4NH3(g) 5O2(g)4NO(g) 6H2O(g) Using standard thermodynamic data at 298K, calculate the free energy change when 1.81 moles of NH3(g) react at standard conditions.
Answer:
-434.14 kJ
Explanation:
Step 1: Write the balanced equation
4 NH₃(g) + 5 O₂(g) ⇒ 4 NO(g) + 6 H₂O(g)
Step 2: Calculate the standard free energy change (ΔG°r) for the reaction
We will use the following expression.
ΔG°r = 4 mol × ΔG°f(NO(g)) + 6 mol × ΔG°f(H₂O(g)) - 4 mol × ΔG°f(NH₃(g)) - 5 mol × ΔG°f(O₂(g))
ΔG°r = 4 mol × (86.55 kJ/mol) + 6 mol × (-228.57 kJ/mol) - 4 mol × (-16.45 kJ/mol) - 5 mol × (0 kJ/mol)
ΔG°r = -959.42 kJ
Step 3: Calculate the standard free energy change for 1.81 moles of NH₃
959.42 kJ are released per 4 moles of NH₃.
[tex]\frac{-959.42 kJ}{4mol} \times 1.81mol = -434.14 kJ[/tex]
Predict the order of acid strengths in the following series of cationic
species: CH3CH2NH3
+, CH3CH=NH2
+ and CH3C-NH+
Answer:
CH3C-NH+> CH3CH=NH2 >CH3CH2NH3+
Explanation:
The acid strength has to do with the ease of the loss of hydrogen ion from the cationic specie. Hydrogen ion will easily be lost from any specie which contains an atom, group of atoms or bond which withdraws electrons along the chain of the N-H bond.
The pi bond system is known to be highly electronegative and withdraws electrons along the chain hence a withdrawal of electron density along the chain which makes the hydrogen ion to be easily lost from a system which contains a pi bond along the chain. A triple bond is more electronegative than a double bond, hence the answer above.
According to the following reaction, how many grams of ammonia will be formed upon the complete reaction of 31.2 grams of hydrogen gas with excess nitrogen gas ? nitrogen(g) + hydrogen(g) ammonia(g)
Answer:
176.8 g of ammonia, NH3.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
N2 + 3H2 —> 2NH3
Next, we shall determine the mass of H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:
Molar mass of H2 = 2x1 = 2 g/mol
Mass of H2 from the balanced equation = 3 x 2 = 6 g
Molar mass of NH3 = 14 + (3x1) = 17 g/mol
Mass of NH3 from the balanced equation = 2 x 17 = 34 g.
From the balanced equation above,
6 g of H2 reacted to produce 34 g of NH3.
Finally, we shall determine the mass of ammonia, NH3 produced by reacting 31.2 g of H2.
This can be obtained as follow:
From the balanced equation above,
6 g of H2 reacted to produce 34 g of NH3.
Therefore, 31.2 g of H2 will react to produce = (31.2 x 34)/6 = 176.8 g of NH3.
Therefore, 176.8 g of ammonia, NH3 were obtained from the reaction.
a boy capable of swimming 2.1m/a in still water is swimming in a river with a 1.8 m/a current. At what angle must he swim in order to end up directly opposite his starting point?
Answer:
The boy must swim at an angle of 59°northwest to get to a position directly opposite his starting point.
Explanation:
To get to a point directly opposite his starting point, the boy must travel at an angle x, in a direction northwest of his starting point. The speed of the boy and the speed of the river current forms a right-angled triangle with an an opposite side of 1.8 m/a and a hypotenuse of 2.1 m/a having an angle x.
Sin x = opp/ hyp
Sin x = 1.8/2.1
x = sin⁻¹ (1.8/2.10
x = 58.99
x = 59°
Therefore, the boy must swim at an angle of 59° in the northwesterly direction to get to a position directly opposite his starting point.
If the heat released during condensation goes only to warming the iron block, what is the final temperature (in ∘C) of the iron block? (Assume a constant enthalpy of vaporization for water of 44.0 kJ/mol and a heat capacity for iron of 0.449 J⋅g−1⋅∘C−1.)
Answer:
[tex]91°C[/tex]
Explanation:
CHECK THE COMPLETE QUESTION BELOW;
Suppose that 0.95 g of water condenses on a 75.0 g block of iron that is initially at 22 °c. if the heat released during condensation is used only to warm the iron block, what is the final temperature (in °c) of the iron block? (assume a constant enthalpy of vaporization for water of 44.0 kj/mol.)
Heat capacity which is the amount of heat required to raise the temperature of an object or a substance by one degree
From the question, it was said that that 0.95 g of water condenses on the block thenwe know that Heat evolved during condensation is equal to the heat absorbed by iron block.
