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 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.
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
Which statements correctly describe the effect of distance in determining the gravitational force and the electrical force? Check all that apply.
Answer:
The gravitational force is inversely proportional to the square of the distance..
The electrical force is inversely proportional to the square of the distance..
Explanation:
From the Newton's law of universal gravity, we understood that the force of attraction between the masses two objects is inversely proportional to the square of the distance apart. Mathematically, it is expressed as:
F = GM1M2/r²
Where:
M1 and M2 are the masses of two objects measured in kilograms (Kg).
G is the gravitational constant with a value of 6.67×10¯¹¹ Nm²/Kg²
r is the distance apart measured in metre (m).
F is the force of attraction measured in Newton (N).
From the equation:
F = GM1M2/r²
The force (F) is inversely proportional to the square of the distance (r) apart. This means that the greater the distance apart, the smaller the force of attraction and the smaller the distance apart, the greater the force of attraction between the masses of the object.
From Coulomb's Law, we understood that the force of attraction between two charged particles is inversely proportional to square of the distance. Mathematically, it expreessed as
F = Kq1q2/ r²
Where:
q1 and q2 are charges of two objects measured in coulomb (C) .
r is the distance apart measured in metre (m)
K is the electrostatic constant with a value of 9×10^9 Nm²/C²
F is the force measured in Newton (N).
From the equation:
F = Kq1q2/ r²
The force (F) is inversely proportional to the square of the distance (r) apart.. This implies that as the distance between the charged particles increase, the smaller the force of attraction between them and as the distance between the charged particles decrease, the greater the force of attraction between the two charged particles.
Answer:
3 and 6
Explanation:
A 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
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 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
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
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
Combustion of 30.42 g of a compound containing only carbon, hydrogen, and oxygen produces 35.21 g CO2 and 14.42 g H2O. What is the empirical formula of the compound
Answer:
C2H4O3
Explanation:
We would have to do some preparations between before solving it the normal way. The main goal is to get the masses of the Individual elements. So here goes;
We can get the mass of C from CO2 using the following steps:
1 mole of CO2 has a mass of 44g (Molar mass) and contains 12g of C.
How did we know the molar mass of CO2 is 44g?
Easy. 1 mole of C = 12, 1 mole of O = 16
But we have two O’s so the total mass of O = (2 * 6) = 32
Total mass of CO2 = mass of C + Mass of O = 12 + 32 = 44
So if 44g of CO2 contains 12g of C, how much of C would be present in 35.21g CO2.
12 = 44
X = 35.21
X = (35.21 * 12) / 44 = 9.603g
We can also get the mass of H from H2O. 1 mole of H2O has a mass of 18g and contains 2g of H.
How did we know the molar mass of H2O is 18g?
Easy. 1 mole of H = 1, 1 mole of O = 16
But we have two H’s so the total mass of H = (2 * 1) = 2
Total mass of H2O = mass of H + Mass of O = 2 + 16 = 18
So how much of H would be present in 14.42g of H2O?
2 = 18
X =14.42
X = (14.42 * 2 ) / 18 = 1.602g
Now we have the masses of C and H. But the question says the compound contains the C, H and O.
So we still have to calculate the mass of Oxygen. We obtain this from;
Mass of Compound = Mass of Carbon + Mass of Oxygen + Mass of Hydrogen
Mass of Oxygen = Mass of compound – (Mass of Carbon + Mass of Hydrogen)
Mass of Oxygen = 30.42 – (9.603 + 1.602)
Mass of Oxygen = 30.42 - 11.205 = 19.215
Now we have all the masses so we are good too go. Let’s have our table.
Elements Carbon (C) Hydrogen (H) Oxygen (O)
Mass 9.603 1.602 19.215
0.800 1.602 1.2001 (Divide by molar mass)
1 2 1.5 (Divide by lowest number)
2 4 3 (Convert to simple integers by * 2)
The Empirical formula of the compound is C2H4O3
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.
Calculate 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]
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.
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.
