Answer:
Atomic mass of Ag-107 = 106.94 amu
Explanation:
Let the mass of Ag-107 be y
Since the mass ratio of Ag-109/Ag-107 is 1.0187, the mass of Ag-109 is 1.0187 times heavier than the mass of Ag-107
Mass of Ag-109 = 1.0187y
Relative atomic mass of Ag = sum of (mass of each isotope * abundance)
Relative atomic mass of Ag = 107.87
Abundance of Ag-107 = 51.839% = 0.51839
Abundance of Ag-109 = 41.161% = 0.48161
107.87 = (y * 0.51839) + (1.018y * 0.48161)
107.87 = 0.51838y + 0.49027898y
107.87 = 1.00865898y
y = 107.87/1.00865898
y = 106.94 amu
Therefore, atomic mass of Ag-107 = 106.94 amu
how are mass and weight affected in chemical reactions?
Answer:
How the chemical reacts
Explanation:
Aspirin (C9H8O4) is produced by the reaction of salicylic acid (C7H6O3, Molar mass = 138.1 g/mol) and acetic anhydride (C4H6O3, Molar mass = 102.1 g/mol) based on the BALANCED equation : C7H6O3(s) + C4H6O3(l ) → C9H8O4(s) + C2H4O2( l) If 63.07 grams of aspirin (Molar mass = 180.2 g/mol) was collected from an experiment when 138.1 grams C7H6O3 reacted with excess C4H6O3, what was the percent yield?
Answer:
35%
Explanation:
Percentage yield = actual yield / theoretical yield × 100.
Given:
Actual yield = 63.07g
Theoretical yield = ?
Mole ratio of C7H6O3 to C4H6O3 = 1 : 1
1 mole of C7H6O3 - 138.1g
Which implies that only 1 mole s[tex]\frac{63.07}{180.2} * 100[/tex]hould be used up in the reaction, yielding 180.2 g of C9H8O4. ⇒ Theoretical yield = 180.2g
∴ % Yield = [tex]\frac{63.07}{180.2} * 100[/tex]
= 35% yield.
Let me know if you found this easy to understand.
Draw the Lewis structure for methane (CH4) and ethane (C2H6) in the box below. Then predict which would have the higher boiling point. Finally, explain how you came to that conclusion.
Answer:
Ethane would have a higher boiling point.
Explanation:
In this case, for the lewis structures, we have to keep in mind that all atoms must have 8 electrons (except hydrogen). Additionally, each carbon would have 4 valence electrons, with this in mind, for methane we have to put the hydrogens around the carbon, and with this structure, we will have 8 electrons for the carbon. In ethane, we will have a bond between the carbons, therefore we have to put three hydrogens around each carbon to obtain 8 electrons for each carbon.
Now, the main difference between methane and ethane is an additional carbon. In ethane, we have an additional carbon, therefore due to this additional carbon, we will have more area of interaction for ethane. If we have more area of interaction we have to give more energy to the molecule to convert from liquid to gas, so, the ethane will have a higher boiling point.
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The Lewis structure shows the valence electrons in a molecule. Ethane will have a higher boiling point than methane.
We can deduce the number of valence electrons in a molecule by drawing the Lewis structure of the molecule. The Lewis structure consists of the symbols of elements in the compound and the valence electrons in the compound.
We know that the higher the molar mass of a compound the greater its boiling point. Looking at the Lewis structures of methane and ethane, we cam see that ethane has a higher molecular mass (more atoms) and consequently a higher boiling point than methane.
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What is a major product of the reaction in the box?
Answer:
Molecule C
Explanation:
In this case, on the first reaction, we will have the production of a Grignard reagent. This molecule will react with [tex]D_2O[/tex] and a deuterium atom will be transferrred to the benzene ring. Then at the top of the molecule, we will have an acetal structure. This acetal can be broken by the action of the acid [tex]DCl[/tex], In the mechanism at the end, we will obtain a carbonyl group bonded to a hydrogen atom. Therefore we will have in the final product the aldehyde group. See figure 1 to further explanations.