Then number of moles =given mass/ molecular mass
Molecular mass of water= 18 g/mol
Given mass= 0.95 g
( 0.95 g/18 g/mol)
= 0.053 moles
Then Heat evolved during condensation = moles of water x Latent heat of vaporization
Q= heat absorbed or released
H=enthalpy of vaporization for water
n= number of moles
Q=nΔH
Q = 0.053 moles x 44.0 kJ/mol
= 2.322 Kj
=2322J
We can now calculate Heat gained by Iron block
Q = mCΔT
m = mass of substance
c = specific heat capacity
=change in temperature
m = 75 g
c = 0.450 J/g/°C
If we substitute into the above formula we have
Q= 75 x 0.450 x ΔT
2322 = 75 x 0.450 x ΔT
ΔT = 68.8°C
Since we know the difference in temperature, we can calculate the final temperature
ΔT = T2 - T1
T1= Initial temperature = 22°C
T2= final temperature
ΔT= change in temperature
T2 = T1+ ΔT
= 68.8 + 22
= 90.8 °C
=91°C
Therefore, final temperature is [tex]91°C[/tex]
The final temperature of the iron block is 91∘C.
Given that;
Heat lost during condensation of the water = Heat gained by iron block
Mass of water(mw) = 0.95 g
Latent heat of vaporization = Latent heat of condensation(L) = 44.0 kJ/mol
Mass of iron(mi) = 75.0 g
Initial temperature of iron(T1) = 22∘C
Final temperature of iron(T2) = ?
Heat capacity of iron(ci) = 0.449 J⋅g−1⋅∘C−1
So;
mwL = mici(T2 - T1)
Substituting values;
(0.95g/18g/mol) × 44.0 × 10^3(J/mol) = 75.0(g) × 0.449 J⋅g−1⋅∘C−1 (T2 - 22∘C)
2322.2 = 33.7T2 - 741.4
2322.2 + 741.4 = 37.4T2
T2 = (2322.2 + 741.4)/ 33.7
T2 =91∘C
Missing parts;
Suppose that 0.95 g of water condenses on a 75.0 g block of iron that is initially at 22 °c. if the heat released during condensation is used only to warm the iron block, what is the final temperature (in °c) of the iron block? (assume a constant enthalpy of vaporization for water of 44.0 kj/mol.)
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g Which ONE of the following molecules and ions has trigonal planar molecular geometry? (NOTE: You must first determine what the Lewis structure of each substance is.) A) PCl3 B) HCN C) CO3 2– D) H3O+ E) NF3
Answer: C) [tex]CO^{2-}_{3}[/tex]
Explanation: The Valence Shell Electron Pair Repulsion Model (VSEPR Model) shows bonding and nonbonding electron pairs present in the valence, outermost, shell of an atom connecting to other atoms. It also gives the molecular geometric shape of a molecule.
To determine molecular geometry:
1) Draw Lewis Structure, i.e., a simplified representation of the valence shell electrons;
2) Count the number of electron pairs (count multiple bonds as 1 pair);
3) Arrange electron pairs to minimise repulsion;
4) Position the atoms to minimise the lone pair;
5) Name the molecular geometry from the atom position;
Trigonal planar molecular geometry is a model which molecule's shape is triangular and in one plane. Such molecule has three regions of electron density extending out from the central atom and the repulsion will be at minimum when angle between any two is 120°.
The Lewis structure of each molecule is shown in the attachment.
Analysing each one, it can be concluded that molecule with trigonal planar geometry is [tex]CO^{2-}_{3}[/tex]
The compound methylamine, CH3NH2, is a weak base when dissolved in water. Write the Kb expression for the weak base equilibrium that occurs in an aqueous solution of methylamine:
Answer:
Kb = [CH₃NH₃⁺] × [OH⁻] / [CH₃NH₂]
Explanation:
According to Brönsted-Lowry acid-base theory:
An acid is a substance that donates H⁺.A base is a substance that accepts H⁺.When methylamine reacts with water, it behaves as a Brönsted-Lowry base, according to the following reaction.
CH₃NH₂(aq) + H₂O(l) ⇄ CH₃NH₃⁺(aq) + OH⁻(aq)
The basic equilibrium constant (Kb) is:
Kb = [CH₃NH₃⁺] × [OH⁻] / [CH₃NH₂]
When 5.58g H2 react by the following balanced equation, 32.8g H2O are formed. What is the percent yield of the reaction? 2H2(g)+O2(g)⟶2H2O(l)
A) 11.7%
B) 17.0%
C) 38.9%
D) 65.7%
Answer:
D) 65.7%
Explanation:
Based on the reaction:
2H2(g)+O2(g)⟶2H2O(l)
2 moles of hydrogen produce 2 moles of water assuming an excess of oxygen.
To find percent yield of the reaction we need to find theoretical yield (The yield assuming all hydrogen reacts producing water). With theoretical yield and actual yield (32.8g H₂O) we can determine percent yield as 100 times the ratio between actual yield and theoretical yield.
