Answer:
Each molecule of O2 is made up of 2 oxygen atoms. So 1 mole of O2 molecules is made up of 2 moles of oxygen atoms. Therefore 1 mole of oxygen gas contains 2 moles of oxygen atoms. And 0.4 moles of oxygen gas contains 0.8 moles of oxygen atoms.
There are 0.8 moles of oxygen atoms in 0.4 moles of oxygen gas.
Oxygen gas (O₂) consists of two oxygen atoms bonded together. Therefore, to determine the number of moles of oxygen atoms present in a given amount of oxygen gas, we can simply multiply the number of moles of oxygen gas by the number of oxygen atoms per molecule, which is 2.
Given that we have 0.4 moles of oxygen gas, we can calculate the number of moles of oxygen atoms as follows:
Number of moles of oxygen atoms = Number of moles of oxygen gas × Number of oxygen atoms per molecule
= 0.4 moles × 2
= 0.8 moles
Therefore, there are 0.8 moles of oxygen atoms present in 0.4 moles of oxygen gas.
This calculation is based on the stoichiometry of oxygen gas, which indicates that each molecule of O₂ contains two oxygen atoms. By considering the mole ratio between oxygen gas and oxygen atoms, we can determine the number of moles of oxygen atoms in a given quantity of oxygen gas.
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What chemical bonds hold atoms?
Consider these metal ion/metal standard reduction potentials Cd2+(aq)|Cd(s) Zn2+(aq)|Zn(s) Ni2+(aq)|Ni(s) Cu2+(aq)|Cu(s) Ag+(aq)|Ag(s) -0.40 V -0.76 V ‑0.25 V +0.34 V +0.80 V Based on the data above, which species is the best reducing agent?
Answer:
The best reducing agent is Zn(s)
Explanation:
A reducing agent must to be able to reduce another compound, by oxidizing itself. Consequently, the oxidation potential must be high. The oxidation potential of a compound is the reduction potential of the same compound with the opposite charge. Given the reduction potentials, the best reducing agent will be the compound with the most negative reduction potential. Among the following reduction potentials:
Cd₂⁺(aq)|Cd(s) ⇒ -0.40 V
Zn²⁺(aq)|Zn(s) ⇒ -0.76 V
Ni²⁺(aq)|Ni(s) ⇒‑0.25 V
Cu²⁺(aq)|Cu(s) ⇒ +0.34 V
Ag⁺(aq)|Ag(s) ⇒ +0.80 V
The most negative is Zn²⁺(aq)|Zn(s) ⇒ -0.76 V
From this, the most reducing agent is Zn. Zn(s) is oxidized to Zn²⁺ ions with the highest oxidation potential (0.76 V).
what is calcium anyone tell plz
Answer:
Calcium is a chemical element with the symbol Ca and atomic number 20.
Calcium is a mineral that is necessary for life. In addition to building bones and keeping them healthy, calcium enables our blood to clot, our muscles to contract, and our heart to beat. About 99% of the calcium in our bodies is in our bones and teeth.
Calculate the amount of heat that must be absorbed by 10.0 g of ice at –20°C to convert it to liquid water at 60.0°C. Given: specific heat (ice) = 2.1 J/g·°C; specific heat (water) = 4.18 J/g·°C; ΔH fus = 6.0 kJ/mol.
Answer:
The amount of heat to absorb is 6,261 J
Explanation:
Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.
The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.
So:
Heat required to raise the temperature of ice from -20 °C to 0 °CBeing the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).
In this case, m= 10 g, specific heat of the ice= 2.1 [tex]\frac{J}{g*C}[/tex] and ΔT=0 C - (-20 C)= 20 C
Replacing: Q= 10 g*2.1 [tex]\frac{J}{g*C}[/tex] *20 C and solving: Q=420 J
Heat required to convert 0 °C ice to 0 °C waterThe heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:
Q= m* ΔHfusion
In this case, being 1 mol of water= 18 grams: Q= 10 g*[tex]6.0 \frac{kJ}{mol} *\frac{1 mol of water}{18 g}[/tex]= 3.333 kJ= 3,333 J (being kJ=1,000 J)
Heat required to raise the temperature of water from 0 °C to 60 °CIn this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 [tex]\frac{J}{g*C}[/tex] and ΔT=60 C - (0 C)= 60 C
Replacing: Q= 10 g*4.18 [tex]\frac{J}{g*C}[/tex] *60 C and solving: Q=2,508 J
Finally, Qtotal= 420 J + 3,333 J + 2,508 J
Qtotal= 6,261 J
The amount of heat to absorb is 6,261 J
The amount of heat to absorb is 6,261 J.
