Answer:
OPTION C is correct
3.86 x 10-19 J
Explanation:
Energy of the line can be calculated using below formula
E= h ν.................(1)
Where E= energy
h= plank constant= 6.626 10-34 J s
c=speed of light=3 x 108 m/s
But we know that Velocity V= = c / λ
Then substitute into equation (1) we have
E = h c / λ.............(2)
We can calculate our( hc ) in nm for unit consistency
h c =( 6.626 ×10^-34)x(3×108)
h c = (1.986 x 10-16 )
hc = 1.986 x 10-16 J nm then since our (hc) and λ are in the same unit , were good to go then substitute into equation(2)
E = h c / λ = (1.986 x 10-16) / 515
E = 3.86 x 10-19 J
Therefore, the Energy is 3.86 x 10-19 J
The intermolecular forces present in CH 3NH 2 include which of the following? I. dipole-dipole II. ion-dipole III. dispersion IV. hydrogen bonding
Answer:
I. dipole-dipole
III. dispersion
IV. hydrogen bonding
Explanation:
Intermolecular forces are weak attraction force joining nonpolar and polar molecules together.
London Dispersion Forces are weak attraction force joining non-polar and polar molecules together. e.g O₂, H₂,N₂,Cl₂ and noble gases. The attractions here can be attributed to the fact that a non -polar molecule sometimes becomes polar because the constant motion of its electrons may lead to an uneven charge distribution at an instant.
Dispersion forces are the weakest of all electrical forces that act between atoms and molecules. The force is responsible for liquefaction or solidification of non-polar substances such as noble gas an halogen at low temperatures.
Dipole-Dipole Attractions are forces of attraction existing between polar molecules ( unsymmetrical molecules) i.e molecules that have permanent dipoles such as HCl, CH3NH2 . Such molecules line up such that the positive pole of one molecule attracts the negative pole of another.
Dipole - Dipole attractions are more stronger than the London dispersion forces but weaker than the attraction between full charges carried by ions in ionic crystal lattice.
Hydrogen Bonding is a dipole-dipole intermolecular attraction which occurs when hydrogen is covalently bonded to highly electronegative elements such as nitrogen, oxygen or fluorine. The highly electronegative elements have very strong affinity for electrons. Hence, they attracts the shared pair of electrons in the covalent bonds towards themselves, leaving a partial positive charge on the hydrogen atom and a partial negative charge on the electronegative atom ( nitrogen in the case of CH3NH2 ) . This attractive force is know as hydrogen bonding.
Answer:
The intermolecular forces present in CH_3NH_2 includes
II. (ion-dipole) and IV. (hydrogen bonding)Explanation:
The intermolecular forces present in CH_3NH_2 includes II. (ion-dipole) and IV. (hydrogen bonding)
It is a polar molecule due to NH polar bond and it can form Hydrogen bond also due to NH bond.
Interaction will be dipole- dipole and Hydrogen dispersion forces can always be taken into account.
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Describe the similarities between H3O and NH3. Compare/contrast their shapes and polarities within the context of your answer. These molecules are called isoelectronic. Why
Answer:
Explanation:
[tex]H_3O^+[/tex] also known as hydronium ion is formed as a result of the reaction between an hydrogen proton and a water molecules.
i.e [tex]\mathtt{H^+ + H_2O \to H_3O^+}[/tex]
(molecular geometry for the hydronium ion shows that the lewis structure of hydronium ion possess a three hydrogen ion bonded to a central atom known as oxygen. The oxygen possess a lone pair with a positive ion. So we have three hydrogen atoms and a lone pair attached to the oxygen. We can now say that there are four groups as the steric number in which one of them is a lone pair. This give rise to the trigonal pyramidal shape of the [tex]H_3O^+[/tex] (hydronium ion) with a bond angle of about 109,5°
Similarly, [tex]NH_3[/tex] on the other hand also known as ammonia has a shape that can be also determined by the Lewis structure.
IN ammonia, there are three hydrogen and a lone pairs of electron spreading out as far away from each other from the centre nitrogen. In essence, the valence shell electron pair around hydrogens tend to repel each other. Hence, giving it a trigonal pyramidal shape.
