenerate themselves, hence, they are considered to be unlimited. They are usually replaced faster than they are used because they have a short regeneration time. A good example is the solar energy.
Nonrenewable resources are those natural resources that cannot naturally regenerate and when they do, it takes a very long time (usually millions of years). They are therefore used at a much faster rate than they are being replaced and their natural deposits are more or less fixed due to the long regeneration time. A good example is the crude oil deposit.
Hence:
Renewable Resources: are considered unlimited, are replaced faster than used.
Nonrenewable Resources: are used more quickly than replaced, have fixed amounts, cannot be replaced in a short time.
Answer: !
Explanation:
A principal constituent of petrol (gasoline) is iso-octane, C8H18. From the following thermodynamic data at
298 K what is the
standard molar enthalpy of combustion of iso-octane in excess oxygen
at 298 K?
C«H;8(1) + 12702() +8C02(g) +91,0(1)
Substance AfHn/kJ mol"}
C8H8(1)
-258.07
02(8)
0
CO2(8)
-393.51
H2O(1)
-285.83
Answer: The enthalpy of combustion of iso-octane in excess oxygen at 298 K is -5462.2kJ/mol
Explanation:
The balanced reaction for combustion of isooctane is:
[tex]C_8H_{18}(l)+\frac{25}{2}O_2(g)\rightarrow 8CO_2(g)+9H_2O(l)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{(CO_2(g))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(1\times \Delta H^o_f_{(C_8H_{18}(g))})+(\frac{25}{2}\times \Delta H^o_f_{(O_2(g))})][/tex]
We are given:
[tex]\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.51kJ/mol\\\Delta H^o_{C_8H_{18}(l)}=-258.07kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(8\times (-393.51))+(9\times (-285.8))]-[(1\times (-258.07))+(\frac{25}{2}\times (0))]\\\Delta H^o_{rxn}=-5462.2kJ/mol[/tex]
The enthalpy of combustion of iso-octane in excess oxygen at 298 K is -5462.2kJ/mol
An aqueous solution was made by dissolving 72.9 grams of glucose, C6H12O6, into 115 grams of water. The vapor pressure of the pure water is 26.4 Torr. The vapor pressure of water over this solution is: (a) 27.9 Torr (b) 24.1 Torr (c) 26.8 Torr (d) 24.8 Torr PLease answer this as quick as possible
Answer:
The correct answer is (d) 24.8 Torr
Explanation:
When a solute is added to a solvent, the water pressure of the solution is lower than the vapor pressure of the pure solvent. This is called vapor pressure lowering and it is given by the following expression:
Psolution= Xsolvent x Pºsolvent
We have to calculate Xsolvent (mole fraction of solvent) which is given by the number of moles of solute divided into the total number of moles.
First, we calculate the number of moles of solute and solvent. The solute is glucose (C₆H₁₂O₆), and its number of moles is calculated from the mass and the molecular weight (MM):
MM (C₆H₁₂O₆)= (12 g/mol x 6) + (1 g/mol x 12) + (16 g/mol x 6) = 180 g/mol
moles of glucose= mass/MM= (72.9 g)/)(180 g/mol)= 0.405 moles
The solvent is water (H₂O) and again we calculate the number of moles as follows:
MM(H₂O)= (1 g/mol x 2) + 16 g/mol = 18 g/mol
moles of water= mass/MM= (115 g)/(18 g/mol)= 6.389 moles
Now, we calculate the total number of moles (nt):
nt= moles of glucose + moles of water= 0.405 moles + 6.389 moles= 6.794 moles
The mole fraction of water (Xsolvent) is given by:
Xsolvent= moles of water/nt= 6.389 moles/6.794 moles= 0.940
Finally, the vapor pressure of water over the solution will be the following:
Psolvent= Xsolvent x Pºsolvent= 0.940 x 26.4 Torr= 24.8 Torr
The Ksp of calcium sulfate, CaSO4, is 9.0 × 10-6. What is the concentration of CaSO4 in a saturated solution? A. 3.0 × 10-3 Molar B. 9.0 × 10-3 Molar C. 3.0 × 10-6 Molar D. 9.0 × 10-6 Molar
Answer: The concentration of [tex]CaSO_4[/tex] in a saturated solution is [tex]3.0\times 10^{-3}M[/tex]
Explanation:
Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as [tex]K_{sp}[/tex]
The equation for the ionization of [tex]CaSO_4[/tex] is given as:
[tex]K_{sp}[/tex] of [tex]CaSO_4[/tex] = [tex]9.0\times 10^{-6}[/tex]
By stoichiometry of the reaction:
1 mole of [tex]CaSO_4[/tex] gives 1 mole of [tex]Ca^{2+}[/tex] and 1 mole of [tex]SO_4^{2-}[/tex]
When the solubility of [tex]CaSO_4[/tex] is S moles/liter, then the solubility of [tex]Ca^{2+}[/tex] will be S moles\liter and solubility of [tex]SO_4^{2-}[/tex] will be S moles/liter.