Calculation for heat:Heat required to raise the temperature of ice from -20 °C to 0 °C.
The formula for specific heat is used to calculate the amount of heat
Q = c * m * ΔT
Where,
Q =heat exchanged by a body,
m= mass of the body
c= specific heat
ΔT= change in temperature
Given:
m= 10 g,
specific heat of the ice= 2.1
ΔT=0 C - (-20 C)= 20 C
On substituting the values:
Q= 10 g*2.1 *20 C
Q=420 J
Heat required to convert 0 °C ice to 0 °C water.
The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:
Q= m* ΔHfusion
Heat required to raise the temperature of water from 0 °C to 60 °C
m= 10 g,
Specific heat of the water= 4.18
ΔT=60 C - (0 C)= 60 C
On substituting:
Q= 10 g*4.18 *60 C
Q=2,508 J
Thus, Qtotal= 420 J + 3,333 J + 2,508 J
Qtotal= 6,261 J
The amount of heat to absorb is 6,261 J
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is the general formula of a certain hydrate. When 256.3 g of the compound is heated to drive off the water, 214.2 g of anhydrous compound is left. Further analysis shows that the percentage composition of the anhydrate is 21.90% Ca, 43.14% Se, and 34.97% O.. (Hint: Treat the anhydrous compound and water just as you have treated elements in calculating in the formula of the hydrate.) (Use an asterisk to enter the dot in the formula. If a subscript is 1, omit it.) Find the empirical formula of the anhydrous compound. Find the empirical formula of the hydrate.
Answer:
The general formula of the hydrate is Caa Seb Oc. nH2O. Based on the given information, the weight of the hydrated compound is 256.3 grams, the weight of the anhydrous compound is 214.2 grams.
Therefore, the weight of water evaporated is 256.3 g - 214.2 g = 42.1 grams
The molecular weight of water is 18 gram per mole. So, the number of moles of water will be,
Moles of water = weight of water/molecular weight
= 42.1 grams / 18 = 2.3
The given composition of calcium is 21.90 %. So, the concentration of calcium in anhydrous compound is,
= 214.2 * 0.2190 = 46.91 grams
The given composition of Se is 43.14 %. So, the concentration of selenium in anhydrous compound is,
= 214.2 * 0.4314 = 92.40 grams
The given composition of oxygen is 34.97%, So, the concentration of oxygen in anhydrous compound is,
= 214.2 * 0.3497 = 74.91 grams
The molecular weight of Ca is 40.078, the obtained concentration is 46.91 grams, stoichiometry will be, 46.91/40.078 = 1.17
The molecular weight of Se is 78.96, the obtained concentration is 92.40, stoichiometry will be,
92.40/78.96 = 1.17
The molecular weight of Oxygen is 15.999, the concentration obtained is 74.91, the stoichiometry will be,
74.91/15.999 = 4.68.
Thus, the formula becomes, Ca1.17. Se1.1e O4.68. 2.3H2O, the closest actual component is CaSeO4.2H2O
A balloon is filled to a volume of 1.50 L with 3.00 moles of gas at 25 °C. With pressure and temperature held constant, what will be the volume (in L) of the balloon if 0.50 moles of gas are released?
Answer:
Volume : 1.25 L
Explanation:
We are given here that the volume ( V[tex]_1[/tex] ) = 1.50 Liters, the initial moles ( held at 25 °C ) = 3.00 mol, and the final moles ( n[tex]_2[/tex] ) = 3.00 - 0.5 = 2.5 mol. The final mol is calculated given that 0.50 mol of gas are released from the prior 3.00 moles of gas.