From above the similarities between H3O and NH3 is in their molecular geometry in which both H3O and NH3 have the same shape.
These molecules are called isoelectronic. Why?
Isoelectronic molecules are molecules having the same number of electrons and same electronic configuration structure. As a result H3O and NH3 possess the same number of electrons in the same orbitals and they also posses the same structure.
Atoms are indivisible spheres. 1.plum pudding model 2.Dalton model 3.Bohr model
Answer: 2. Dalton Model
Explanation:
John Dalton proposed that atoms are indivisible spheres. Although his model of an atom was not entirely new to the scientific world since the ancient Greeks has made a similar statement in the past ( all matter are made up of small indivisible particle called atom).
As of when Dalton proposed his model of an atom, electrons and nucleus where yet to be discovered.
If one pound is the same as 454 grams, then convert the mass of 78 grams to pounds.
Answer:
0.17 lb
Explanation:
78 g * (1 lb/454 g)=0.17 lb
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Please Help! Use Hess’s Law to determine the ΔHrxn for: Ca (s) + ½ O2 (g) → CaO (s) Given: Ca (s) + 2 H+ (aq) → Ca2+ (aq) + H2 (g) ΔH = 1925.9 kJ/mol 2 H2 (g) + O2 (g) → 2 H2O (l) ΔH = −571.68 kJ/mole CaO (s) + 2 H+ (aq) → Ca2+ (aq) + H2O (l) ΔH = 2275.2 kJ/mole ΔHrxn =
Answer:
ΔHrxn = -635.14kJ/mol
Explanation:
We can make algebraic operations of reactions until obtain the desire reaction and, ΔH of the reaction must be operated in the same way to obtain the ΔH of the desire reaction (Hess's law). Using the reactions:
(1)Ca(s) + 2 H+(aq) → Ca2+(aq) + H2(g) ΔH = 1925.9 kJ/mol
(2) 2H2(g) + O2 g) → 2 H2O(l) ΔH = −571.68 kJ/mole
(3) CaO(s) + 2 H+(aq) → Ca2+(aq) + H2O(l) ΔH = 2275.2 kJ/mole
Reaction (1) - (3) produce:
Ca(s) + H2O(l) → H2(g) + CaO(s)
ΔH = 1925.9kJ/mol - 2275.2kJ/mol = -349.3kJ/mol
Now this reaction + 1/2(2):
Ca(s) + ½ O2(g) → CaO(s)
ΔH = -349.3kJ/mol + 1/2 (-571.68kJ/mol)
ΔHrxn = -635.14kJ/molA student ran the following reaction in the laboratory at 242 K: 2NOBr(g) 2NO(g) Br2(g) When she introduced 0.143 moles of NOBr(g) into a 1.00 liter container, she found the equilibrium concentration of NOBr(g) to be 0.108 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc
Answer:
1.84 × 10⁻³
Explanation:
Step 1: Write the balanced equation
2 NOBr(g) ⇄ 2 NO(g) + Br₂(g)
Step 2: Calculate the initial concentration of NOBr
0.143 moles of NOBr(g) are introduced into a 1.00 liter container. The molarity is:
M = 0.143 mol / 1.00 L = 0.143 M
Step 3: Make an ICE chart
2 NOBr(g) ⇄ 2 NO(g) + Br₂(g)
I 0.143 0 0
C -2x +2x +x
E 0.143-2x 2x x
Step 4: Find the value of x
The equilibrium concentration of NOBr(g) was 0.108 M. Then,
0.143-2x = 0.108
x = 0.0175
Step 5: Calculate the concentrations at equilibrium
[NOBr] = 0.108 M
[NO] = 2x = 0.0350 M
[Br₂] = x = 0.0175 M
Step 6: Calculate the equilibrium constant (Kc)
Kc = [0.0350]² × [0.0175] / [0.108]²
Kc = 1.84 × 10⁻³
How many atoms of oxygen are in one molecule of water (H2O)? one two four three
Answer:
there is one atom of oxygen and two atoms of hydrogen
Explanation:
Determine which set of properties correctly describes copper (Cu)?