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Nylon 88 is made from the monomers H2N(CH2)8NH2 and HOOC(CH2)6COOH. So, would you characterize nylon 88 as rather an addition or a condensation polymer? Please explain your answer.
Answer:
Combination of H2N(CH2)8NH2 and HOOC(CH2)6COOH leads to the loss of water molecules at each linkage position.
Explanation:
A condensation polymer is a polymer formed when two monomers combine with the elimination of a small molecule such as water. The removal of the small molecule occurs at the point where the two monomers are joined to each other.
Nylon is known to form condensation polymers. This is because it involves the linkage of an -OH group to an -NH2 group. Water is eliminated in the process.
In the case of H2N(CH2)8NH2 and HOOC(CH2)6COOH, linkage of the both monomers at the 8 position of each chain leads to the formation of nylon- 8,8 with loss of water molecules at each linkage position. This stepwise loss of water molecules at each linkage makes it a condensation polymer.
If the equilibrium constant of the reaction is 0.85, then which statement is true if the mass of A is 10.5 grams; the density of B is 0.82 g/ml; the concentration of C is 0.64 M; and the concentration of D is 0.38 M.
A(s) + 3 B(l) _____ 2(aq) + D(aq)
Pick the correct statement about this system.
A. Q < K and reaction shifts left
B. Q > K and reaction shifts left
C. Q > K and reaction shifts right
D. Q = K and reaction does not shift
E. Q < K and reaction shifts right
Answer:
E. Q < K and reaction shifts right
Explanation:
Step 1: Write the balanced equation
A(s) + 3 B(l) ⇄ 2(aq) + D(aq)
Step 2: Calculate the reaction quotient (Q)
The reaction quotient, as the equilibrium constant (K), only includes aqueous and gaseous species.
Q = [C]² × [D]
Q = 0.64² × 0.38
Q = 0.15
Step 3: Compare Q with K and determine in which direction will shift the reaction
Since Q < K, the reaction will shift to the right to attain the equilibrium.
11. In TLC analysis of ferrocene and acetylferrocene (on silica TLC plate) which prediction is correct: A) ferrocene is more polar and moves higher up the plate (higher Rf value) B) Acetylferrocene is more polar and moves higher up the plate (higher Rf value) C) ferrocene is less polar and moves higher up the plate (higher Rf value) D) Acetylferrocene is less polar and moves higher up the plate (higher Rf value)
Answer:
Alternative C would be the correct choice.
Explanation:
The dual compounds were evaluated on something like a TLC plate through three separate additives in conducting a TLC study of ferrocene versus acetylferrocene.The polar as well as nonpolar ferrocene where nonpolar is about 0.63 with the maximum [tex]R_f[/tex] value, and indeed the polar is somewhere around 0.19 with [tex]R_f[/tex].TLC plate (30:1 toluene/ethanol) established with.The other three choices are not related to the given circumstances. So that option C would be the appropriate choice.
Determine the volume occupied by 10 mol of helium at
27 ° C and 82 atm
Answer:
3.00 L
Explanation:
PV = nRT
(82 atm × 101325 Pa/atm) V = (10 mol) (8.314 J/mol/K) (27 + 273) K
V = 0.00300 m³
V = 3.00 L
Which is most likely to happen during a precipitation reaction?
A. A solid substance will break down into two new substances that
are gases.
B. An insoluble solid will form when ions in dissolved compounds
switch places.
C. A substance will react with oxygen to form water and carbon
dioxide.
D. A gas will form when positive ions switch places to form new
compounds.
Answer:
I think its B
Explanation:
Precipitation reactions leave a solid behind. The solid is called a precipitate.
Answer:
B
Explanation:
An insoluble solid will form when ions in dissolved compounds switch places.