Volume ( V[tex]_1[/tex] ) = 1.50 L,
Initial moles ( n[tex]_1[/tex] ) = 3.00 mol,
Final Volume ( n[tex]_2[/tex] ) = 3.00 - 0.5 = 2.5 mol
Applying the combined gas law, we can calculate the final volume ( V[tex]_2[/tex] ).
P[tex]_1[/tex]V[tex]_1[/tex] / n[tex]_1[/tex]T[tex]_1[/tex] = P[tex]_2[/tex]V[tex]_2[/tex] / n[tex]_2[/tex]T[tex]_2[/tex] - we know that the pressure and temperature are constant, and therefore we can apply the following formula,
V[tex]_1[/tex] / n[tex]_1[/tex] = V[tex]_2[/tex] / n[tex]_2[/tex] - isolate V[tex]_2[/tex],
V[tex]_2[/tex] = V[tex]_1[/tex] n[tex]_2[/tex] / n[tex]_1[/tex] = 1.50 L [tex]*[/tex] 2.5 mol / 3.00 mol = ( 1.5 [tex]*[/tex] 2.5 / 3 ) L = 1.25 L
The volume of the balloon will be 1.25 L.
What is meant by functional group in an organic compound?
Functional group for
A) keton
B) acid anhydride
C) aldehyde
D) amide
C) alcohol
Answer:
i THINK the answer to this is C) aldehyde
Answer:
The answer is C, the guy above is correct.
Explanation:
is a polyprotic acid. Write balanced chemical equations for the sequence of reactions that carbonic acid can undergo when it's dissolved in water.
Answer:
H₂CO₃ H₂O ⇄ HCO₃⁻ + H₃O⁺ Ka1
HCO₃⁻ + H₂O ⇄ CO₃⁻² + H₃O⁺ Ka2
CO₃⁻² + H₂O ⇄ HCO₃⁻ + OH⁻ Kb1
HCO₃⁻ + H₂O ⇄ H₂CO₃ + OH⁻ Kb2
Explanation:
Formula for carbonic acid is: H₂CO₃
It is a dyprotic acid, because it can release two protons. We can also mention that is a weak one. The equilibrums are:
H₂CO₃ H₂O ⇄ HCO₃⁻ + H₃O⁺ Ka1
HCO₃⁻ + H₂O ⇄ CO₃⁻² + H₃O⁺ Ka2
When the conjugate strong bases, carbonate and bicarbonate take a proton from water, the reactions are:
CO₃⁻² + H₂O ⇄ HCO₃⁻ + OH⁻ Kb1
HCO₃⁻ + H₂O ⇄ H₂CO₃ + OH⁻ Kb2
Notice, that bicarbonate anion can release or take a proton to/from water. This is called amphoteric,
A 635 mL NaCl solution is diluted to a volume of 1.13 L and a concentration of 5.00 M . What was the initial concentration C1?
Answer:
8.90 M
Explanation:
Step 1: Given data
Initial concentration (C₁): ?Initial volume (V₁): 635 mL = 0.635 LFinal concentration (C₂): 5.00 MFinal volume (V₂): 1.13 LStep 2: Calculate the initial concentration
We have a concentrated NaCl solution and we want to prepare a diluted one. We will use the dilution rule.
C₁ × V₁ = C₂ × V₂
C₁ = C₂ × V₂ / V₁
C₁ = 5.00 M × 1.13 L / 0.635 L
C₁ = 8.90 M
Answer:
[tex]\large \boxed{\text{8.90 mol/L}}[/tex]
Explanation:
We can use the dilution formula to calculate the concentration of the original solution.