A. Giant structure, conducts electricity, high melting point, soluble in water, malleable
B. Malleable, brittle, soluble in oil or gasoline, high melting point, simple structure
C. Ionic lattice, conducts electricity, soluble in oil or gasoline, low melting point, ductile
D. Malleable, conducts electricity, high melting point, giant structure, metallic lattice
Answer:
D. Malleable, conducts electricity, high melting point, giant structure, metallic lattice
Explanation:
Copper is a metal with an atomic number of 29. This metal is soft and reddish in color which explains why it is very malleable(beaten to form various shapes without breaking).
All metals are good conductors of electricity including copper which is also a metal. Metals generally are insoluble in water. Copper also has a high melting point which is a characteristic of metals due to their giant structure and metallic lattice which makes it difficult to be broken down.
1.) A sample of neon gas at a pressure of 0.646 atm and a temperature of 242 °C, occupies a volume of 515 mL. If the gas is cooled at constant pressure until its volume is 407 mL, the temperature of the gas sample will be ________°C.
2.) A sample of argon gas at a pressure of 0.633 atm and a temperature of 261 °C, occupies a volume of 694 mL. If the gas is heated at constant pressure until its volume is 796 mL, the temperature of the gas sample will be___________°C.
3.) 0.962 mol sample of carbon dioxide gas at a temperature of 20.0 °C is found to occupy a volume of 21.5 liters. The pressure of this gas sample ismm ____________ Hg.
Answer:1 )T2=134°C 2) T2=339.48°C. 3)
P=817.59 mmHg.
Explanation:
1.Given ;
pressure, P1 of neon gas = 0.646 atm
temperature, T1 =242oC + 273=515oC
Volume, V1 =515ml
Volume V2= 407ml
temperature , T 2= ?
Solution;
And at constant pressure, the volume cools at V2=407 mL at T2=?
From ideal gas equation, PV=nRT
V/T=constant
therefore
V1/V2=T1/T2 = T2=(V2 xT1)/V1
T2=(407 mL x 515 K)/515 mL= 407K.
T2= 407K -273= 134°C. recall 0°C=273 K)
2..Given ;
pressure, P1 of neon gas = 0.633 atm
temperature, T1 =261oC + 273=534oC
Volume, V1 =694ml
Volume V2= 796ml
temperature , T 2= ?
Solution;
And at constant pressure, the volume expands at V2=796mL at T2=?
From ideal gas equation, PV=nRT
V/T=constant
therefore
V1/V2=T1/T2 = T2=(V2 xT1)/V1
T2=(796 mL x 534 K)/694mL= 612.48K.
T2= 612.48K -273= 339.48°C. recall 0°C=273 K
3
Given;
moles of CO2= n=0.962 mol,
temperature T=20°C=20+273 K =293 K,
volume V=21.5 L,
gas constant R at L·mmHg/mol·K= 62.3637 L mmHg mol^-1 K^-1
Using ideal gas equation PV=nRT
P=nRT/V
P=(0.962 mol)x(62.3637mmHg mol^-1 K^-1)x(293 K)/(21.5L)
P=817.59 mmHg.
By heating a 93% pure kclo3 sample, what percentage of its mass is reduced?
2KCLO3---->2KCL+3O2
Explanation:
free your mind drink water and go outside take fresh air you will get answers
Which of these species would you expect to have the lowest standard entropy (S°)?
a. CH4(g)
b. H2O(g)
c. NH3(g)
d. HF(g)
Answer:
d. HF(g)
Explanation:
Hello,
In this case, the standard entropy S° could be predicted by looking at the amount of bonds the compound has, thus, the fewer the number bonds, the lower the standard entropy, it means that d. HF(g) has lowest value as it has one bond only whereas methane has four bonds, water two bonds and ammonia three bonds.
Best regards.
A small amount of solid calcium hydroxide is shaken vigorously in a test tube almost full of water until no further change occurs and most of the solid settles out. The resulting solution is:______.
Answer:
Lime water, [tex]Ca(OH)_{2}_({aq} )[/tex] is formed.
Explanation:
Lime-water is a clear and colourless dilute solution of aqueous calcium hydroxide salt.