[tex]K_{sp}=[Ca^{2+}][SO_4^{2-}][/tex]
[tex]9.0\times 10^{-6}=[s][s][/tex]
[tex]9.0\times 10^{-6}=s^2[/tex]
[tex]s=3.0\times 10^{-3}M[/tex]
Thus concentration of [tex]CaSO_4[/tex] in a saturated solution is [tex]3.0\times 10^{-3}M[/tex]
How many moles of sulfur trioxide will be produced when the complete combustion of 100.0 g of sulfur dioxide takes place
Answer:
1.563 moles of SO3.
Explanation:
We begin by calculating the number of mole present in 100g of sulphur dioxide, SO2. This can be obtained as follow:
Molar mass of SO2 = 32 + (16x2) = 64g/mol
Mass of SO2 = 100g
Mole of SO2 =..?
Mole = mass/Molar mass
Mole of SO2 = 100/64
Mole of SO2 = 1.563 mole
Now, we can obtain the number of mole of sulphur trioxide, SO3 produce from the reaction as follow:
2SO2 + O2 —> 2SO3
From the balanced equation above,
2 moles of SO2 reacted to produce 2 moles of SO3.
Therefore, 1.563 moles of SO2 will also react to produce 1.563 moles of SO3.
Therefore, 1.563 moles of SO3 is obtained from the reaction.
For the reaction Ca(s)+Cl2(g)→CaCl2(s) calculate how many grams of the product form when 14.4 g of Ca completely reacts. Assume that there is more than enough of the other reactant.
Answer:
[tex]m_{CaCl_2}=39.96gCaCl_2[/tex]
Explanation:
Hello,
In this case, for the undergoing reaction, we can compute the grams of the formed calcium chloride by noticing the 1:1 molar ratio between calcium and it (stoichiometric coefficients) and using their molar mass of 40 g/mol and 111 g/mol by using the following stoichiometric process:
[tex]m_{CaCl_2}=14.4gCa*\frac{1molCa}{40gCa} *\frac{1molCaCl_2}{1molCa} *\frac{111gCaCl_2}{1molCaCl_2}\\ \\m_{CaCl_2}=39.96gCaCl_2[/tex]
Clearly, chlorine is not used since it is said there is enough for the reaction to go to completion.
Best regards.
8. How many grams of SO2 are there in 2.5 mol of SO2? (Show Work)
Answer:
160g
Explanation:
Mass in grams is equal to product of moles and molar mass of compound.
the molar solubility of Zn(OH)2 is 5.7x 10^-3 mol/L at a certain temperature. Calculate the value of Ksp for Zn(OH)2 at this temperataure
Answer:
Ksp = 7.4x10⁻⁷
Explanation:
Molar solubility of a substance is defined as the amount of moles of that can be dissolved per liter of solution.
Ksp of Zn(OH)₂ is:
Zn(OH)₂(s) ⇄ Zn²⁺ + 2OH⁻
Ksp = [Zn²⁺] [OH⁻]²
And the molar solubility, X, is:
Zn(OH)₂(s) ⇄ Zn²⁺ + 2OH⁻
⇄ X + 2X
Because X are moles of substance dissolved.
Ksp = [X] [2X]²
Ksp = 4X³
As molar solubility, X, is 5.7x10⁻³mol/L:
Ksp = 4X³
Ksp = 4 (5.7x10⁻³mol/L)³
Ksp = 7.4x10⁻⁷A sample of an unknown substance has a mass of 0.158kg. If 2,520.0 j of heat is required to heat the substance from 32.0C to 61.0C what is the specific heat of the substance
2. Write the chemical equations for the neutralization reactions that occurred when HCL and NaOH were added to the buffer solution.