[tex]\begin{array}{rcl}V_{1}c_{1} & = & V_{2}c_{2}\\\text{635 mL }\times c_{1} & = & \text{1130 mL} \times \text{5.00 mol/L}\\635 c_{1}&=& \text{5650 mol/L}\\c_{1}& = & \dfrac{5650}{635}\text{ mol/L}\\\\& = & \textbf{8.90 mol/L}\\\end{array}\\\text{The initial concentration was $\large \boxed{\textbf{8.90 mol/L }}$}[/tex]
Calculate the percentage of O in CH₃COOH
Answer:
53.28%
(you could also write 55.33%, but I rounded it so just go with the answer above ---> 53.28%)
Explanation:
Akeem cut his finger during an investigation, and it is bleeding slightly. Before helping him bandage the wound,
which precaution should the teacher take?
O Tell someone to call 911,
O Put on protective gloves.
O Wash Akeem's finger in the shower.
O Apply disinfectant before cleaning.
Answer:
b.) Put on protective gloves
Answer:
2020 Put on protective gloves.
Explanation:
Which of the examples is potassium?
es )
A)
B)
B
C)
Answer:
examples of things which contain potassium are:
green vegetables
root vegetables
fruits
potassium chloride
potassium sulphate
Explanation:
if you need a specific answer please send the options
Answer:
C
Explanation:
The answer is the one with 20 protons, 20 neutrons, and 6-8-8-2 electrons.
Find the standard enthalpy of formation of iodine atoms. (Round your answer to one decimal place.) Standard enthalpy of formation
Answer:
Enthalpy of formation is the energy change when one mole of a substance is formed from its constituent atoms under standard conditions
Write an equation to show how the base NaOH(s) behaves in water. Include states of matter in your answer. Click in the answer box to open the symbol palette.
Answer:
The reaction is given as:
[tex]NaOH(s)\rightarrow Na^+(aq)+OH^-(aq)[/tex]
Explanation:
Bases are defined as those chemical substances which give hydroxide ions in their aqueous solutions.
[tex]BOH(s)\rightarrow B^+(aq)+OH^-(aq)[/tex]
When sodium hydroxide is added to water it gets dissociated into two ions that are sodium ions and hydroxide ions. Along with this heat energy also releases during this reaction.
The reaction is given as:
[tex]NaOH(s)\rightarrow Na^+(aq)+OH^-(aq)[/tex]
The equation to show how NaOH behaves in water is NaOH → Na⁺ + (OH)⁻
The compound that produce negative hydroxide (OH−) ions when dissolved
in water are called bases .
This compounds NaOH (sodium hydroxide) is an example of a base.
When it dissolves in water it dissociate to form negative hydroxide (OH−)
ions and positive sodium (Na+) ions.
It can be represented by the following equation:
NaOH → Na⁺ + (OH)⁻
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243
Am
95
1. The atomic symbol of americium-243 is shown. Which of the following is correct?
• A. The atomic mass is 243 amu, and the atomic number is 95.
B. The atomic mass is 338 amu, and the atomic number is 95.
• C. The atomic mass is 95 amu, and the atomic number is 243.
D. The atomic mass is 243 amu, and the atomic number is 338.
Answer:
A. The atomic mass is 243 amu, and the atomic number is 95.
Calculate the pH of a solution formed by mixing 250.0 mL of 0.15 M NH4Cl with 200.0 mL of 0.12 M NH3. The Kb for NH3 is 1.8 × 10-5.
Answer:
The pH of the solution is 9.06.
Explanation:
The reaction of the dissociation of NH₃ in water is:
NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq) (1)
[NH₃] - x [NH₄⁺] + x x
The concentration of NH₃ and NH₄⁺ is:
[tex] [NH_{3}] = \frac{n_{NH_{3}}}{V_{T}} = \frac{C_{i}_{(NH_{3})}*Vi_{NH_{3}}}{V_{NH_{3}} + V_{NH_{4}^{+}}} = \frac{0.12 M*0.2 L}{0.2 L + 0.25 L} = 0.053 M [/tex]
[tex] [NH_{4}^{+}] = \frac{C_{i}_{(NH_{4}^{+})*V_{NH_{4}^{+}}}}{V_{NH_{3}} + V_{NH_{4}^{+}}} = \frac{0.15 M*0.25 L}{0.2 L + 0.25 L} = 0.083 M [/tex]
From equation (1) we have:
[tex]Kb = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]}[/tex]
[tex] 1.8 \cdot 10^{-5} = \frac{(0.083 + x)*x}{0.053 - x} [/tex]
[tex] 1.8 \cdot 10^{-5}(0.053 - x) - (0.083 + x)*x = 0 [/tex]
By solving the above equation for x we have:
x = 1.15x10⁻⁵ = [OH⁻]
The pH of the solution is:
[tex] pOH = -log([OH^{-}]) = -log(1.15 \cdot 10^{-5}) = 4.94 [/tex]
[tex] pH = 14 - pOH = 14 - 4.94 = 9.06 [/tex]
Therefore, the pH of the solution is 9.06.
