A molecule of aluminum fluoride has one aluminum atom. How many fluorine atoms are present?
Answer:
3 fluorine atoms will be present
Answer:
3
Explanation:
The chemical formula of aluminum fluoride is AlF3. As you can see, there is a 1:3 ratio of aluminum atoms to fluorine atoms. Therefore, if a molecule of AlF3 has one aluminum atom, you know there must be 3 fluorine atoms present.
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This pluton occurs deep in Earth and does not cause any changes to the surface of Earth . True or False
Answer:
The given statement is false.
Explanation:
However, if the pluton exists beneath the ground, this could be conveniently shown in the illustration something from the peak such pluton appears convex in form resembling a lopolith and perhaps diapir, which would be a particular form of statistically significant pluton recognized as the sill.Mostly from the figure it could also be shown that subsurface sheets are lined or curved, throughout the pluton mold. And therefore it is inferred that such a pluton creates adjustment to something like the ground atmosphere by altering the form of the levels above it.So that the given is incorrect.
Un globo lleno de helio tenia un volumen de 8.5 L en el suelo a 20°C y a una presión de 750 torr. Cuando se le soltó, el globo se elevo a una altitud donde la temperatura era de -20°C y la presión de 425 torr, ¿Cuál era el volumen del gas del globo en estas condiciones?
Answer:
El volumen del gas era 12.95 L
Explanation:
Se relaciona la presión y el volumen mediante la ley de Boyle, que dice:
“El volumen ocupado por una determinada masa gaseosa a temperatura constante, es inversamente proporcional a la presión”
La ley de Boyle se expresa matemáticamente como: P*V=k
Por otro lado, la Ley de Charles consiste en la relación que existe entre el volumen y la temperatura absoluta de una cierta cantidad de gas ideal, el cual se mantiene a una presión constante. Esta ley dice que cuando la cantidad de gas y de presión se mantienen constantes, el cociente que existe entre el volumen y la temperatura siempre tendrán el mismo valor:
[tex]\frac{V}{T}=k[/tex]
Por último, la Ley de Gay Lussac dice que la temperatura absoluta y la presión son directamente proporcionales. Es decir, cuando se mantiene todo lo demás constante, mientras suba la temperatura de un gas subirá también su presión. Y mientras la temperatura del gas baje, lo mismo ocurrirá con la presión:
[tex]\frac{P}{T}=k[/tex]
Combinado las mencionadas tres leyes se obtiene:
[tex]\frac{P*V}{T} =k[/tex]
Cuando se desean estudiar dos diferentes estados, uno inicial y una final de un gas, se puede aplicar:
[tex]\frac{P1*V1}{T1} =\frac{P2*V2}{T2}[/tex]
Recordando que la temperatura debe usarse en grados Kelvin, conoces los siguientes datos:
P1: 750 torrV1: 8.5 LT1: 20°C= 293°K (siendo 0°C=273°K)P2: 425 torrV2: ?T2: -20°C= 253 °KReemplazando:
[tex]\frac{750 torr*8.5 L}{293K} =\frac{425 torr*V2}{253 K}[/tex]
Resolviendo:
[tex]V2=\frac{750 torr*8.5 L}{293K} *\frac{253 K}{425 torr}[/tex]
V2= 12.95 L
El volumen del gas era 12.95 L
The equilibrium between carbon dioxide gas and carbonic acid is very important in biology and environmental science. CO2 ( aq) + H2O ( l) H2CO3 ( aq) Which one of the following is the correct equilibrium constant expression (K c) for this reaction?
a) K =[H2CO3]/ [CO2]
b) K=[CO2]/ [H2CO3]
c) K=[H2CO3]/ [CO2][H2O]
d) K=[CO2][H2O]/ [H2CO3]
e) K=1/[H2CO3]
Answer:
Kc = [H₂CO₃] / [CO₂]
Explanation:
Equilibrium constant expression (Kc) of any reaction is defined as the ratio between molar concentrations in equilibrium of products over reactants.