The reaction, 2 SO3(g) <--> 2 SO2(g) + O2(g) is endothermic. Predict what will happen if the temperature is increased.
Explanation:
This reaction is in equilibrium and would hence obey lechatelier's principle. This principle states that whenever a system at equilibrium undergoes a change, it would react in way so as to annul that change.
Since it is an endothermic reaction, increasing the temperature would cause the reaction to shift towards the right.
This means that it favours product formation and more of the product would be formed.
Why are antiparallel beta sheets more stable than parallel beta sheets?
Answer:
The side chains of the amino acids alternate above and below the sheet
Explanation:
Hydrogen bonds are formed between the amine and carbonyl groups across strands. ... Antiparallel ß sheets are slightly more stable than parallel ß sheets because the hydrogen bonding pattern is more optimal.
2
22. A sodium chloride solution is 15.0% m/m%. Calculate mass of sodium chloride in 219 g solution.
14.2g
80.38
11.2 g
32.9 g
Answer: The mass of sodium chloride in 219 g solution is 32.9 g
Explanation:
To calculate the mass percent of element in a given compound, we use the formula:
[tex]\text{Mass percent of A}=\frac{\text{Mass of A}}{\text{mass of A +mass of B}}\times 100[/tex]
To find mass of sodium chloride in solution:
[tex]\text{Mass percent of sodium chloride}=\frac{\text{Mass of sodium chloride}}{\text{mass of solution}}\times 100[/tex]
Mass percent of sodium chloride= 15.0 %
Mass of solution = 219g
[tex]15=\frac{\text{Mass of sodium chloride}}{219}\times 100[/tex]
[tex]{\text{Mass of sodium chloride}=32.9g[/tex]
Thus mass of sodium chloride in 219 g solution is 32.9 g
Calculate the equilibrium concentrations of N2O4 and NO2 at 25 ∘C in a vessel that contains an initial N2O4 concentration of 0.0655 M . The equilibrium constant Kc for the reaction N2O4(g)⇌2NO2(g) is 4.64×10−3 at 25 ∘C. Express your answers using four decimal places separated by a comma.
Answer:
[N2O4] = 0.0573M
[NO2] = 0.0163M
Explanation:
The equilibrium of N2O4 is:
N2O4(g)⇌2NO2(g)
Where Kc is defined as:
Kc = 4.64x10⁻³ = [NO2]² / [N2O4]
When you add just N2O4, the reaction will occurs until [NO2]² / [N2O4] = 4.64x10⁻³. Here, the system reaches equilibrium.
That means if 0.0655M N2O4 begin reaction, in equilibrium we will have:
[N2O4] = 0.0655M - X
[NO2] = 2X
Where X is defined as reaction coordinate
Replacing in Kc:
4.64x10⁻³ = [NO2]² / [N2O4]
4.64x10⁻³ = [2X]² / [0.0655-X]
3.0392x10⁻⁴ - 4.64x10⁻³X = 4X²
3.0392x10⁻⁴ - 4.64x10⁻³X - 4X² = 0
Solving for X:
X = -0.0093 → False solution. there is no negative concentrations
X = 0.008156M → Right solution.
Replacing X, equilibrium concentrations are:
[N2O4] = 0.0655M - X
[NO2] = 2X
[N2O4] = 0.0573M[NO2] = 0.0163MThe accepted value of the number of Liters of gas in a mole is 22.4. List two possible reasons on why our experiment yielded a different value for the number of Liters in a mole of a gas.
Hint: Our experiment was conducted in July, in St. Paul, Minnesota.
Answer:
- Pressure in St. Paul, Minnesota
- Temperature in St. Paul, Minnesota
Explanation:
22.4 L or dm³ is the volume for a gas under Standard pressure and temperature conditions.
It is logically to say, that tempereature value at the day of the experiment was not 273.15 K, which is 32°F
We can say, that the pressure was not 1 atm. St Paul Minnesota has a minimum, but a little height, so the pressure differs by few figures from the standard pressure values.