I hope it helps you!
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.
Select True or False: Pi bonds are covalent bonds in which the electron density is concentrated above and below a plane containing the nuclei of the bonding atoms and occurs by sideways overlap of p orbitals.
Answer:
True
Explanation:
In pi bonds, the electron density concentrates itself between the atoms of the compound but are present on either side of the line joining the atoms. Electron density is found above and below the plane of the line joining the internuclear axis of the two atoms involved in the bond.
Pi bonds usually occur by sideways overlap of atomic orbitals and this leads to both double and triple bonds.
Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 25°C. (The equation is balanced.) 3 Cl2(g) + 2 Fe(s) → 6 Cl-(aq) + 2 Fe3+(aq) Cl2(g) + 2 e- → 2 Cl-(aq) E° = +1.36 V Fe3+(aq) + 3 e- → Fe(s) E° = -0.04 V
The cell potential for the electrochemical cell has been 1.40 V.
The standard reaction for the cell will be:
[tex]\rm 3\;Cl_2\;+\;2\;Fe\;\rightarrow\;6\;Cl^-\;+\;2\;Fe^3^+[/tex]
The half-reaction of the cells has been:
[tex]\rm Fe^3^+\;+\;3\;e^-\;\rightarrow\;Fe[/tex]
The potential for this reduction has been -0.04 V.
[tex]\rm Cl_2\;+\;2\;e^-\;\rightarrow\;2\;Cl^-[/tex]
The potential for the reduction has been 1.36 V.
The cell potential has been: Potential of reduction - Potential of oxidation
Cell potential = 1.36 - (-0.04) V
Cell potential = 1.40 V.
The cell potential for the electrochemical cell has been 1.40 V.
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The last group of elements on the periodic table are called _____. noble gases halogens metals noble solids
Answer:
The answer is noble gases
Explanation:
Here is your explanation The vertical columns are called groups. There are eighteen groups. The last group on the far right is called the noble, or inert gases. Elements in a group have similar chemical properties. For example, elements in the noble gas group are all gases under. This is the thing from the passege bye god bless you
The last group of elements on the periodic table are called "noble gases."
The noble gases are a group of elements located in Group 18 of the periodic table. They are also known as Group 0 or the "inert gases." The noble gases include helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn).
The noble gases are unique because they have a full complement of valence electrons in their outermost energy level. This full electron configuration gives them exceptional stability, making them chemically unreactive or inert under normal conditions. In other words, noble gases are less likely to form chemical bonds with other elements.
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A mixture of 50ml of 0.1M HCOOH and 50ml of 0.05M NaOH is equivalent to
Answer:
d) a solution that is 0.025M in HCOOH and 0.025M in HCOONa
Explanation:
The reaction of a weak acid (HOOH) with NaOH is as follows:
HCOOH + NaOH → HCOONa + H₂O
Based on the reaction, 1 mole of the acid reacts with 1 mole of the base (Ratio 1:1).
The initial moles of both species are:
HCOOH: 0.050L × (0.1mol / L) = 0.0050 moles of HCOOH
NaOH: 0.050L × (0.05 mol / L) = 0.0025 moles NaOH
After the reaction, all NaOH reacts with HCOOH producing HCOONa (Because moles of NaOH < moles HCOOH).