Pure solids and liquids don't affect the equilibrium and you don't have to take its concentrations in the equilibrium.
Also, each specie must be powered to its reactant coefficient.
For example, for the reaction:
aA(s) + bB(aq) ⇄ cC(l) + nD(g) + xE(aq)
The equilibrium constant, kc is:
Kc = [D]ⁿ / [B]ᵇ[E]ˣ
You don't take A nor C species because are pure solids and liquids. b, n and x are the reactant coefficients of each substance. Ratio of products over reactants
Thus, for the reaction:
CO₂(aq) + H₂O(l) ⇄ H₂CO₃(aq)
The Kc is:
Kc = [H₂CO₃] / [CO₂]
Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this list of thermodynamic properties.HCl(g)+NaOH(s)⟶NaCl(s)+H2O(l)
Answer:
-179.06 kJ
Explanation:
Let's consider the following balanced reaction.
HCl(g) + NaOH(s) ⟶ NaCl(s) + H₂O(l)
We can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.
ΔH°r = 1 mol × ΔH°f(NaCl(s)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(NaOH(s))
ΔH°r = 1 mol × (-411.15 kJ/mol) + 1 mol × (-285.83 kJ/mol) - 1 mol × (-92.31 kJ/mol) - 1 mol × (-425.61 kJ/mol)
ΔH°r = -179.06 kJ
A 40.80 gram sample of copper is heated in the presence of excess sulfur. A metal sulfide is formed with a mass of 51.09 g. Determine the empirical formula of the metal sulfide.
Answer:
Cu₂S
Explanation:
From the question,
Cu S
Mass: 40.80 g 51.09-40.80 = 10.29 g
Mole ratio: 40.80/63.5 10.29/32.1
0.64 : 0.32
Divide by the smallest,
0.64/0.32 : 0.32/0.32
2 : 1
Therefore,
Empirical formula = Cu₂S.
Which Carbon is the triple bound attached to in 6-ethyl-2-octyne?
-first
-fourth
-third
-second
Answer:
-second
Explanation:
6-ethyl-2-octyne is an unsaturated compound with a triple bond.
6-ethyl-2-octyne will have a triple bound attached to the second carbon. The suffix -yne suggests that compound carry a triple bond and the number "2" before suffix refers to the position of triple bond that is second carbon.
Hence, the correct option is "-second ".
Write electron configurations for the following ion: Cd2 Cd2 . Express your answer in order of increasing orbital energy. For example, the electron configuration of LiLi would be entered in complete form as 1s^22s^1 or in condensed form as [He]2s^1.
Answer:
Cd2+ : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 or [Kr] 4d¹⁰
Explanation:
Before proceeding to write out the electron configuration of Cd2+, we have to obtain the electron configuration of Cadmium (Cd),
Cadmium has an atomic number of 48, this means that a neutral cadmium atom will have a total of 48 electrons surrounding its nucleus.
The electronic configuration of Cadmium is;
Cd: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10
The shorthand notation is given as;
Cd: [Kr] 4d¹⁰5s²
Cd2+ means that it has two less electrons, hence it's electron configuration is given as;
Cd2+ : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 or [Kr] 4d¹⁰
what is chemical equation of Braium chloride?
Answer:
BaCl2
Explanation:
Barium = Ba
Chloride => Cl-
Chemical Equation:
Ba + Cl => BaCl2
Note:
The valency of barium is 2 and valency of chloride is 1 (i.e. chlorine). The formula formed by the combination of these elements is BaCl2 (there's exchange of valencies when these two elements combine).
Based on their molecular structure, identify the stronger acid from each pair of oxyacids. Match the words in the left column to the appropriate blanks in the sentences on the right.
1) HI is a stronger acid than H2Te because iodine____than tellurium.
2) H2Te is a stronger acid than H2S because the H-Te bond is_____.
3) NaH is not acidic because hydrogen____than sodium.
a. has a more negative electron afflity
b. is more electronegative
c. has a larger atomic radius
d. stronger
e. is harder to ionize
Answer:
1)is more electronegative
2)
3) is more electronegative
Explanation:
1) for the first question, iodine is more electronegative than tellurium hence we naturally expect that HI should be more acidic than H2Te since electronegativities play a role in the acidity of chemical species.