Answer:
HCI(aq)+CH3COONa(s) ----> CH3COOH(aq)+NaCl(s)
NaOH(aq)+CH3COOH(aq) ----> CH3COONa(s)+H2O(l)
Explanation:
A buffer is a solution that resists changes in acidity or alkalinity. A buffer is able to neutralize a little amount of acid or base thereby maintaining the pH of the system at a steady value.
A buffer may be an aqueous solution of a weak acid and its conjugate base or a weak base and its conjugate acid.
The equations for the neutralizations that occurred upon addition of HCl or NaOH are;
HCI(aq)+CH3COONa(s) ----> CH3COOH(aq)+NaCl(s)
NaOH(aq)+CH3COOH(aq) ----> CH3COONa(s)+H2O(l)
1. List the conjugate acid or conjugate base for each chemical. a. The acid HF b. The base KOH c. The base NH3 d. The acid HNO3 e. The acid HCOOH f. The base CH3NH2
Answer:
a) Conjugate base F– b) Conjugate acid K+ c) Conjugate acid NH4+ d) Conjugate base NO2- e) Conjugate base HCOO– f) Conjugate acid CH4+
Explanation:
Acid will produce Conjugate base
Base will produce Conjugate acid.
Answer:
a. The acid HF: F-
b. The base KOH: H2O
c. The base NH3: NH4+
d. The acid HNO3: NO3-
e. The acid HCOOH: COOH-
f. The base CH3NH2: CH3NH3+
Explanation:
130I decays by emission of beta particles to form stable 130Xe. A 3.00 g iodine sample containing some I-130 was recorded as having 9487 disintegrations per min. k = 0.00094 min -
1. How many radioactive I-130 atoms are present in the sample?
The number of radioactive atoms present in the sample is 1.01 × 10⁷ atoms.
Emission of Beta Particles takes place when there is a numerous amount of neutron to the proton in the nucleus. As a result, the neutron is transformed to proton (which is retained in the nucleus) and electron which leaves the nucleus under high energy.
Given that:
[tex]\mathbf{^{130} I \ \ \ \to ^{- \beta } \ \ \ ^{130}Xe }[/tex]
The rate of disintegration of the radioactive sample per minute is expressed by using the formula:
A = kN
where;
A = number of decay per unitk = decay constantN = total number of atomic particles present in the sample∴
[tex]\mathbf{N = \dfrac{A}{k}}[/tex]
[tex]\mathbf{N = \dfrac{9487}{0.00094}}[/tex]
N = 10092553.19 atoms
N = 1.01 × 10⁷ atoms
Therefore, we can conclude that the number of radioactive atoms present in the sample is 1.01 × 10⁷ atoms.
Learn more about emission of beta particles here:
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Which of the following cannot have hydrogen bonds? Select one: A. NH3 B. H2O C. HF D. CH3NH2 E. Which of the following cannot have hydrogen bonds? Select one: A. NH3 B. H2O C. HF D. CH3NH2 E. HCl
Answer:
E. HCl
Explanation:
Cl atom does not have enough electronegativity to make enough positive charge on H.
HCl is the compound which doesn't have hydrogen bonds. This is because of
the higher size of the chlorine atom.
There is no hydrogen bond because of the high size of the chlorine.
Chlorine have electrons with a very low density. It is also very
electronegative which explains why the formation of hydrogen bonds in the
compound HCl is not possible.
Instead, HCl has covalent bonds in which electron is shared between the
hydrogen and chlorine to achieve a stable configuration.