We also have to mention, that 22.4 L is the value for the Ideal gases at standards conditions. Ideal gases does not exisist on practice, we always talk about real gases. Don't forget the Ideal Gases Law equation:
P . V = n . R . T
Pressure . Volume = number of moles . 0.082 L.atm /mol. K . 273.15K
Number of moles must be 1 at STP, to determine a volume of 22.4L
Given the initial rate data for the reaction being A + B + C --> D determine the rate expression for the reaction and the (k) rate constant. (The units of [A] [B] and [C] are all moles/liter and the units of IRR is moles/liter seconds). If the [A]=[B]=[C]=.30M, what would the IRR be? [A] [B] [C] IRR 0.20 0.10 0.40 .20 0.40 0.20 0.20 1.60 0.20 0.10 0.20 .20 0.20 0.20 0.20 .80
Answer:
k = 100 mol⁻² L² s⁻¹, r= k[A][B]²
Explanation:
A + B + C --> D
[A] [B] [C] IRR
0.20 0.10 0.40 .20
0.40 0.20 0.20 1.60
0.20 0.10 0.20 .20
0.20 0.20 0.20 .80
Comparing the third and fourth reaction, the concentrations of A and C are constant. Doubling the concentration of B causes a change in the rate of the reaction by a factor of 4.
This means the rate of reaction is second order with respect to B.
Comparing reactions 2 and 3, the concentrations of B and C are constant. Halving the concentration of A causes a change in the rate of the reaction by a factor of 2.
This means the rate of reaction is first order with respect to A.
Comparing reactions 1 and 3, the concentrations of A and B are constant. Halving the concentration of A causes no change in the rate of the reaction.
This means the rate of reaction is zero order with respect to C.
The rate expression for this reaction is given as;
r = k [A]¹[B]²[C]⁰
r= k[A][B]²
In order to obtain the value of the rate constant, let's work with the first reaction.
r = 0.20
[A] = 0.20 [B] = 0.10
k = r / [A][B]²
k = 0.20 / (0.20)(0.10)²
k = 100 mol⁻² L² s⁻¹
A saturated solution was formed when 5.16×10−2 L of argon, at a pressure of 1.0 atm and temperature of 25 ∘C, was dissolved in 1.0 L of water.
Calculate the Henry's law constant for argon. it must be im M/atm
Answer:
The Henry's law constant for argon is [tex]k=2.11*10^{-3}\frac{ M}{atm}[/tex]
Explanation:
Henry's Law indicates that the solubility of a gas in a liquid at a certain temperature is proportional to the partial pressure of the gas on the liquid.
C = k*P
where C is the solubility, P the partial pressure and k is the Henry constant.
So, being the concentration [tex]C=\frac{ngas}{V}[/tex]
where ngas is the number of moles of gas and V is the volume of the solution, you must calculate the number of moles ngas. This is determined by the Ideal Gas Law: P*V=n*R*T where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. So [tex]n=\frac{P*V}{R*T}[/tex]
In this case:
P=PAr= 1 atmV=VAr= 5.16*10⁻² LR=0.082 [tex]\frac{atm*L}{mol*K}[/tex]T=25 °C=298 °KThen:
[tex]n=\frac{1 atm*5.16*10^{-2} L}{0.082 \frac{atm*L}{mol*K} *298K}[/tex]
Solving:
n= 2.11 *10⁻³ moles
So: [tex]C=\frac{ngas}{V}=\frac{2.11*10^{-3} moles}{1 L} =2.11*10^{-3} \frac{moles}{L}= 2.11*10^{-3} M[/tex]
Using Henry's Law and being C=CAr and P =PAr:
2.11*10⁻³ M= k* 1 atm
Solving:
[tex]k=\frac{2.11*10^{-3} M}{1 atm}[/tex]
You get:
[tex]k=2.11*10^{-3}\frac{ M}{atm}[/tex]
The Henry's law constant for argon is [tex]k=2.11*10^{-3}\frac{ M}{atm}[/tex]
The Henry's law constant for argon gas in 1 litre of water is 2.1 × 10⁻³M/atm.