If 8.23 g of magnesium chloride react completely with sodium phosphate, how many grams of magnesium phosphateare produced
Answer:
The correct answer is 7.57 grams of magnesium phosphate.
Explanation:
Based on the given question, the chemical reaction taking place is:
2Na₃PO₄ (aq) + 3MgCl₂ (aq) ⇒ Mg₃(PO₄)2 (s) + 6NaCl (aq)
From the given reaction, it is evident that two moles of sodium phosphate reacts with three moles of magnesium chloride to produce one mole of magnesium phosphate.
Based on the given information, 8.23 grams of magnesium chloride reacts completely with sodium phosphate, therefore, magnesium chloride in the given case is the limiting reagent.
In the given case, 3 moles of magnesium chloride produce 1 mole of magnesium phosphate. Therefore, 1 mole of magnesium chloride will produce 1/3 mole of magnesium phosphate.
The molecular mass of magnesium chloride is 95.21 grams per mole. So, 1 mole of magnesium chloride is equivalent to 95.21 grams of magnesium chloride.
On the other hand, the molecular mass of magnesium phosphate is 262.85 grams per mole. Therefore, 1 mole of magnesium phosphate is equal to 262.85 grams of magnesium phosphate.
As seen earlier that 1 mole of magnesium chloride = 1/3 moles of magnesium phosphate. So,
95.21 grams of magnesium chloride = 1/3 × 262.85 grams of magnesium phosphate
= 262.85 / 3 grams of magnesium phosphate
1 gram of magnesium chloride = 262.85 / 3 × 95.21 grams of magnesium phosphate
8.23 grams of magnesium chloride = 262.85 / 3 × 95.21 × 8.23 grams of magnesium phosphate
= 7.57 grams of magnesium phosphate
Hence, when 8.23 grams of magnesium chloride when reacts completely with sodium phosphate, it produces 7.57 grams of magnesium phosphate.
How are the oxygen atoms balanced for redox equations in basic solutions
Answer: H2O and OH^- are added to balance the oxygen atoms.
Explanation:
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:
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.
a. If the temperature change of the dissolution of urea from your experiment were determined much lower than the actual temperature change, What is the effect on the value of entropy calculated from the incorrect data?
Explanation:
Before proceeding, we have to first understand the relationship between temperature and entropy.
From the equation of entropy which is given as;
ΔS = ΔQ / T
We can tell that entropy is inversely proportional to temperature.
This means that at constant heat, as the temperature increases, the entropy decreases and vice versa.
Inn this question, the temperature is much lower - a decrease. The effect that this would have on the value of the entropy is that it would be increased. A higher value of entropy would be recorded than the actual value of the entropy.
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]
Please help me solve this it’s very important I get this right
Answer:
D. exothermic reaction
Explanation:
In an exothermic reaction, the reactants are at a higher energy level than the products.
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 = 7.4x10⁻⁷➔ 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.
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.
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.
A chemist has a block of aluminum metal (density is 2.7 g/mL). The block weighs 1.5. What is the volume of the aluminum block?
Answer:
0.56 mL
Explanation:
Volume = mass ÷ density
Volume = 1.5 ÷ 2.7 g/mL
Volume = 0.5555555556 = 0.56 mL
The volume of the aluminum block is 0.56 mL.
Hope this helps. :)
The volume of aluminum block is 0.556 mL.
The density of a substance is its mass per unit volume.
It given by formula:
[tex]\text{Density}=\frac{\text{Mass}}{\text{Volume}} [/tex]
Given:
Density = 2.7 g/mL
Mass= 1.5 g
To find:
Volume=?
On substituting the values in the above formula:
[tex]\text{Density}=\frac{\text{Mass}}{\text{Volume}} \\\\\text{Volume}=\frac{\text{Mass}}{\text{Density}} \\\\\text{Volume}=\frac{1.5}{2.7} \\\\\text{Volume}=0.556mL[/tex]
Thus, the volume of the aluminum block is 0.556mL.
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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 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]
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:
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