Theoretical yield:
Moles of 5.58g H₂:
5.58g H₂ ₓ (1 mol / 2.016g) = 2.768 moles H₂
As 2 moles of H₂ produce 2 moles of H₂O, if all hydrogen reacts will produce 2.768 moles H₂O. In grams:
2.768 moles H₂O ₓ (18.015g / mol) =
49.86g H₂O is theoretical yield
Percent yield:
Percent yield = Actual yield / Theoretical yield ₓ 100
32.8g H₂O / 49.86g ₓ 100 =
65.7% is percent yield of the reaction
D) 65.7%The respiration rate of a goldfish is measured. The goldfish is then placed in cold water and the respiration rate is measured again. What is the INDEPENDENT variable?
Answer:
Temperature of the water
Explanation:
In every study, there must be independent and dependent variables. An independent variable is the variable that is changed in order to obtain a response. In this case, the temperature of the water is being changed, the response in this experiment is the respiration rate of the goldfish.
Thus the respiration rate of the goldfish is the dependent variable because it is controlled by the temperature of the water and changes accordingly.
Summarily, the independent variable is the temperature of the water while the dependent variable is the respiration rate of the goldfish.
16. The concentration of a solution of potassium hydroxide is determined by titration with nitric
acid. A 30.0 mL sample of KOH is neutralized by 42.7 mL of 0.498 M HNO3. What is the
concentration of the potassium hydroxide solution?
Answer:
[tex]M_{base}=0.709M[/tex]
Explanation:
Hello,
In this case, since the reaction between potassium hydroxide and nitric acid is:
[tex]KOH+HNO_3\rightarrow KNO_3+H_2O[/tex]
We can see a 1:1 mole ratio between the acid and base, therefore, for the titration analysis, we find the following equality at the equivalence point:
[tex]n_{acid}=n_{base}[/tex]
That in terms of molarities and volumes is:
[tex]M_{acid}V_{acid}=M_{base}V_{base}[/tex]
Thus, solving the molarity of the base (KOH), we obtain:
[tex]M_{base}=\frac{M_{acid}V_{acid}}{V_{base}} =\frac{0.498M*42.7mL}{30.0mL}\\ \\M_{base}=0.709M[/tex]
Regards.
From the following balanced equation, CH4(g)+2O2(g)⟶CO2(g)+2H2O(g) how many grams of H2O can be formed when 1.25g CH4 are combined with 1.25×10^23 molecules O2? Use 6.022×10^23 mol−1 for Avogadro's number.
Answer:
2.81 g of H2O.
Explanation:
We'll begin by calculating mass of O2 that contains 1.25×10²³ molecules O2.
This can be obtained as follow:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.022×10²³ molecules. This implies that 1 mole of O2 also contains 6.022×10²³ molecules.
1 mole of O2 = 16x2 = 32 g.
Thus 6.022×10²³ molecules is present in 32 g of O2,
Therefore, 1.25×10²³ molecules will be present in =
(1.25×10²³ × 32) / 6.022×10²³ = 6.64 g of O2.
Therefore, 1.25×10²³ molecules present in 6.64 g of O2.
Next, the balanced equation for the reaction. This is given below:
CH4(g) + 2O2(g) —> CO2(g) + 2H2O(g)
Next, we shall determine the masses of CH4 and O2 that reacted and the mass of H2O produced from the balanced equation.
This can be obtained as follow:
Molar mass of CH4 = 12 + (4x1) = 16 g/mol.
Mass of CH4 from the balanced equation = 1 x 16 = 16 g
Molar mass of O2 = 16x2 = 32 g/mol.
Mass of O2 from the balanced equation = 2 x 32 = 64 g
Molar mass of H2O = (2x1) + 16 = 18 g/mol.
Mass of H2O from the balanced equation = 2 x 18 = 36 g
From the balanced equation above,
16 g of CH4 reacted with 64 g of O2 to produce 36 g if H2O.
Next, we shall determine the limiting reactant.
This can be obtained as follow:
From the balanced equation above,
16 g of CH4 reacted with 64 g of O2.
Therefore, 1.25 g of CH4 will react with = (1.25 x 64)/16 = 5 g of O2.
From the above calculations, we can see that only 5 g out of 6.64 g of O2 is needed to react completely with 1.25 g of CH4.
Therefore, CH4 is the limiting reactant.
Finally, we shall determine the mass of H2O produced from the reaction.
In this case, the limiting reactant will be used because it will give the maximum yield of H2O.
The limiting reactant is CH4 and the mass of H2O produced from the reaction can be obtained as follow:
From the balanced equation above,
16 g of CH4 reacted to produce produce 36 g if H2O.
Therefore, 1.25 g of CH4 will react to produce = (1.25 x 36)/16 = 2.81 g of H2O.
Therefore, 2.81 g of H2O were obtained from the reaction.
The mass in grams of H₂O which can be formed when 1.25g CH₄ are combined with 1.25×10²³ molecules O₂ is 2.8 grams.
What is stoichiometry?Stoichiometry of any reaction tells about the amount of species present before and after the completion of the reaction.