What is Henry's law?Henry's law of gas states that solubility of a gas in any liquid at particular temperature is directly proportional to the partial pressure of the gas.
C∝P
C = kP, where
k = Henry's constant
P = partial pressure of gas
C is the solubility and it is present in the form of concentration and will be calculated as:
C = n/V
n = no. of moles
V = volume
And moles of the gas will be calculated by using the ideal gas equation as:
PV = nRT
n = (1)(5.16×10⁻²) / (0.082)(298) = 2.1 × 10⁻³ moles
And Concentration in liquid will be:
C = 2.1 × 10⁻³mol / 1L = 2.1 × 10⁻³ M
Now we put all these values in the first equation to calculate the value of k as:
k = (2.1 × 10⁻³M) / (1atm) = 2.1 × 10⁻³M/atm
Hence required value of k is 2.1 × 10⁻³M/atm.
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Balance the following redox reaction in acidic solution: H+(aq)+Zn(s)→H2(g)+Zn2+(aq) Express your answer as a chemical equation. Identify all of the phases in your answer. nothing
Answer:
The balanced equation is: Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)
Explanation:
Zn(s) is a simple substance (its oxidation number is zero) and it is oxidized to Zn²⁺. It loses two electrons, so the half reaction is the following:
Zn(s) → Zn²⁺(aq) + 2 e- (oxidation reaction)
Hydrogen ion (H⁺) is reduced to hydrogen gas (H₂). The oxidation number is decreased from +1 to 0 (because H₂ is a simple substance). H⁺ gains 1 electron per H atom, so the half reaction is the following:
2H⁺(aq) + 2 e- → H₂(g) (reduction reaction)
We obtain the overall reaction from the addition of the two half reactions. We write the reduction reaction first and then the oxidation reaction, as follows:
2H⁺(aq) + 2 e- → H₂(g)
+
Zn(s) → Zn²⁺(aq) + 2 e-
---------------------------------
Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)
The two electrons at both sides of the equation (2 e-) are canceled. The overall reaction is in acidic solution due to the presence of H⁺ ions. The net charge at both sides is the same : +2, so the mass and the charge are balanced.
How many atoms of hydrogens are found in 3.21 mol of
C3H8?
Answer:
1.55 × 10²⁵ atoms of H
Explanation:
3.21mol C₃H₈ × 8mol H × (6.022×10²³)
The second-order decomposition of HI has a rate constant of 1.80 · 10-3 M-1s-1. How much HI remains after 27.3 s if the initial concentration of HI is 4.78 M?
Answer: 3.87M of HI remains after 27.3 s
Explanation:
Using the Second order decomposition equation of
1/[H]t =K x t +1/[A]o
Given initial concentration ,[A]o = 4.78M
time, t = 27.3 s
rate of constant , k= 1.80 x 10^-3 M-1s-1
1/[H] t= 1/[A] t= concentration after time, t=?
SOLUTION
1/[A] t =kt +1/[A]o
1/[A] t =(1.80 x 10^-3 (27.3)+1/4.78
0.04914+0.2092=0.2583
1/[A] t =0.2583
[A] t =1/0.2583= 3.87M
Write the electron configuration for the element titanium, TiTi. Express your answer in order of increasing orbital energy as a string without blank space between orbitals. For example, the electron configuration of LiLi could be entered as 1s^22s^1 or [He]2s^1.
Answer:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d² or [Ar] 3d² 4s²
Explanation:
Electron configuration can basically be referred to as the location of electron; how the electrons are arranged in the orbitals of the atoms.
Following Aufbau principle, electrons are arranged in the following order of orbitals.
1s 2s 2p 3s 3p 4s 3d and so on.