Small amounts of calcium hydroxide salt, [tex]Ca(OH)_{2}_(s)[/tex] is sparsely soluble at room temperature when dispersed vigorously. if in excess, a white suspension called 'milk of lime'is formed.
I hope this explanation is helpful.
Write the equation for the reaction described: A solid metal oxide, , and hydrogen are the products of the reaction between metal and steam. (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.)
Answer:
Pb + 2H2O --> PbO2 + 2H2
Explanation:
Products:
Solid metal; PbO2
Hydrogen; H
Reactants:
Metal; Pb
Steam; H2O
Reactants --> Products
Pb + H2O --> PbO2 + H2
Upon balancing we have;
Pb + 2H2O --> PbO2 + 2H2
An actacide tablet containing Mg(OH)2 (MM = 58.3g / (mol)) is titrated with a 0.100 M solution of HNO3. The end point is determined by using an indicator. Based on 20.00mL HNO3 being used to reach the endpoint, what was the mass of the Mg * (OH) in the antacid tablet? * 0.0583 g 0.583 5.83 g 58.3 g
Answer:
0.0583g
Explanation:
The equation of the reaction is;
2HNO3(aq) + Mg(OH)2(aq) -------> Mg(NO3)2(aq) + 2H2O(l)
From the question, number of moles of HNO3 reacted= concentration × volume
Concentration of HNO3= 0.100 M
Volume of HNO3 = 20.00mL
Number of moles of HNO3= 0.100 × 20/1000
Number of moles of HNO3 = 2×10^-3 moles
From the reaction equation;
2 moles of HNO3 reacts with 1 mole of Mg(OH)2
2×10^-3 moles reacts with 2×10^-3 moles ×1/2 = 1 ×10^-3 moles of Mg(OH)2
But
n= m/M
Where;
n= number of moles of Mg(OH)2
m= mass of Mg(OH)2
M= molar mass of Mg(OH)2
m= n×M
m= 1×10^-3 moles × 58.3 gmol-1
m = 0.0583g
A balloon has an initial volume of 2.954 L containing 5.50 moles of helium. More helium is added so that the balloon expands to 4.325 L. How much helium (moles) has been added if the temperature and pressure stay constant during this process.
Answer:
8.05 moles
Explanation:
5.50 / 2.954 = x / 4.325
x = 8.05
According to ideal gas equation, if the temperature and pressure stay constant during the process 0.520 moles have been added so that the balloon expands to 4.325 L.
What is ideal gas equation?The ideal gas equation is a equation which is applicable in a hypothetical state of an ideal gas.It is a combination of Boyle's law, Charle's law,Avogadro's law and Gay-Lussac's law . It is given as, PV=nRT where R= gas constant whose value is 8.314.The law has several limitations.The law was proposed by Benoit Paul Emile Clapeyron in 1834.
In the given example if pressure and temperature are constant then V=nR substituting V=4.325 l and R=8.314 so n=V/R=4.325/8.314=0.520 moles.
Thus, 0.520 moles of helium are added if the temperature and pressure stay constant during this process.
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How has the work of chemists affected the environment over the years?
Answer:
Chemistry is one of the causes for global warming, and in some cases it can even cause certain illnesses.
Answer:
Chemists have both hurt the environment and helped the environment by their actions.
Explanation:
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Testbank Question 47 Consider the molecular orbital model of benzene. In the ground state how many molecular orbitals are filled with electrons?
Answer:
There are fifteen molecular orbitals in benzene filled with electrons.
Explanation:
Benzene is an aromatic compound. Let us consider the number of bonding molecular orbitals that should be present in the molecule;
There are 6 C-C σ bonds, these will occupy six bonding molecular orbitals filled with electrons.
There are 6 C-H σ bonds, these will occupy another six molecular orbitals filled with electrons
The are 3 C=C π bonds., these will occupy three bonding molecular pi orbitals.
All these bring the total number of bonding molecular orbitals filled with electrons to fifteen bonding molecular orbitals.