Given chemical reaction is:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
Moles of CH₄ will b calculate as:
n = W/M, where
W = given mass = 1.25g
M = molar mass = 16g/mol
n = 1.25/16 = 0.078 moles
Molecues of CH₄ in 0.078 moles = 0.078×6.022×10²³ = 0.46×10²³
Given molecules of O₂ = 1.25×10²³
Required molecules of CH₄ is less as compared to the molecules of O₂, so here CH₄ is the limiting reagent and formation of water is depends on it only.
From the stoichiometry of the reaction it is clear that:
1 mole of CH₄ = will produce 2 moles of H₂O
0.078 moles of CH₄ = will produce 2×0.078=0.156 moles of H₂O
Mass of H₂O will be calculated by using its moles as:
W = (0.156)(18) = 2.8g
Hence required mass of H₂O is 2.8g.
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A hypothetical metal crystallizes with the face-centered cubic unit cell. The radius of the metal atom is 198 picometers and its molar mass is 195.08 g/mol. Calculate the density of the metal in g/cm3.
Answer:
7.38 g/cm³ is the density of the metal
Explanation:
In a Face-centered cubic unit cell you have 4 atoms. Also, the edge length is √8×r (r is radius of the atom).
To solve this problem, we need first to calculate the volume of the unit cell and then, with molar mass calculate the mass of 4 atoms. As density is the ratio between mass and volume we can obtain this value.
Volume of the unit cellVolume = a³
a = √8×r
(r = 198x10⁻¹²m)
a = 5.6x10⁻¹⁰ m
Volume = 1.756x10⁻²⁸ m³
1m = 100cm → 1m³ = (100cm)³:
1.756x10⁻²⁸ m³× ((100cm)³ / 1m³) =
1.756x10⁻²² cm³ → Volume of the unit cell in cm³Mass of the unit cell:There are 4 atoms of gold:
4 atoms × (1mol / 6.022x10²³ atoms) = 6.64x10⁻²⁴ moles of gold
As 1 mole weighs 195.08g:
6.64x10⁻²⁴ moles of gold × (195.08g / mol) =
1.296x10⁻²¹g is the mass of the unit cellDensity of the metal:1.296x10⁻²¹g / 1.756x10⁻²² cm³ =
7.38 g/cm³ is the density of the metalThe density of the metal is 7.40 g/cm³
In cubic crystal system, face-centered cubic FFC is the name given to sort of atom arrangement observed in which structure is made up of atoms organized in a cube with a portion of an atom in each corner and six extra atoms in the center of each cube face.
It is expressed by using the formula:
[tex]\mathbf{\rho = \dfrac{Z \times M}{N_A\times a^}}[/tex]
where;
[tex]\rho[/tex] = density of the metalZ = atoms coordination no = 4 (for FCC)Molar mass (M) = 195.8 g/molAvogadro's constant (NA) = 6.022 × 10²³ /mola = edge lengthFor face-centered cubic FFC;
The edge length [tex]\mathbf{a =2 \sqrt{2}\times r }[/tex]
[tex]\mathbf{a =2 \sqrt{2}\times 198 \ pm }[/tex]
[tex]\mathbf{a =560.0285 \ pm }[/tex]
a = 5.60 × 10⁻⁸ cm
Replacing it into the previous equation, we have:
[tex]\mathbf{\rho = \dfrac{4 \times 195.8}{6.022 \times 10^{23} \times( 5.60 \times 10^{-8} )^3}}[/tex]
[tex]\mathbf{\rho = 7.40\ g/cm^3 }[/tex]
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21. What are the two main ways of working with clay?
Answer:
Diferentes tipos de arcilla
ARCILLA DE LADRILLOS. Contiene muchas impurezas. ...
ARCILLA DE ALFARERO. Llamada también barro rojo y utilizada en alfarería y para modelar. ...
ARCILLA DE GRES. Es una arcilla con gran contenido de feldespato. ...
ARCILLAS “BALL CLAY” O DE BOLA. ...
CAOLIN. ...
ARCILLA REFRACTARIA. ...
BENTONITA.
Explanation:
Answer:
Coil method and the slab method.
Explanation:
Enter the complementary strand of this DNA strand. Give your answer as a string.
Answer:
atagcca
Explanation:
t goes with a, and c goes with g
atagcca matches with
ta tcggt
conversion of 35 mL to ML
Answer:
1000ml=1l
35ml. = ?
Explanation:
35×1/1000
0.035litres
Advantages of using a resource person in handling the first aid lesson
The advantage of a resource person would be that it will provide a hands-on activity that will allow the students to experience spacing between organs and on the body of the person.
It will also allow them to identify challenges when doing this and will engage them more in the activity and lesson.
Answer:A resource person add knowledge to the course
Explanation:
Data is collected for the gas phase reaction 2 A + B + 3 C → Products at 470 K.What is the order of the reaction with respect to A?