Final moles:
HCOOH: 0.0050 moles - 0.0025 moles (After reaction) = 0.0025 moles
HCOONa: Moles HCOONa = Initial Moles NaOH: 0.0025 moles
As volume of the mixture is 100mL (50 from the acid + 50 from NaOH), molarity of both HCOOH and HCOONa is:
0.0025 moles / 0.100L = 0.025M of both HCOOH and HCOONa
Thus, the initial mixture is equivalent to:
d) a solution that is 0.025M in HCOOH and 0.025M in HCOONaDraw Lewis structures to show how H+ is transferred when HNO₂ and NH₃ react with each other. The Lewis structure of HNO₂ is:
Attached are the Lewis structures:
Hope it helps...
[tex] \: \: \: [/tex]
Stibnite, Sb2S3, is a dark gray mineral from which antimony metal is obtained. What is the mass percent of antimony in the sulfide? If you have 1.00 kg of an ore that contains 10.6% antimony, what mass of Sb2S3 (in grams) is in the ore?
Answer:
duw8 Wert gsi
Explanation:
pues pues y 8y1rcuecisscfjfj3eoeu xv cihskdkkd HD jekifeuifkeñ elijo eh fh FC eh SSH DJ djdvheshdhs
The mass percent of antimony in the sulfide is 71.7% and the mass of Sb2S3 is 0.148 g.
What is mass percent?Mass percent is defined as a way of expressing a concentration or describing a component in a particular mixture.
To calculate the mass percent of an element in a compound, divide the mass of the element in 1 mole of the compound by the molar mass of the compound and multiply the result by 100.
Mass percent = Mass of chemical / Total mass of compound x 100
Mass of ore = 10 kg
% antimony in ore = 10.6%
= 10.6 / 100 = 0.106 g
Mass of antimony = 0.106 g
Mass percent = Mass of antimony / Mass of stibnite x 100
71.7% = 0.106 / X x 100
X = 0.106 x 1000 / 71.7 x 1000
= 106 / 717 = 0.148 g
Thus, the mass percent of antimony in the sulfide is 71.7% and the mass of Sb2S3 is 0.148 g.
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what is the valency of element sulphur
sorry not sure wish I can help u
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:
The Lewis structure of N2H2 shows ________. Group of answer choices a nitrogen-nitrogen single bond each hydrogen has one nonbonding electron pair each nitrogen has one nonbonding electron pair each nitrogen has two nonbonding electron pairs a nitrogen-nitrogen triple bond
Answer:
one bond between nitrogen and hydrogen and a double bond between the nitrogen atoms.
Explanation:
H-N=N-H
At what temperature in K will 0.750 moles of oxygen gas occupy 10.0 L and exert 2.50 atm of pressure
Answer:
406 K.
Explanation:
The following data were obtained from the question:
Number of mole (n) = 0.750 mole
Volume (V) = 10.0 L
Pressure (P) = 2.50 atm
Temperature (T) =.?
Note: Gas constant (R) = 0.0821 atm.L/Kmol
The temperature, T can be obtained by using the ideal gas equation as follow:
PV = nRT
2.5 x 10 = 0.75 x 0.0821 x T
Divide both side by 0.75 x 0.0821
T = (2.5 x 10) /(0.75 x 0.0821 )
T = 406 K.
Therefore, the temperature is 406 K.
Answer: 406 K
Explanation:
We can rewrite the ideal gas law to solve for T:
PV = nRT
T=PV / nR
We are given the following from the problem:
n=0.750 mol P=2.50 atm V=10.0 L
Plugging in our values and using R=0.08206 L⋅atm / K⋅mol we get:
T=(2.50 atm)(10.0 L) / (0.750 mole)(0.08206L ⋅ atm ⋅ mole K) = 406 K
Calculate the molar solubility of CaF2 at 25°C in a solution that is 0.010 M in Ca(NO3)2. The Ksp for CaF2 is 3.9 x 10-11.