Using the data: C2H4(g), = +51.9 kJ mol-1, S° = 219.8 J mol-1 K-1 CO2(g), = ‑394 kJ mol-1, S° = 213.6 J mol-1 K-1 H2O(l), = ‑286.0 kJ mol-1, S° = 69.96 J mol-1 K-1 O2(g), = 0.00 kJ mol-1, S° = 205 J mol-1 K-1 calculate the maximum amount of work that can be obtained, at 25.0 °C, from the process: C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(l)
Answer:
The correct answer is 1332 KJ.
Explanation:
Based on the given information,
ΔH°f of C2H4 is 51.9 KJ/mol, ΔH°O2 is 0.0 KJ/mol, ΔH°f of CO2 is -394 KJ/mol, and ΔH°f of H2O is -286 KJ/mol.
Now the balanced equation is:
C2H4 (g) + 3O2 (g) ⇔ 2CO2 (g) + 2H2O (l)
ΔH°rxn = 2 × ΔH°f CO2 + 2 × ΔH°fH2O - 1 × ΔH°fC2H4 - 3×ΔH°fO2
ΔH°rxn = 2 (-394) + 2(-286) - 1(51.9) - 3(0)
ΔH°rxn = -1411.9 KJ
Now, the given ΔS°f of C2H4 is 219.8 J/mol.K, ΔS°f of O2 is 205 J/mol.K, ΔS°f of CO2 is 213.6 J/mol.K, and ΔS°f of H2O is 69.96 J/mol.K.
Now based on the balanced chemical reaction,
ΔS°rxn = 2 × ΔS°fCO2 + 2 ΔS°fH2O - 1 × ΔS°f C2H4 - 3 ΔS°fO2
ΔS°rxn = 2 (213.6) + 2(69.96) - 1(219.8) -3(205)
ΔS°rxn = -267.68 J/K or -0.26768 KJ/K
T = 25 °C or 298 K
Now putting the values of ΔH, ΔS and T in the equation ΔG = ΔH-TΔS, we get
ΔG = -1411.9 - 298.0 × (-0.2677)
ΔG = -1332 KJ.
Thus, the maximum work, which can obtained is 1332 kJ.
Why can gasses change volume?
A. The forces holding the gas particles together are
stronger than gravity.
B. The gas particles have no mass, so they can change volume.
C. Gravity has no effect on gas particles, so they can float away.
O D. There are no forces holding the gas particles together.
Answer:
There are no forces holding the gas particles together.
Explanation:
Identify four general properties that make an NSAID unique as compared to the NSAID aspirin. List specific properties that make aspirin, naproxen, and ibuprofen unique from one another
Answer:
NSAIDs are steroidal anti-inflammatories, their action is on the phospholipase A2 enzyme, this enzyme is responsible for breaking down the phospholipids of the membrane to trigger an inflammatory response. This is how steroidal anti-inflammatory drugs inhibit ALL inflammatory pathways (not like NSAIDs that they only inhibit the COX pathway).
These corticosteroid drugs cannot exceed the systemic mineralocorticoid value 1 in the body, since this corticosteroid hormone is also released by the adrenal cortex.
The NSAIDs generate: sporadic peaks in blood glucose, hypertension, fluid retention, increase in body fat mass, possible suppression of the adrenal cortex over time, inhibiting endogenous synthesis of corticosteroids.
On the other hand, naproxen and ibuprofen are NSAIDs, that is, non-steroidal anti-inflammatory drugs that do not influence both routes of inflammation, but only COX, this enzyme is abbreviated as COX but is called cyclooxygenase, and is responsible for a single route of inflammation.
NSAIDs such as naproxen and ibuprofen can cause gastric disorders such as ulcers or gastritis if they are consumed in a very repetitive manner.
In addition, both drugs are anti-inflammatory, analgesic and antipyretic. Although its two main functions are the first two, it was shown to have an effect in lowering body temperature.
That they are anti-inflammatory means that they inhibit the path of inflammation and analgesics the path of pain.
Explanation:
Both types of drugs generate the same effect but by different mechanisms.
Some are steroids and others are not, although steroids are considered to have a greater risk of benefit that is why they are administered against more systematically compromised instances such as anaphylactic shock.
NSAIDs such as naproxen and ibuprofen are the most prescribed today, since they have few risks and very good benefits, meaning that their adverse effects are not lethal or highly relevant and have a good effect on symptoms.