Explanation:
The equation is given as;
2 A + B + 3 C → Products
The order of the reaction refers to the extent at which the rate depends n the concentration of the reactant.
The order of reaction is experimentally obtained. It can also be obtained from the rate law of the reaction.
If the rate law is given as;
rate law = k [A]²[B][C]³
Then the order is second order with respect to A.
The order is second order with respect to A.
Reaction series;Given that;
2A + B + 3C → Products at 470 K
Find:
Order of reaction with respect to A
Computation:
The reaction that takes place refers to how much the rate is influenced by the reactant concentration.
The reaction order is determined empirically. This can also be derived from the reaction's rate law.
Rate law = k[A]²[B][C]³
So, The order is second order with respect to A.
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A 3.140 molal solution of NaCl is prepared. How many grams of NaCl are present in a sample containing 2.692 kg of water
Answer:
494.49 g of NaCl.
Explanation:
Data obtained from the question include the following:
Molality of NaCl = 3.140 m
Mass of water = 2.692 kg
Mass of NaCl =.?
Next, we shall determine the number of mole of NaCl in the solution.
Molality is simply defined as the mole of solute per unit kilogram of solvent. Mathematically, it is expressed as
Molality = mole of solute /Kg of solvent
With the above formula, we can obtain the number of mole NaCl in the solution as follow:
Molality of NaCl = 3.140 m
Mass of water = 2.692 kg
Mole of NaCl =..?
Molality = mole of solute /Kg of solvent
3.140 = mole of NaCl /2.692
Cross multiply
Mole of NaCl = 3.140 x 2.692
Mole of NaCl = 8.45288 moles
Finally, we shall covert 8.45288 moles of NaCl to grams. This can be obtained as follow:
Mole of NaCl = 8.45288 moles
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl =.?
Mole = mass /Molar mass
8.45288 = mass of NaCl /58.5
Cross multiply
Mass of NaCl = 8.45288 × 58.5
Mass of NaCl = 494.49 g.
Therefore, 494.49 g of NaCl are present in the solution.
The mass of NaCl in 3.140 molal NaCl solution has been 494.493 grams.
Molality can be defined as the mass of solute present in 1000 grams of solvent.
Molality = [tex]\rm \dfrac{moles\;of\;solute}{mass\;of\;solvent\;(kg)}[/tex]
The moles of NaCl has been calculated as;
3.140 = [tex]\rm \dfrac{moles\;of\;NaCl}{mass\;of\;water\;(kg)}[/tex]
3.140 = [tex]\rm \dfrac{moles\;of\;NaCl}{2.692\;kg}[/tex]
Moles of NaCl = 3.140 [tex]\times[/tex] 2.692
Moles of NaCl = 8.45288 mol
The moles can be expressed as;
Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
Molecular weight of NaCl = 58.5 g/mol
The mass of NaCl can be calculated as:
8.45288 mol = [tex]\rm \dfrac{mass\;of\;NaCl}{58.5\;g/mol}[/tex]
Mass of NaCl = 58.5 g/mol [tex]\times[/tex] 8.45288 mol
Mass of NaCl = 494.493 grams.
The mass of NaCl in 3.140 molal NaCl solution has been 494.493 grams.
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1. What's the concentration of hydronium ions if a water-base solution has a temperature of 25°C (Kw = 1.0x10-14), with a concentration of
hydroxide ions of 2.21x10-6 M? A. 3.1x10-6 M
B. 4.52 X10-9 M
C. 2.8x10-8 M
D. 1.6x10-9 M
Answer:
B. 4.52 X10-9 M
Explanation:
Our goal for this question is to calculate the concentration of hydronium ions [tex]H^+[/tex] produced by water in a vessel with a concentration of hydroxide ions of [tex]2.21X10^-^6~M[/tex]. So, our first approach can be the ionization reaction of water:
[tex]H_2O_(_l_)~->~H^+~_(_a_q_)~+~OH^-~_(_a_q_)[/tex]
If we write the Keq expression for this reaction we will have:
[tex]Keq=[H^+][OH^-][/tex]
Now, water is the universal solvent, so, Keq has a special name. In the equilibrium problems for water we have to use "Kw" instead of "Keq":
[tex]Kw=[H^+][OH^-][/tex]
From this equation, we know the Kw value () and the concentration of the hydroxide ions ([2.21X10^-^6~M]). If we replace these values into the equation we can solve for [tex][H^+][/tex]:
[tex]1.0X10^-^1^4=[H^+][2.21X10^-^6~M][/tex]
[tex][H^+]=\frac{1.0X10^-^1^4}{2.21X10^-^6}=4.52X^-^9[/tex]
I hope it helps!