2) the correct option is not listed because the H2Te bond is weaker than the H2S bond. This makes it easier for H2Te to dissociate releasing H^+ , thereby being more acidic than H2S.
3) Hydrogen is more electronegative than sodium hence it cannot be ionized thus NaH is not acidic.
18. Sucralose contains which two functional groups: (2 points)
A) benzene
B) halogen
C) carboxyl
D) hydroxy!
Answer:
The correct answer is option B and D, that is, halogen (chlorine) and hydroxyl.
Explanation:
An artificial sweetener and sugar substitute is sucralose. It is noncaloric as the majority of the sucralose ingested does not get dissociated within the body. The generation of sucralose takes place by the chlorination of sucrose. It is about 300 to 1000 times sweeter in comparison to sucrose.
The consumption of sucralose is safe for both nondiabetics and diabetics, it is used in various food and beverage components due to non-caloric sweetener characteristics. It does not affect the levels of insulin and does not affect dental health. As it is produced by chlorination of sucrose, thus, the functional groups present in it are a halogen (chlorine) and a hydroxyl.
Which is the electron configuration for bromine?
Answer:
The answer below would be written in a straight line from left to right but I wrote it as a list to make it easier to read.
Explanation:
1s^2
2s^2
2p^6
3s^2
3p^6
4s^2
3d^10
4p^5
6. Potassium hydrogen phthalate (KHP, KHC8H4O4) is also a good primary standard. 20 mL of NaOH was titrated with 0.600 M KHC8H4O4 solution. The data was graphed and the equivalence point was found when 15.5 mL of the standard 0.600 M KHP solution was added. The reaction equation is: a. What is the molar ratio of NaOH:KHC8H4O4? b. What is the molarity of the NaOH solution?
Answer:
a. 1
b. 0.465M NaOH
Explanation:
KHP reacts with NaOH as follows:
KHP + NaOH → KP⁻ + Na⁺ + H₂O
a. Molar ratio represents how many moles of NaOH reacts per mole of KHP. As you can see in the reaction, 1 mole of NaOH reacts with 1 mole of KHP. Molar ratio is:
1/1 = 1
b. With volume and molar concentration of the KHP solution you can find how many moles of KHP were added until equivalence point, thus:
15.5mL = 0.0155L ₓ (0.600 moles KHP / L) = 0.0093 moles of KHP
In equivalence point, moles of NaOH = Moles KHP. That means moles of NaOH titrated are 0.0093 moles NaOH.
The volume of the NaOH solution was 20mL = 0.020L. Molarity of the solution is:
0.0093 moles NaOH / 0.020L =
0.465M NaOHa. The balanced equation shows a 1:1 molar ratio between NaOH and KHC₈H₄O₄. This means that for every 1 mole of NaOH, we require 1 mole of KHC₈H₄O₄. Therefore, the molar ratio of NaOH:KHC₈H₄O₄ is 1:1.
The balanced equation for the reaction:
NaOH + KHC₈H₄O₄ → NaKC₈H₄O₄ + H₂O
b. Molarity of KHP solution × volume of KHP solution = Molarity of NaOH solution × volume of NaOH solution at the equivalence point
Molarity of KHP solution = 0.600 M
Volume of KHP solution = 15.5 mL = 0.0155 L
Volume of NaOH solution at the equivalence point = 20 mL = 0.0200 L
Molarity of NaOH solution = (Molarity of KHP solution × volume of KHP solution) / volume of NaOH solution at the equivalence point
Molarity of NaOH solution = (0.600 M × 0.0155 L) / 0.0200 L
Molarity of NaOH solution ≈ 0.465 M
To learn more about the balanced equation, follow the link:
https://brainly.com/question/12192253?referrer=searchResults
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A four carbon chain; the second carbon is also single bonded to CH3. Spell out the full name of the compound
Answer:
This description shows a methyl group.