Both must be administered with care when treating a diabetic patient since corticosteroids generate glycemic peaks or increase in blood glucose, and NSAIDs compete for plasma protein with oral hypoglycemic agents, thus generating that these are in higher free concentrations. high diffusing better through the tissues and increases the potency of the adverse effects of these.
The heat of vaporization of water is 40.66 kJ/mol. How much heat is absorbed when 3.11 g of water boils at atmospheric pressure?
Answer:
The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.
Explanation:
A molar heat of vaporization of 40.66 kJ / mol means that 40.66 kJ of heat needs to be supplied to boil 1 mol of water at its normal boiling point.
To know the amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure, the number of moles represented by 3.11 g of water is necessary. Being:
H: 1 g/moleO: 16 g/molethe molar mass of water is:
H₂O= 2* 1 g/mole + 16 g/mole= 18 g/mole
So: if 18 grams of water are contained in 1 mole, 3.11 grams of water in how many moles are present?
[tex]moles of water=\frac{3.11 grams*1 mole}{18 gramos}[/tex]
moles of water= 0.1728
Finally, the following rule of three can be applied: if to boil 1 mole of water at its boiling point it is necessary to supply 40.66 kJ of heat, to boil 0.1728 moles of water, how much heat is necessary to supply?
[tex]heat=\frac{0.1728 moles*40.66 kJ}{1 mole}[/tex]
heat= 7.026 kJ
The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.
An aqueous solution of potassium bromide, KBr, contains 4.34 grams of potassium bromide and 17.4 grams of water. The percentage by mass of potassium bromide in the solution is 20 %.
Answer:
True
Explanation:
The percentage by mass of a substance in a solution can be calculated by dividing the mass of the substance dissolved in the solution by the total mass of the solution. This can be expressed mathematically as:
Percentage by mass = mass of substance in solution/mass of solution x 100
In this case;
mass of KBr = 4.34 grams
mass of water = 17.4 grams
mass of solution = mass of KBr + mass of water = 4.34 + 17.4 = 21.74
Percentage by mass of KBr = 4.34/21.74 x 100
= 19.96 %
19.96 is approximately 20%.
Hence, the statement is true.
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!
A student determines the value of the equilibrium constant to be 1.5297 x 107 for the following reaction: HBr(g) + 1/2 Cl2(g) --> HCl(g) +1/2 Br2(g) Based on this value of Keq, calculate the Gibbs free energy change for the reaction of 2.37 moles of HBr(g) at standard conditions at 298 K.
Answer:
[tex]\Delta G=-97.14kJ[/tex]
Explanation:
Hello,
In this case, the relationship between the equilibrium constant and the Gibbs free energy of reaction is:
[tex]\Delta G=-RTln(K)[/tex]
Hence, we compute it as required:
[tex]\Delta G=-8.314\frac{J}{mol\times K}*298K*ln(1.5297x10^7)\\\\\Delta G=-40.99kJ/mol[/tex]
And for 2.37 moles of hydrogen bromide, we obtain:
[tex]\Delta G=-40.99kJ/mol*2.37mol\\\\\Delta G=-97.14kJ[/tex]
Best regards.
Fill in the blanks with the words given below- [Atoms, homogeneous, metals, true, saturated, homogeneous, colloidal, compounds, lustrous] 1.An element which are sonorous are called................ 2.An element is made up of only one kind of .................... 3.Alloys are ............................. mixtures. 4.Elements chemically combines in fixed proportion to form ........................ 5. Metals are................................... and can be polished. 6. a solution in which no more solute can be dissolved is called a .................... solution. 7. Milk is a .............. solution but vinegar is a .................. solution. 8. A solution is a ................... mixture. pls help, could not get these answers
Answer:
1. metals
2. atom
3. homogeneous
4. compounds
5. lustrous
6. saturated
7. colloidal
8. homogeneous
Explanation:
A. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
Br(g)
Cl2(g)
I2(g)
F2(g)
B. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
H2S(g)
H2O(g)
H2O2(g)
C. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
C(s, amorphous)
C(s, diamond)
C(s, graphite)
Answer:
A. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy
I2(g)>Br2(g)>Cl2(g)>F2(g)
B. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
H2O2(g)>H2S(g) >H2O(g)
C. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
C(s, amorphous) >C(s, graphite)>C(s, diamond)
Explanation:
Hello,
In this case, we can apply the following principles to explain the order:
- The greater the molar mass, the larger the standard molar entropy.