Answer:
B. 4.52 * 10^-9M
Explanation:
did the test
Each energy sublevel contains __________ number of electrons. For example, sublevel D can hold up to _______ electrons. A. the same, 10 B. the same, 14 C. a different, 6 D. a different, 10
Answer:
Each energy sublevel contains a different number of electrons. For example, sublevel D can contain up to 10 electrons
Explanation:
The atoms are surrounded by propellers that within each propeller there is a certain number of electrons, these electrons jump from orbit to orbit according to the amount of energy they have. The four levels that make up the electronic cloud that surrounds an atom are: s p d f.
When these electrons change orbit or level they release energy in the form of light, which is known as a photon.
what is ammonium nitrate
Answer:
Ammonium nitrate is a chemical compound with the chemical formula NH₄NO₃. It is a white crystalline solid consisting of ions of ammonium and nitrate.
10. For the following isotopes that have missing information, fill in the missing informatic
complete the notation: 36P
Answer:
Krypton.
Explanation:
Krypton is an atom which has 36 protons in its nucleus. There are 31 isotopes of Krypton which have same number of protons i. e. 36, same number of electrons i. e. 36 but different number of neutrons. Isotope refers to those atoms having same atomic number i. e. number of proton but different mass number i. e. number of neutron. For example, in Krypton-78, there 36 protons and 42 neutrons.
After combining hydrogen ions and hydroxide ions combine to form water molecules, the next step in balancing a redox reaction under basic conditions is to: ______
Explanation:
After combining hydrogen ions and hydroxide ions combine to form water molecules, the next step is the last step when balancing a redox reaction under basic condition.
The last step is to cancel all common terms that arises as a result of the formation of the water molecule. Usually, there's need to balance the water molecules in the reactant and product side of the reaction.
Nitric oxide (NO) can be formed from nitrogen, hydrogen and oxygen in two steps. In the first step, nitrogen and hydrogen react to form ammonia: (g) (g) (g) In the second step, ammonia and oxygen react to form nitric oxide and water: (g) (g) (g) (g) Calculate the net change in enthalpy for the formation of one mole of nitric oxide from nitrogen, hydrogen and oxygen from these reactions. Round your answer to the nearest .
Answer: [tex]\Delta H = -272.25kJ[/tex] for 1 mole of NO.
Explanation: Hess' Law of Constant Summation or Hess' Law states that the total enthalpy change of a reaction with multiple stages is the sum of the enthalpies of all the changes.
For this question:
1) [tex]N_{2}_{(g)} + 3H_{2}_{(g)}[/tex] => [tex]2NH_{3}_{(g)}[/tex] [tex]\Delta H=-92kJ[/tex]
2) [tex]4NH_{3}_{(g)}+5O_{2}_{(g)}[/tex] => [tex]4NO_{(g)}+6H_{2}O_{(g)}[/tex] [tex]\Delta H=-905kJ[/tex]
Amonia ([tex]NH_{3}_{(g)}[/tex]) appeares as product in the first equation and as reagent in the 2 reaction, so when adding both, there is no need to inverse reactions. However, in the 2nd, there are 4 moles of that molecule, so to cancel it, you have to multiply by 2 the first chemical equation and enthalpy:
[tex]2N_{2}_{(g)} + 6H_{2}_{(g)}[/tex] => [tex]4NH_{3}_{(g)}[/tex] [tex]\Delta H=-184kJ[/tex]
Now, adding them:
[tex]2N_{2}_{(g)} + 6H_{2}_{(g)}[/tex] => [tex]4NH_{3}_{(g)}[/tex] [tex]\Delta H=-184kJ[/tex]
[tex]4NH_{3}_{(g)}+5O_{2}_{(g)}[/tex] => [tex]4NO_{(g)}+6H_{2}O_{(g)}[/tex] [tex]\Delta H=-905kJ[/tex]
[tex]2N_{2}_{(g)}+6H_{2}_{(g)}+5O_{2}_{(g)}=>4NO_{(g)}+6H_{2}O_{(g)}[/tex] [tex]\Delta H = -185-905[/tex]
[tex]2N_{2}_{(g)}+6H_{2}_{(g)}+5O_{2}_{(g)}=>4NO_{(g)}+6H_{2}O_{(g)}[/tex] [tex]\Delta H = -1089kJ[/tex]
Note net enthalpy is for the formation of 4 moles of nitric oxide.
For 1 mole:
[tex]\Delta H = \frac{-1089}{4}[/tex]
[tex]\Delta H=-272.25kJ[/tex]
To form 1 mol of nitric oxide from nitrogen, oxygen and hydrogen, net change in enthalpy is [tex]\Delta H=-272.25kJ[/tex].