Explanation:
Diluting sulfuric acid with water is highly exothermic:
(Use data from the Appendix to find for diluting 1.00 mol of H2SO4(l) (d = 1.83 g/mL) to 1 L of 1.00 MH2SO4(aq) (d = 1.060 g/mL). )
Suppose you carry out the dilution in a calorimeter. The initial T is 25.2°C, and the specific heat capacity of the final solution is 3.458 J/gK. What is the final T in °C ?
Answer:
The correct answer is 51.2 degree C.
Explanation:
The standard enthalpy for H₂SO₄ (l) is -814 kJ/mole and the standard enthalpy for H₂SO₄ (aq) is -909.3 kJ/mole.
Now the dHreaction = dHf (product) - dHf (reactant)
= -909.3 - (-814)
dHreaction or q = -95.3 kJ of energy will be used for dissociating one mole of H₂SO₄.
The heat change in calorimetry can be determined by using the formula,
q = mass * specific heat capacity * change in temperature -----------(i)
Based on the given information, the density of H₂SO₄ is 1.060 g/ml
The volume of H₂SO₄ is 1 Liter
Therefore, the mass of H₂SO₄ will be, density/Volume = 1.060 g/ml / 1 × 10⁻³ ml = 1060 grams
The initial temperature given is 25.2 degrees C, or 273+25.2 = 298.2 K, let us consider the final temperature to be T₂.
ΔT = T₂ -T₁ = T₂ - 298.2 K
Now putting the values in equation (i) we get,
95.3 kJ = 1060 grams × 3.458 j/gK (T₂ - 298.2 K) (the specific heat capacity of the final solution is 3.458 J/gK)
(T₂ - 298.2 K) = 95300 J / 1060 × 3.458 = 26 K
T₂ = 298.2 K + 26 K
T₂ = 324.2 K or 324.2 - 273 = 51.2 degree C.
A hot lump of 27.4 g of aluminum at an initial temperature of 69.5 °C is placed in 50.0 mL H2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the aluminum and water, given that the specific heat of aluminum is 0.903 J/(g·°C)? Assume no heat is lost to surroundings.
Answer:
[tex]\large \boxed{29.7 \,^{\circ}\text{C}}[/tex]
Explanation:
There are two heat transfers involved: the heat lost by the aluminium and the heat gained by the water.
According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.
Let the Al be Component 1 and the H₂O be Component 2.
Data:
For the Al:
[tex]m_{1} =\text{27.4 g; }T_{i} = 69.5 ^{\circ}\text{C; }\\C_{1} = 0.903 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}[/tex]
For the water:
[tex]m_{2} =\text{50.0 g; }T_{i} = 25.0 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}[/tex]
Calculations
(a) The relative temperature changes
[tex]\begin{array}{rcl}\text{Heat lost by Al + heat gained by water} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{27.4 g}\times 0.903 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{50.0 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\24.74\Delta T_{1} + 209.2\Delta T_{2} & = & 0\\\end{array}[/tex]
(b) Final temperature
[tex]\Delta T_{1} = T_{\text{f}} - 69.5 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 25.0 ^{\circ}\text{C}[/tex]
[tex]\begin{array}{rcl}24.74(T_{\text{f}} - 69.5 \, ^{\circ}\text{C}) + 209.2(T_{\text{f}} - 25.0 \, ^{\circ}\text{C}) & = & 0\\24.74T_{\text{f}} - 1719 \, ^{\circ}\text{C} + 209.2T_{\text{f}} -5230 \, ^{\circ}\text{C} & = & 0\\233.9T_{\text{f}} - 6949\, ^{\circ}\text{C} & = & 0\\233.9T_{\text{f}} & = & 6949 \, ^{\circ}\text{C}\\T_{\text{f}}& = & \mathbf{29.7 \, ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature is $\large \boxed{\mathbf{29.7 \,^{\circ}}\textbf{C}}$}[/tex]
Check:
[tex]\begin{array}{rcl}27.4 \times 0.903 \times (29.7 - 69.5) + 50.0 \times 4.184 (29.7 - 25.0)& = & 0\\24.74(-39.8) +209.2(4.7) & = & 0\\-984.6 +983.2 & = & 0\\-985 +983 & = & 0\\0&=&0\end{array}[/tex]
The second term has only two significant figures because ΔT₂ has only two.