The s can hold a maximum of 2 electrons, p can hold a maximum of 6 electrons and d can hold a maximum of 10 electrons.
Titanium has an atomic number of 22. So the arrangement is given as;
Ti = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d²
The short hand notation is given as;
[Ar] 3d² 4s²
The electron configuration of Ti is
[tex]Ti: 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{2}[/tex] or [tex]Ti: [Ar] 4s^{2} 3d^{2}[/tex]
The electron configuration of an element is the distribution of electrons in atomic orbitals.
According to Aufbau's principle, the orbitals with lower energies are filled before the orbitals with higher energies.
We can know this order, using the diagonal rule (attached image).
The maximum number of electrons in each sublevel is:
s = 2p = 6d = 10f = 14Considering all these facts, and that Titanium has 22 electrons, the electron configuration of Ti is:
[tex]Ti: 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{2}[/tex]
Since [tex]1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6}[/tex] is the electron configuration of Argon, this can also be written as:
[tex]Ti: [Ar] 4s^{2} 3d^{2}[/tex]
The electron configuration of Ti is
[tex]Ti: 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{2}[/tex] or [tex]Ti: [Ar] 4s^{2} 3d^{2}[/tex]
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Review the reversible reactions given, along with the associated equilibrium constant Kat room temperature. In each case, determine whether the forward or reverse reaction is favored.
CH3COOH → CH3C00^- + H^+
Ka=1.8 x 10^-5
AgCl → Ag^+ + Cl^-
Ksp=1.6 x 10^-10
Al(OH)3 → Al^3+ + 3OH^-
Ksp=3.7 x 10^-15
A+B → C
K=4.9 x 10^3
Answer:
The answers to your questions are given below
Explanation:
The following data were obtained from the question:
CH3COOH → CH3C00^- + H^+
Equilibrium constant, Ka = 1.8 x 10^-5
AgCl → Ag^+ + Cl^-
Equilibrium constant, Ksp = 1.6 x 10^-10
Al(OH)3 → Al^3+ + 3OH^-
Equilibrium constant, Ksp = 3.7 x 10^-15
A+B → C
Equilibrium constant, K = 4.9 x 10^3
When the value of the equilibrium constant is grater than 1, it shows that the concentration of product is higher than that of the reactant and it implies that the forward reaction is favored.
When the value of the equilibrium constant is 1, it shows that the the concentration of the product and reactant are the same. Therefore neither the forward nor the reverse reaction is favored.
When the value of the equilibrium constant is lesser than 1, it shows that the concentration of the reactant is higher than the concentration of the product. Therefore, the reversed reaction is favored.
Now, we shall the question given above as follow:
A. CH3COOH → CH3C00^- + H^+
Equilibrium constant, Ka = 1.8 x 10^-5
Since the value of the equilibrium constant is lesser than 1, it means that the reverse reaction is favored.
B. AgCl → Ag^+ + Cl^-
Equilibrium constant, Ksp = 1.6 x 10^-10
Since the value of the equilibrium constant is lesser than 1, it means that the reverse reaction is favored.
C. Al(OH)3 → Al^3+ + 3OH^-
Equilibrium constant, Ksp = 3.7 x 10^-15
Since the value of the equilibrium constant is lesser than 1, it means that the reverse reaction is favored.
D. A+B → C
Equilibrium constant, K = 4.9 x 10^3
Since the value of the equilibrium constant is greater than 1, it means that the forward reaction is favored.