- The greater the molar mass and the structural complexity, the larger the standard molar entropy.
- The greater the structural complexity, the larger the standard molar entropy.
A. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy
I2(g)>Br2(g)>Cl2(g)>F2(g)
This is due to the fact that the greater the molar mass, the larger the standard molar entropy.
B. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
H2O2(g)>H2S(g) >H2O(g)
This is due to the fact that the greater the molar mass and the structural complexity, the larger the standard molar entropy as the hydrogen peroxide has four bonds and weights 34 g/mol as well as hydrogen sulfide that has two bonds only.
C. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
C(s, amorphous) >C(s, graphite)>C(s, diamond)
Since the molecular complexity is greater in the amorphous carbon (messy arrangement), mid in the graphite and lower in the diamond (well organized).
Regards.
What is the mass number of an element
Answer:
A (Atomic mass number or Nucleon number)
Explanation:
The mass number is the total number of protons and nucleons in an atomic nucleus.
Hope this helps.
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When balancing redox reactions under basic conditions in aqueous solution, the first step is to:________.
a. balance oxygen
b. balance hydrogen
c. balance the reaction as though under acidic conditions
d. none of the above
Answer:
When balancing redox reactions under basic conditions in aqueous solution, the first step is to balance oxygen.
Explanation:
Oxidation-reduction reactions or redox reactions are those in which an electron transfer occurs between the reagents. An electron transfer implies that there is a change in the number of oxidation between the reagents and the products.
The gain of electrons is called reduction and the loss of electrons oxidation. That is to say, there is oxidation whenever an atom or group of atoms loses electrons (or increases its positive charges) and in the reduction an atom or group of atoms gains electrons, increasing its negative charges or decreasing the positive ones.
The oxidation and reduction half-reactions, in a basic medium, adjust the oxygens and hydrogens as follows:
In the member of the half-reaction that presents excess oxygen, you add as many water molecules as there are too many oxygen. Then, in the opposite member, the necessary hydroxyl ions are added to fully adjust the half-reaction. Normally, twice as many hydroxyl ions, OH-, are required as water molecules have previously been added.
In short, you first adjust the oxygens with OH-, then you adjust the H with H₂O, and finally you adjust the charge with e-
So, when balancing redox reactions under basic conditions in aqueous solution, the first step is to balance oxygen.
Answer:
c. balance the reaction as though under acidic conditions
Explanation:
When balancing redox reactions under basic conditions, a good technique is to first balance the reaction as though under acidic conditions. We then adjust the result to reflect the basic conditions.
A sample is found to contain 1.29×10-11 g of salt. Express this quantity in picograms
Answer:12.9e-12g or in short 12.9pg
Explanation:as p=1e-12
g What is the molarity of hydrochloric acid if 40.95 mL of HCl is required to neutralize 0.550 g of sodium oxalate, Na2C2O4
Answer:
0.0002 M
Explanation:
The molarity of the HCl required would be 0.0002 M.
First, let us consider the balanced equation of the reaction:
[tex]Na_2C_2O_4 + 2HCl = 2NaCl + H_2 + 2CO_2[/tex]
Stoichiometrically, 1 mole of [tex]Na_2C_2O_4[/tex] reacts with 2 moles of [tex]HCl[/tex] for a complete neutralization reaction.
Recall that: mole = [tex]\frac{mass}{molar mass}[/tex]
Mole of 0.550 g sodium oxalate = 0.550/134 = 0.0041 mole
If 1 mole [tex]Na_2C_2O_4[/tex] requires 2 moles HCl, then 0.0041 mole will require:
0.0041 x 2 = 0.0082 mole HCl
Volume of the HCl = 40.95 L
Molarity = mole/volume
Hence, molarity of the HCl = 0.0082/40.95 = 0.0002 M