It agrees to two significant figures
1. Methanol is a high-octane fuel used in high performance racing engines. 2 CH3OH(l) + 3O2(g) → 2CO2(g) + 4 H20(g) a) Calculate ∆H० and ∆S० using thermodynamic data, and then ∆G
Answer:
The reaction given in the question is:
2CH₃OH (l) + 3O₂ (g) ⇒ 2CO₂ (g) + 4H₂O (g)
The values of ΔH°formation and ΔS° of the reactants and products given in the reaction based on the thermodynamics data is:
ΔH°formation values of CH3OH (l) is -238.4 kJ/mol, CO2(g) is -393.52 kJ/mol, H2O (g) is -241.83 kJ/mol and O2 (g) is 0.
The S° values of CH3OH (l) is 127.19 J/molK, CO2(g) is 213.79 J/molK, H2O (g) is 188.84 J/moleK, and O2 (g) is 205.15 J/molK.
Now the values of ΔH° and ΔS° are,
ΔH°rxn = 2 * ΔH°formation CO2 (g) + 4 * ΔH°formation H2O (g) - 2*ΔH°formation CH3OH (l)
ΔH°rxn = 2 * (-393.52) + 4 (-241.83) -2 * (-238.4)
ΔH°rxn = -1277.56 kJ/mole
ΔS°rxn = 2 * S° CO2 (g) + 4 * S° H2O (g) - 2*S° CH3OH (l) - 3 * S° O2 (g)
ΔS°rxn = 2 * 213.79 + 4 * 188.84 - 2 * 127.19 - 3*205.15
ΔS°rxn = 313.11 J/mole/K
Now the formula for calculating ΔG°rxn is,
ΔG°rxn = ΔH°rxn - TΔS°rxn
ΔG°rxn = -1277.56 * 1000 J/mole - 298 * 313.11 J/mole
ΔG°rxn = -1370.86 kJ/mol
A. Identify the structure drawn below.
Answer:
Hexane
Explanation:
You have a carbon structure with only single bonds. This means that the name will end in -ane.
There are 6 carbon atoms. This means that the name will begin with hex-.
The structure is hexane.
1)The average lethal dose of Valium is 1.52 mg/kg of body weight. Estimate how many grams of Valium would be lethal for a 200.-lb woman. Show all your calculations. (1lb = 453.6 g)
2) A patient in hospital is receiving the antibiotic amoxcillin IV at the rate of 50. mL/h. The IV contains 1.5 g of the antibiotic in 1000. mL. (IV stands for intravenous). Calculate the mg/min of the drip. Show all your calculations
Answer:
1. 0.138g of valium would be lethel in the woman
2. 125mg/min is the drip of the patient
Explanation:
1. In a body, an amount of Valium > 1.52mg / kg of body weight would be lethal.
A person that weighs 200lb requires:
200lb × (453.6g / 1lb) × (1kg / 1000g) = 90.72kg (Weight of the woman in kg)
90.72kg × (1.52mg / kg) =
137.9mg ≡
0.138g of valium would be lethel in the woman2. The IV contains 1.5g = 1500mg/mL.
If the patient is receiving 5.0mL/h, its rate in mg/h is:
5.0mL/h × (1500mg/mL) = 7500mg/h
Now as 1h = 60min:
7500mg/h × (1h / 60min) =
125mg/min is the drip of the patientIf the average rate of the reaction A --->2B C is 1M/s, what is the average rate of formation (in M/s) of B over that same period of time
Answer:
[tex]r_B=2M/s[/tex]
Explanation:
Hello,
In this case, since the average rate of reaction is related with the consumption of A which has an stoichiometric coefficient of 1, the rate of formation of B will be:
[tex]r_B=2*1M/s\\\\r_B=2M/s[/tex]
By cause of the stoichiometric coefficient of B which doubles the average rate.