Read more on https://brainly.com/question/10462894
g The "Coulomb barrier" is defined to be the electric potential energy of a system of two nuclei when their surfaces barely touch. The probability of a nuclear reaction greatly increases if the energy of the system is above this barrier. What is the Coloumb barrier (in MeV) for the absorption of an alpha particle by a lead-208 nucleus
Answer:
The Coulomb Barrier U is 25.91 MeV
Explanation:
Given that:
Atomic Mass of lead nucleus A = 208
atomic mass of an alpha particle A = 4
Radius of an alpha particle [tex]R_\alpha = R_o A^{^{\dfrac{1}{3}}[/tex]
where;
[tex]R_\alpha = 1.2 \times 10 ^{-15} \ m[/tex]
[tex]R_\alpha = R_o A^{^{\dfrac{1}{3}}[/tex]
[tex]R_\alpha = 1.2 \times 10 ^{-15} \ m \times (4) ^{^{\dfrac{1}{3}}[/tex]
[tex]R_\alpha = 1.905 \times 10^{-15} \ m[/tex]
Radius of the Gold nucleus
[tex]R_{Au}= R_o A^{^{\dfrac{1}{3}}[/tex]
[tex]R_{Au}= 1.2 \times 10 ^{-15} \ m \times (208) ^{^{\dfrac{1}{3}}[/tex]
[tex]R_{Au} = 7.11 \times 10^{-15} \ m[/tex]
[tex]R = R_\alpha + R_{Au}[/tex]
[tex]R = 1.905 \times 10^{-15} \ m + 7.11 \times 10^{-15} \ m[/tex]
[tex]R = 9.105 \times 10 ^{-15} \ m[/tex]
The electric potential energy of the Coulomb barrier [tex]U = \dfrac{Ke \ q_{\alpha} q_{Au}}{R}[/tex]
[tex]U = \dfrac{8.99 \times 10^9 \ N.m \ ^2/C ^2 \ \times 2 ( 82) \times \(1.60 \times 10^{-19} C \ \ e } {9.105 \times 10^{-15} \ m }[/tex]
U = 25908577.7eV
U = 25.908577 × 10⁶ eV
U = 25.91 MeV
The Coulomb Barrier U is 25.91 MeV
How are the oxygen atoms balanced for redox equations in basic solutions
Answer: H2O and OH^- are added to balance the oxygen atoms.
Explanation:
What is the mass number of an atom with 24 protons anſ 30 neutrons?
Answer:
54
Explanation:
The mass number of the atom is 54
How much energy in joules will be required to raise the temperature of 50.0 g of water from 20 degrees C to 60 degree C
Answer: 8368 Joules
Explanation:
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.
[tex]Q=m\times c\times \Delta T[/tex]
Q = Heat absorbed or released =?
c = specific heat capacity of water = [tex]4.184J/g^0C[/tex]
Initial temperature of water = [tex]T_i[/tex] = [tex]20^0C[/tex]
Final temperature of water = [tex]T_f[/tex] = [tex]60^0C[/tex]
Change in temperature ,[tex]\Delta T=T_f-T_i=(60-20)^0C=40^0C[/tex]
Putting in the values, we get:
[tex]Q=50.0g\times 4.184J/g^0C\times 40^0C=8368J[/tex]
Thus energy in Joules required is 8368.
Given the information about each pair of acids fill in the correct answer.
a. Acid A has a lower % ionization than B:_______ is a stronger acid.
b. Acid B has a larger K_a than acid A._______ will have a larger percent ionization.
c. A is a stronger acid than B. Acid B will have________ percent ionization than A.
Answer:
a. Acid B
b. Acid B
c. lower
Hope this helps you
If you needed a 1.5 x 1 0-4 M solution of a compound that has a molar mass of 760 g/mol, what would it concentration be in parts per million?
Answer:
114 ppm
Explanation:
Data obtained from the question include:
Conc. of compound in mol/L = 1.5×10¯⁴ mol/L
Molar mass of compound = 760 g/mol
Conc. in ppm =..?
Next, we shall determine the concentration of the compound in grams per litre (g/L) . This is illustrated below:
Conc. in mol/L = conc. in g/L / Molar mass
1.5×10¯⁴ = conc. In g/L / 760
Cross multiply
Conc. in g/L = 1.5×10¯⁴ x 760
Conc. in g/L = 0.114 g/L
Next, we shall convert 0.114 g/L to milligrams per litre (mg/L). This is illustrated below:
1 g/L = 1000 mg/L
Therefore, 0.114 g/L = 0.114 x 1000 = 114 mg/L
Finally, we shall convert 114 mg/L to parts per million (ppm). This is illustrated below:
1 mg/L = 1 ppm
Therefore, 114 mg/L = 114 ppm
From the calculations made above,
1.5×10¯⁴ mol/L Is equivalent to 114 ppm.
The complete combustion of ethanol, C2H5OH(l), to form H2O(g) and CO2(g) at constant pressure releases 1235 kJ of heat per mole of C2H5OH.
Write a balanced equation for this reaction.
Express your answer as a chemical equation. Identify all of the phases in your answer.