The reaction conditions are:
A. The reverse reaction is favored.
B. The reverse reaction is favored.
C. The reverse reaction is favored.
D. The forward reaction is favored.
Chemical reaction:
A. [tex]CH_3COOH[/tex] → [tex]CH_3COO^- + H^+[/tex]
Equilibrium constant, Ka = [tex]1.8 * 10^{-5}[/tex]
B. [tex]AgCl[/tex] → [tex]Ag^+ + Cl^-[/tex]
Equilibrium constant, Ksp = [tex]1.6 * 10^{-10}[/tex]
C. [tex]Al(OH)_3[/tex] → [tex]Al^{3+} + 3OH^-[/tex]
Equilibrium constant, Ksp = [tex]3.7 * 10^{-15}[/tex]
D. A+B → C
Equilibrium constant, K = [tex]4.9 * 10^3[/tex]
Conditions for Equilibrium constant:When the value of the equilibrium constant is greater than 1, it shows that the concentration of product is higher than that of the reactant and it implies that the forward reaction is favored.
When the value of the equilibrium constant is 1, it shows that the the concentration of the product and reactant are the same. Therefore neither the forward nor the reverse reaction is favored.
When the value of the equilibrium constant is lesser than 1, it shows that the concentration of the reactant is higher than the concentration of the product. Therefore, the reversed reaction is favored.
Thus, the reactions will be:
A. The reverse reaction is favored.
B. The reverse reaction is favored.
C. The reverse reaction is favored.
D. The forward reaction is favored.
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Please help me out ASAP!
What is meant by concentration?
Answer:
concentration is the abundance of a constituent divided by the total volume of a mixture.
Balance the following
Na+02-→ Na20
Al+O2 ->Al2O3
H2+12+ ->HI
Mg+H2O → Mg(OH)2+H2
Ca+O2 -> Cao
Answer:
1. Na + O2 → Na2O (Balanced)
2. 4Al + 3O2 → 2(Al2O3) (Balanced)
3. H2 + i2 → 2HI (Balanced)
4. Mg + 2H2O → Mg(OH)2+ H2 (Balanced)
5. 2Ca +O2 → 2CaO (Balanced)
Draw a Lewis structure for one important resonance form of HBrO4 (HOBrO3). Include all lone pair electrons in your structure. Do not include formal charges in your structure.
Answer:
The Lewis structure is attached with the answer -
Explanation:
Lewis structure or Lewis dot diagram are diagrams or representation of showing the bonding between different or same atoms of a molecule in any and also shows lone pairs of electrons that may exist in the molecule as dots.
HBrO₄ is bromine oxoacid which is also known as perbromic acid. It is a unstable inorganic compound.
The Lewis structure is attached in form of image with representation of lone pairs of electrons.
A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol of H2, 0.17 mol of CO, 0.74 mol of H2O, and some graphite. Some O2 is added to the system and a spark is applied so that the H2 reacts completely with the O2.
Find the amount of CO in the flask when the system returns to equilibrium.
Express your answer to two significant figures and include the appropriate units.
Answer:
0.44 moles
Explanation:
Given that :
A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol of H2, 0.17 mol of CO, 0.74 mol of H2O, and some graphite.
The equilibrium constant [tex]K_c= \dfrac{[CO][H_2]}{[H_2O]}[/tex]
The equilibrium constant [tex]K_c= \dfrac{(0.17 )(0.17)}{0.74}[/tex]
The equilibrium constant [tex]K_c= 0.03905[/tex]
Some O2 is added to the system and a spark is applied so that the H2 reacts completely with the O2.
The equation for the reaction is :
[tex]H_2 + \dfrac{1}{2}O_2 \to H_2O \\ \\ 0.17 \ \ \ \ \ \ \ \ \ \to0.17[/tex]
Total mole of water now = 0.74+0.17
Total mole of water now = 0.91 moles
Again:
[tex]K_c= \dfrac{[CO][H_2]}{[H_2O]}[/tex]
[tex]0.03905 = \dfrac{[0.17+x][x]}{[0.91 -x]}[/tex]
0.03905(0.91 -x) = (0.17 +x)(x)
0.0355355 - 0.03905x = 0.17x + x²
0.0355355 +0.13095 x -x²
x² - 0.13095 x - 0.0355355 = 0
By using quadratic formula
x = 0.265 or x = -0.134
Going by the value with the positive integer; x = 0.265 moles
Total moles of CO in the flask when the system returns to equilibrium is :
= 0.17 + x
= 0.17 + 0.265
= 0.435 moles
=0.44 moles (to two significant figures)
Identify the elements that have the following abbreviated electron configurations.