Best regards.
In the laboratory you are asked to make a 0.694 m copper(II) iodide solution using 455 grams of water. How many grams of copper(II) iodide should you add
Answer:
100.2g of CuI₂ you must add
Explanation:
Molality, m, is defined as the ratio between moles of solute and kg of solvent.
In the problem, you have a 0.694m of copper (II) iodide -CuI₂, molar mass: 317.35 g/mol-. That means there are 0.694 moles of CuI₂ per kg of water.
As you have 455g = 0.455kg of water -solvent-, moles of CuI₂ are:
0.455kg ₓ (0.694 moles CuI₂ / kg) = 0.316 moles of CuI₂
Using molar mass, grams of CuI₂ in the solution are:
0.316moles CuI₂ ₓ (317.35g / mol) =
100.2g of CuI₂ you must addidentify the correct acid/conjugate base pair in this equation:
NaHCO3 + H20 = + H2CO3 + OH
+ Na
H20 is an acid and H2CO3 is its conjugate base.
HCO3 is an acid and OH is its conjugate base.
H20 is an acid and HCO3 is its conjugate base.
H20 is an acid and OH is its conjugate base.
Answer:
H20 is an acid and OH is its conjugate base.
Explanation:
Chemical reactions involving acids and bases occur. An acid is a substance that dissociates in water i.e. lose an hydrogen ion/proton. According to the Bronsted-Lowry acid-base theory, when an acid dissociates in water and loses its hydrogen ion, the resulting substance that forms is the CONJUGATE BASE. A conjugate base is the compound formed as a result of the removal of an H+ ion from an acid.
Based on the chemical reaction in the question, NaHCO3 + H20 = H2CO3 + OH- + Na+
The H20 loses its hydrogen ion (H+) to form an anion OH-. This anion formed is the conjugate base while H20 is its acid.
Which of the following statements about water is not true?
Answer:
Water has a low specific heat capacity and so large bodies of water moderate temperatures on Earth.
Explanation:
Water has a very high specific heat capacity, meaning that it has to absorb a lot of energy to raise the temperature by one degree. Because water has a high specific heat capacity, large bodies of water can moderate the temperature of nearby land.
Hope this helps.
The NMR spectrum of your final compound will contain extra peaks that were not present in your starting material. For what hydrogen nuclei do those peaks occur?
Answer:
The peaks are registered from tetramethyl silane (TMS)
Explanation:
Tetramethyl silane (TMS) is used as internal reference in proton nmr (H NMR) spectrometry.
Its peak is usually registered at about a 2.0 chemical shift means that the hydrogen atoms which caused that peak need a magnetic field two millionths less than the field needed by TMS to produce resonance. This is not affected by the chemical shift of the sample analysed.
I hope this helped.
Solid sodium oxide and gaseous water are formed by the decomposition of solid sodium hydroxide (NaOH) .
Write a balanced chemical equation for this reaction.
Answer:
2NaOH(s) → Na₂O(s) + H₂O(g)
Hope that helps.
Consider the reaction 2N2(g) O2(g)2N2O(g) Using the standard thermodynamic data in the tables linked above, calculate Grxn for this reaction at 298.15K if the pressure of each gas is 22.20 mm Hg.