Given that H2(g)+F2(g)⟶2HF(g)ΔH∘rxn=−546.6 kJ 2H2(g)+O2(g)⟶2H2O(l)ΔH∘rxn=−571.6 kJ calculate the value of ΔH∘rxn for
Answer:
ΔH∘rxn of the reaction is -521.6kJ
Explanation:
Complete question: "Calculate the value of ΔH°rxn for 2F2(g)+2H2O(l)⟶4HF(g)+O2(g)"
You can find the ΔH of a reaction by the algebraic sum of similar reactions (Hess's law) as follows:
(1) H₂(g) + F₂(g) ⟶ 2HF(g) ΔH∘rxn=−546.6 kJ
(2) 2H₂(g)+O₂(g)⟶2H₂O(l) ΔH∘rxn=−571.6 kJ
Subtracting 2ₓ(1) - (2)
2ₓ(1) 2H₂(g) + 2F₂(g) ⟶ 4HF(g) ΔH∘rxn=2ₓ−546.6 kJ = -1093.2kJ
-(2) 2H₂O(l) ⟶ 2H₂(g)+O₂(g) ΔH∘rxn=- (-571.6 kJ) = 571.6kJ
2ₓ(1) - (2) 2F₂(g)+ 2H₂O(l) ⟶ 4HF(g) + O₂(g)
H₂(g) are cancelled because are the same in products and reactants
ΔH∘rxn = -1093.2kJ + 571.6kJ
ΔH∘rxn = -521.6
ΔH∘rxn of the reaction is -521.6kJWhat element is being reduced in the following redox reaction? MnO4-(aq) + H2C2O4(aq) → Mn2+(aq) + CO2(g) What element is being reduced in the following redox reaction? MnO4-(aq) + H2C2O4(aq) → Mn2+(aq) + CO2(g) H O Mn C
Answer: Mn is getting reduced.
Explanation:
Oxidation-reduction reaction or redox reaction is defined as the reaction in which oxidation and reduction reactions occur simultaneously.
Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.
Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.
[tex]MnO_4^-(aq)+H_2C_2O_4(aq)\rightarrow Mn^{2+}(aq)+CO_2(g)[/tex]
Oxidation : As Manganese has an oxidation state of +7 in [tex]MnO_4^-[/tex] and +2 in [tex]Mn^{2+}[/tex], the oxidation state is decreasing and hence it is getting reduced.
Reduction : As carbon has an oxidation state of +3 in [tex]H_2C_2O_4[/tex] and +4 in [tex]CO_2[/tex], the oxidation state is increasing and hence it is getting oxidized.
The element which is reduced in the redox reaction is Mn.
The redox reaction is shown below:
MnO₄⁻(aq) + H₂C₂O₄(aq) → Mn²⁺(aq) + CO₂(g)
A redox reaction is the type of reaction which involves oxidation processes
occurring with a corresponding reduction process.
Oxidation reaction involves the reaction in which a substance loses its
electrons to become positively charged with an increase in the oxidation
state.
Reduction reaction is defined as the reaction in which a substance gains
electrons to become positively charged with a decrease in the oxidation
state.
MnO₄⁻ has an oxidation state of +7 and Mn²⁺ having an oxidation state of +2
signifies a decrease in the oxidation state.
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A compound is found to contain 26.94 % nitrogen and 73.06 % fluorine by weight. To answer the questions, enter the elements in the order presented above. 1. What is the empirical formula for this compound? 2. The molecular weight for this compound is 104.02 g/mol. What is the molecular formula for this compound?
Answer:
THE EMPIRICAL FORMULA OF THE COMPOUND IS NF2
THE MOLECULAR FORMULA OF THE COMPOUND IS N2F4
Explanation:
To calculate the empirical formula for the compound, we:
1. Write out the percentage weight of each elements
N = 26.94%
F = 73.06 %
2. Divide each by its atomic mass
( N= 14, F = 19)
N = 26.94 / 14 = 1.924
F = 73.06 / 19 = 3.845
3. Divide each by the smaller of the values
N = 1.924 / 1.924 = 1
F = 3.845 / 1.924 = 1,998
4. Round up to a whole number and write the empirical formula
N= 1
F = 2
So the empirical formula of the compound is N F2
To calculate the molecular formula, we:
(N F2 )n = molecular weight
( 14 + 19*2) n = 104.02
52 n = 104.02
n = 2.000
The molecular formula of the compound will be:
(N F2)2 = N2F4
In conclusion, the empirical formula of the compound is NF2 and the molecular formula of the compound is N2F4
At 2000°C the equilibrium constant for the reaction 9_1.gif is 9_2.gif If the initial concentration of 9_3.gif is 0.200 M, what are the equilibrium concentrations of 9_4.gif and 9_5.gif?
Answer:
[tex][N_2]_{eq}=[H_2]_{eq}=0.09899M[/tex]
[tex][NO]_{eq}=0.00202M[/tex]
Explanation:
Hello,
In this case, for the given chemical reaction:
[tex]2NO\rightleftharpoons N_2+O_2[/tex]
We know the equilibrium constant and equilibrium expression:
[tex]Kc=2.4x10^3=\frac{[N_2][O_2]}{[NO]^2}[/tex]
That in terms of the reaction extent [tex]x[/tex] (ICE procedure) we can write:
[tex]2.4x10^3=\frac{x*x}{(0.2M-2*x)^2}[/tex]
In such a way, solving for [tex]x[/tex] by using a quadratic equation or solver, we obtain:
[tex]x_1=0.09899M\\x_2=0.1010M[/tex]
Clearly the solution is 0.09899M since the other value will result in a negative equilibrium concentration of NO. In such a way, the equilibrium concentrations of all the species are:
[tex][N_2]_{eq}=[H_2]_{eq}=x=0.09899M[/tex]
[tex][NO]_{eq}=0.2M-2*0.09899M=0.00202M[/tex]
Regards.