Answer:
[tex]Molar \ solubility=3.12x10^{-5}M[/tex]
Explanation:
Hello,
In this case, for the dissociation of calcium fluoride:
[tex]CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-[/tex]
The equilibrium expression is:
[tex]Ksp=[Ca^{2+}][F^-]^2[/tex]
In such a way, via the ICE procedure, including an initial concentration of calcium of 0.01 M (due to the calcium nitrate solution), the reaction extent [tex]x[/tex] is computed as follows:
[tex]3.9x10^{-11}=(0.01+x)(2*x)^2\\\\x=0.0000312M[/tex]
Thus, the molar solubility equals the reaction extent [tex]x[/tex], therefore:
[tex]Molar \ solubility=3.12x10^{-5}M[/tex]
Regards.
The molar solubility of Calcium fluoride has been calculated as [tex]3.12\;\times\;10^-^5\;\rm M[/tex].
The dissociation of calcium fluoride has been given by:
[tex]\rm CaF_2\;\rightarrow\;Ca^2^+\;+\;2\;F^-[/tex]
The solubility constant, ksp has been given as:
[tex]ksp=\rm[Mg^2^+]\;[F^-]^2[/tex]
From the dissociation of Calcium nitrate, the concentration of Ca ion in the solution has been 0.01 M.
The dissociation of Calcium fluoride x M has been resulted in x M Ca and 2x M F ions.
The concentration of Ca in the solution has been resulted as x + 0.01 M.
The solubility product can be given as:
[tex]3.9\;\times\;10^-^1^1=[x+0.01]\;[2x]^2\\3.9\;\times\;10^-^1^1=[x+0.01]\;4x^2\\x=3.12\;\times\;10^-^5[/tex]
The molar solubility of Calcium fluoride has been calculated as [tex]3.12\;\times\;10^-^5\;\rm M[/tex].
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A student sets up the following equation to convert a measurement. The (?) Stands for a number the student is going to calculate. Fill in the missing part of this equation. (0.030 cm^3) x ? =m^3
Answer:
\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 3.0 \times 10^{-7} \text{ m}^{3}
Explanation:
0.030 cm³ × ? = x m³
You want to convert cubic centimetres to cubic metres, so you multiply the cubic centimetres by a conversion factor.
For example, you know that centi means "× 10⁻²", so
1 cm = 10⁻² m
If we divide each side by 1 cm, we get 1 = (10⁻² m/1 cm).
If we divide each side by 10⁻² m, we get (1 cm/10⁻² m) = 1.
So, we can use either (10⁻² m/1 cm) or (1 cm/10⁻² m) as a conversion factor, because each fraction equals one.
We choose the former because it has the desired units on top.
The "cm" is cubed, so we must cube the conversion factor.
The calculation becomes
[tex]\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 0.30 \times 10^{-6}\text{ m}^{3} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}\\\\\textbf{0.30 cm}^{\mathbf{3}} \times \left (\dfrac{\mathbf{10^{-2}}\textbf{ m}}{\textbf{1 cm}}\right )^{\mathbf{3}} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}[/tex]
How does the spontaneity of the process below depend on temperature? PCl5(g)+H2O(g)→POCl3(g)+2HCl(g) ΔH=−126 kJ mol−1, ΔS=146 J K−1mol−
The given question is incomplete, the complete question is:
How does the spontaneity of the process below depend on temperature? PCI5(9)+H2O(g)POCI3(g) +2HCI(g) -126 kJ mol1, AS = 146 J K-'mol1 ΔΗ Select the correct answer below: nonspontaneous at all temperatures spontaneous at all temperatures spontaneous at high temperatures and nonspontaneous at low temperatures spontaneous at low temperatures and nonspontaneous at high temperatures
Answer:
The correct answer is spontaneous at all the temperatures.
Explanation:
Gibbs Free energy is an essential relation that determines the spontaneity of any reaction, that is, ΔG = ΔH - TΔS
When ΔG is less than zero, that is, negative, the reaction is considered to be in spontaneous state. Based on the given information, ΔH = -126 kJ/mol
= -126000 J/mol, it is negative
ΔS = 146 J/K/mol, it is positive
Now, ΔG = ΔH-TΔS
= (-ve) - T (+ve), Thus, when ΔH, is -ve, ΔS is +ve, -TΔS is -ve, the ΔG will be -ve. Therefore, reaction will be spontaneous at all the temperatures.