Answer:
[tex]\Delta G^0 _{rxn} = 207.6\ kJ/mol[/tex]
ΔG ≅ 199.91 kJ
Explanation:
Consider the reaction:
[tex]2N_{2(g)} + O_{2(g)} \to 2N_2O_{(g)}[/tex]
temperature = 298.15K
pressure = 22.20 mmHg
From, The standard Thermodynamic Tables; the following data were obtained
[tex]\Delta G_f^0 \ \ \ N_2O_{(g)} = 103 .8 \ kJ/mol[/tex]
[tex]\Delta G_f^0 \ \ \ N_2{(g)} =0 \ kJ/mol[/tex]
[tex]\Delta G_f^0 \ \ \ O_2{(g)} =0 \ kJ/mol[/tex]
[tex]\Delta G^0 _{rxn} = 2 \times \Delta G_f^0 \ N_2O_{(g)} - ( 2 \times \Delta G_f^0 \ N_2{(g)} + \Delta G_f^0 \ O_{2(g)})[/tex]
[tex]\Delta G^0 _{rxn} = 2 \times 103.8 \ kJ/mol - ( 2 \times 0 + 0)[/tex]
[tex]\Delta G^0 _{rxn} = 207.6\ kJ/mol[/tex]
The equilibrium constant determined from the partial pressure denoted as [tex]K_p[/tex] can be expressed as :
[tex]K_p = \dfrac{(22.20)^2}{(22.20)^2 \times (22.20)}[/tex]
[tex]K_p = \dfrac{1}{ (22.20)}[/tex]
[tex]K_p[/tex] = 0.045
[tex]\Delta G = \Delta G^0 _{rxn} + RT \ lnK[/tex]
where;
R = gas constant = 8.314 × 10⁻³ kJ
[tex]\Delta G =207.6 + 8.314 \times 10 ^{-3} \times 298.15 \ ln(0.045)[/tex]
[tex]\Delta G =207.6 + 2.4788191 \times \ ln(0.045)[/tex]
[tex]\Delta G =207.6+ (-7.687048037)[/tex]
[tex]\Delta G =[/tex] 199.912952 kJ
ΔG ≅ 199.91 kJ
Which accurately describes one impact of the atmosphere on Earth’s cycles?
Answer:
Produces Wind Currents
Explanation:
Answer:
produces wind currents
Explanation:
i just took the test and got it right :}
If phosphorus (P) has 4 naturally occurring isotopes, phosphorus-29(32.7.%), phosphorus-30(48.03%), phosphorus-31(18.4%), and phosphorus-33 (0.87%), what is its average r.a.m.?
The Average atomic mass of phosphorus is 29.9.
What is Average atomic mass ?The average atomic mass (sometimes called atomic weight) of an element is the weighted average mass of the atoms in a naturally occurring sample of the element.
Average masses are generally expressed in unified atomic mass units (u), where 1 u is equal to exactly one-twelfth the mass of a neutral atom of carbon-12.
An element can have differing numbers of neutrons in its nucleus, but it always has the same number of protons.
The versions of an element with different neutrons have different masses and are called isotopes.
The average atomic mass for an element is calculated by summing the masses of the element’s isotopes, each multiplied by its natural abundance on Earth i.e,
Average atomic mass of P = ∑(Isotope mass) (its abundance)
∴ Average atomic mass of P = (P-29 mass) (its abundance) + (P-30 mass)(its abundance) + (P-31 mass) (its abundance) + (P-33 mass) (its abundance)
Abundance of isotope = % of the isotope / 100.
∴ Average atomic mass of P = (29)(0.327) + (30)(0.4803) + (31)(0.184) + (33)(0.0087) = 29.88 a.m.u ≅ 29.9 a.m.u.
Hence , The Average atomic mass of phosphorus is 29.9.
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What is the name of this molecule?
Answer:
[tex]\boxed{Butyne}[/tex]
Explanation:
Triple Bonds => So it is an alkyne
The suffix used will be "-yne"
4 Carbons => The prefix used will be "But-"
Combining the prefix and suffix, we get:
=> Butyne
Answer:
[tex]\boxed{\mathrm{Butyne}}[/tex]
Explanation:
Alkynes have triple bonds ≡. The molecule has one triple bond.
Suffix ⇒ yne
The molecule has 4 carbon atoms and 6 hydrogen atoms.
Prefix ⇒ But (4 carbons)
The molecule is Butyne.
[tex]\mathrm{C_4H_6}[/tex]
Which of the following elements is in the same family as fluorine?
a. silicon
b. antimony
O c. iodine
O d. arsenic
e. None of these.
Answer:
c iodine
Explanation:
fluorine is a halogen group element like Bromine, Iodine,Astatine,Chloride