A) [Ne] 3s23p5.
B) [Ar] 4s23d7.
C) [Xe] 6s1.
Answer:
A) Chlorine (Cl)
B) Cobalt (Co)
C) Caesium (Cs)
Hope this helps.
The abbreviated electron configurations that was given in the question belongs to
Chlorine (Cl)
Cobalt (Co)
Caesium (Cs) respectively.
Electronic configurations can be regarded as the electronic structure, which is the way an electrons is arranged in energy levels towards an atomic nucleus.The electron configurations is very useful when describing the orbitals of an atom in its ground state.To calculate an electron configuration, we can put the periodic table into sections, and this section will represent the atomic orbitals which is the regions that house the electrons. Groups one of the period table and two belongs to s-block, group 3 through 12 belongs to the d-block, while 13 to 18 can be attributed to p-block ,The rows that is found at bottom are the f-blockTherefore, electron configurations explain orbitals of an atom when it is in it's ground state.
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1. Define the Law of Conservation of Mass (via text). Now that you’ve defined this law, explain what it means in your own words using an example.
Explanation:
The law of conservation of mass states that mass can neither be created nor be destroyed.
Explanation in own words = this means that in this universe no one can create or destroy mass.
No physical or chemical force.
Solid iron(II) oxide reacts with oxygen gas to produce solid iron(III) oxide. Balance the equation for this reaction (in lowest multiple integers). Write the unbalanced equation for this reaction.
Answer
Hello
I think the reaction is like this FeO+OFe²O³
And the balance reaction is 2Fe+OFe²O³
Explanation:
At first we should find sth that has more atoms than the other then for example we realized that we have two atoms of Fe in Fe²O³ then put 2 before FeO and now we have 2 atoms of Fe in right side and 2 atoms of Fe in left then Oxygen in FeO change to 2 atoms of Oxygen and we have an other one in right side that they become 3 atoms of Oxygen and now we have 3 atoms of Oxygen in both right and left side.
Finally our reaction balanced.
Good luck
230g sample of a compound contains 136.6g carbon, 26.4g hydrogen, and 31.8g nitrogen. What is masspercentif oxygen
Answer:
15.3 %
Explanation:
Step 1: Given data
Mass of the sample (ms): 230 gMass of carbon (mC); 136.6 gMass of hydrogen (mH): 26.4 gMass of nitrogen (mN): 31.8 gStep 2: Calculate the mass of oxygen (mO)
The mass of the sample is equal to the sum of the masses of all the elements.
ms = mC + mH + mN + mO
mO = ms - mC - mH - mN
mO = 230 g - 136.6 g - 26.4 g - 31.8 g
mO = 35.2 g
Step 3: Calculate the mass percent of oxygen
%O = (mO / ms) × 100% = (35.2 g / 230 g) × 100% = 15.3 %
Iron(II) is available to bond with chloride ion. How many of each type of ion will bond to form an ionic compound?
A) 3 iron(II), 1 chloride
B) 2 iron(II), 3 chloride
C) 2 iron(II), 1 chloride
D) 1 iron(II), 2 chloride
Answer:
D) 1 iron(II), 2 chloride
Explanation:
Iron II chloride is the compound; FeCl2. It is formed as follows, ionically;
Fe^2+(aq) + 2Cl^-(aq) -----> FeCl2
The formation of one mole of FeCl2 involves the reaction one mole of iron and two moles of chloride ions. This means that in FeCl2, the ratio of iron to chlorine is 1:2 as seen above.
Therefore there is one iron II ion and two chloride ions in each mole of iron II chloride, hence the answer.