Define the following terms - you may need to consult your lecture text or other suitable resource:
a. monomer,
b. repeating unit,
c. condensation polymerization,
d. cross-linked polymer
Answer:
a) Monomers: monomers are unit molecules, that can react together with other monomers, to form a long chain molecule called a polymer. Th polymer formed can also be in a three dimensional network. The process of this conversion of monomers to polymers is called polymerization.
b) Repeating unit: A repeating unit is a unit of the polymer formed, whose repetition would produce a long complete polymer chain. A polymer is made up of these repeating links of molecules that form a long chain of molecules.
c) Condensation polymerization: This is a form of condensation reaction, that involves the combination of molecules into polymers with the loss of small molecules such as water or methanol as by products.
d) Cross-linked polymer: This is a polymer formed from a type of bonding of molecules. The bonding is usually in the form of covalent bonds or ionic bonds and the polymers can be either synthetic polymers or natural polymers. The cross-links leads to an alteration in the physical properties of the polymer.
The definition of following terms are :
a) Monomers:
The monomers are unit atoms, that can respond in conjunction with other monomers, to create a long chain molecule called a polymer.
The polymer shaped can too be in a three dimensional arrange.
b) Repeating unit:
A rehashing unit may be a unit of the polymer shaped, whose reiteration would produce a long total polymer chain.
A polymer is made up of these rehashing joins of atoms that shape a long chain of molecules.
c) Condensation polymerization:
This is often a frame of condensation response, that includes the combination of particles into polymers with the misfortune of little particles such as water or methanol as by products.
d) Cross-linked polymer:
This can be a polymer shaped from a sort of holding of particles.
The cross-links leads to an modification within the physical properties.
DefinitionsDefinition is a rhetorical style that uses various techniques to impress upon the reader the meaning of a term, idea, or concept.
Definition may be used for an entire essay but is often used as a rhetorical style within an essay that may mix rhetorical styles.
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Rectangular cube 3.2 m length 1.2 m in height and 5 m in length is split into two parts. The container has a movable airtight divider that divides its length as necessary. Part A has 58 moles of gas and part B has 165 moles of a gas.
Required:
At what length will the divider to equilibrium?
Answer:
The length the divider is to equilibrium from Part A = 1.30 m and from Part B = 3.70 m
Explanation:
Given that:
A rectangular cube with 3.2 m breadth, 1.2 m height and 5 m in length is splitted into two parts.
The diagrammatic expression for the above statement can be found in the attached diagram below.
The container has a movable airtight divider that divides its length as necessary.
Part A has 58 moles of gas
Part B has 165 moles of a gas.
Thus, the movable airtight divider will stop at a length where the pressure on it is equal on both sides.
i.e
[tex]\mathtt{P = P_A = P_B}[/tex]
Using the ideal gas equation,
PV = nRT
where, P,R,and T are constant.
Then :
[tex]\mathsf{\dfrac{V_A}{n_A}= \dfrac{V_B}{n_B}}[/tex]
[tex]\mathsf{\dfrac{L_A \times B \times H}{n_A}= \dfrac{L_B \times B \times H}{n_B}}[/tex] --- (1)
since Volume of a cube = L × B × H
From the question; the L = 5m
i,e
[tex]\mathsf{L_A +L_B}[/tex] = 5
[tex]\mathsf{L_A = 5 - L_B}[/tex]
From equation (1) , we divide both sides by (B × H)
Then :
[tex]\mathsf{\dfrac{L_A }{n_A}= \dfrac{L_B }{n_B}}[/tex]
[tex]\mathsf{\dfrac{5-L_B}{58}= \dfrac{L_B }{165}}[/tex]
By cross multiplying; we have:
165 ( 5 - [tex]\mathsf{L_B}[/tex] ) = 58 (
825 - 165[tex]\mathsf{L_B}[/tex] = 58
825 = 165[tex]\mathsf{L_B}[/tex] +58
825 = 223[tex]\mathsf{L_B}[/tex]
[tex]\mathsf{L_B}[/tex] = 825/223
[tex]\mathsf{L_B}[/tex] = 3.70 m
[tex]\mathsf{L_A = 5 - L_B}[/tex]
[tex]\mathsf{L_A = 5 - 3.70}[/tex]
[tex]\mathsf{ L_A}[/tex] = 1.30 m
The length the divider is to equilibrium from Part A = 1.30 m and from Part B = 3.70 m
