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 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.
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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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:
Which of the following processes have a ΔS < 0? Which of the following processes have a ΔS < 0? carbon dioxide(g) → carbon dioxide(s) water freezes propanol (g, at 555 K) → propanol (g, at 400 K) methyl alcohol condenses All of the above processes have a ΔS < 0.
Answer:
All of the above processes have a ΔS < 0.
Explanation:
ΔS represents change in entropy of a system. Entropy refers to the degree of disorderliness of a system.
The question requests us to identify the process that has a negative change of entropy.
carbon dioxide(g) → carbon dioxide(s)
There is a change in state from gas to solid. Solid particles are more ordered than gas particles so this is a negative change in entropy.
water freezes
There is a change in state from liquid to solid. Solid particles are more ordered than liquid particles so this is a negative change in entropy.
propanol (g, at 555 K) → propanol (g, at 400 K)
Temperature is directly proportional to entropy, this means higher temperature leads t higher entropy.
This reaction highlights a drop in temperature which means a negative change in entropy.
methyl alcohol condenses
Condensation is the change in state from gas to liquid. Liquid particles are more ordered than gas particles so this is a negative change in entropy.
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.
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.
Enter the balanced chemical equation for the reaction of each of the following carboxylic acids with KOH.Part Aacetic acidExpress your answer as a chemical equation. Assume that there is no dissociation (i.e., enter only whole compounds, not ions).Part B2-methylbutanoic acid (CH3CH2CH(CH3)COOH)Express your answer as a chemical equation. Assume that there is no dissociation (i.e., enter only whole compounds, not ions).Part C4-chlorobenzoic acid (ClC6H4COOH)Express your answer as a chemical equation. Assume that there is no dissociation (i.e., enter only whole compounds, not ions).
Answer:
Explanation:
Answer in attached file .
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
In a reversible reaction, the endothermic reaction absorbs ____________ the exothermic reaction releases. A. less energy than B. None of these, endothermic reactions release energy C. the same amount of energy as D. more energy than
Answer: C. the same amount of energy as
Explanation:
A reversible reaction is a chemical reaction where the reactants form products that, in turn, react together to give the reactants back.
Reversible reactions will reach an equilibrium point where the concentrations of the reactants and products will no longer change.
[tex]A+B\rightleftharpoons C+D[/tex]
Thus if forward reaction is exothermic i.e. the heat is released , the backward reaction will be endothermic i.e. the heat is absorbed and in same amount.
The amount of energy released will be equal and opposite in sign to the energy absorbed in that reaction.
Answer:
C.) the same amount of energy as
Explanation:
I got it correct on founders edtell
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.
When methane is burned with oxygen, the products are carbon dioxide and water. If you produce 9 grams of water and 11 grams of carbon dioxide from 16 grams of oxygen, how many
The given question is incomplete.
The complete question is:
When methane is burned with oxygen, the products are carbon dioxide and water. If you produce 9 grams of water and 11 grams of carbon dioxide from 16 grams of oxygen, how many grams of methane were needed for the reaction?
Answer: 4 grams of methane were needed for the reaction
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
{tex]CH_4+2O_2\rightarrow CO_2+H_2O[/tex]
Given: mass of oxygen = 16 g
Mass of carbon dioxide = 11 g
Mass of water = 9 g
Mass of products = Mass of carbon dioxide + mass of water = 11 g +9 g = 20 g
Mass or reactant = mass of methane + mass of oxygen = mass of methane + 16 g
As mass of reactants = mass of products
mass of methane + 16 g= 20 g
mass of methane = 4 g
Thus 4 grams of methane were needed for the reaction
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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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/molThe tosylate of (2R,3S)-3-phenylbutan-2-ol undergoes an E2 elimination on treatment with sodium ethoxide. Draw the structure of the alkene that is produced.
Answer:
(R)-but-3-en-2-ylbenzene
Explanation:
In this reaction, we have a very strong base (sodium ethoxide). This base, will remove a hydrogen producing a double bond. We know that the reaction occurs through an E2 mechanism, therefore, the hydrogen that is removed must have an angle of 180º with respect to the leaving group (the "OH"). This is known as the anti-periplanar configuration.
The hydrogen that has this configuration is the one that placed with the dashed bond (red hydrogen). In such a way, that the base will remove this hydrogen, the "OH" will leave the molecule and a double bond will be formed between the methyl and the carbon that was previously attached to the "OH", producing the molecule (R) -but-3- en-2-ylbenzene.
See figure 1
I hope it helps!
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
what are the monomers of bakelite
Answer:
Bakelite is a polymer made up of the monomers phenol and formaldehyde. This phenol-formaldehyde resin is a thermosetting polymer.
Answer: The monomers of bakelite are formaldehyde and phenol
Explanation:
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.
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
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.
A 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
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.
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.
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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A sample of ice absorbs 15.6kJ of heat as it undergoes a reversible phase transition to form liquid water at 0∘C. What is the entropy change for this process in units of JK? Report your answer to three significant figures. Use −273.15∘C for absolute zero.
Answer:
Entropy change of ice changing to water at 0°C is equal to 57.1 J/K
Explanation:
When a substance undergoes a phase change, it occurs at constant temperature.
The entropy change Δs, is given by the formula below;
Δs = q/T
where q is the quantity of heat absorbed or evolved in Joules and T is temperature in Kelvin at which the phase change occur
From the given data, T = 0°C = 273.15 K, q = 15.6 KJ = 15600 J
Δs = 15600 J / 273.15 K
Δs = 57.111 J/K
Therefore, entropy change of ice changing to water at 0°C is equal to 57.1 J/K
The entropy change of ice changing to water will be "57.1 J/K".
Entropy changeThe shift in what seems like a thermodynamic system's condition of confusion is caused by the transformation of heat as well as enthalpy towards activity. Entropy seems to be greater mostly in a network with a high quantity or measure of chaos.
According to the question,
Temperature, T = 0°C or,
= 273.15 K
Heat, q = 15.6 KJ or,
= 15600 J
We know the formula,
Entropy change, Δs = [tex]\frac{q}{T}[/tex]
By substituting the values, we get
= [tex]\frac{15600}{273.15}[/tex]
= 57.11 J/K
Thus the above answer is correct.
Find out more information about Entropy change here:
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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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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.
What is the energy of a photon of electromagnetic radiation with a wavelength of 963.5 nm? (c = 3.00 × 108 m/s, h = 6.63 × 10–34 J · s
Answer:
[tex]E=2.06\times 10^{-19}\ J[/tex]
Explanation:
Given that,
The wavelength of electromagnetic radiation is 963.5 nm.
We need to find the energy of a photon with this wavelength.
The formula used to find the energy of a photon is given by :
[tex]E=\dfrac{hc}{\lambda}\\\\E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{963.5\times 10^{-9}}\\\\E=2.06\times 10^{-19}\ J[/tex]
So, the energy of a photon is [tex]2.06\times 10^{-19}\ J[/tex].
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:
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
A 50.0 L cylinder of oxygen gas is stored at 150. atm. What volume would the oxygen gas occupy if the cylinder were opened into a hot air balloon (completely deflated) until the final pressure is 735 torr
Answer:
THE VOLUME OF THE OXYGEN GAS AFTER DEFLATION TILL A PRESSURE OF 735 TORR IS ATTAINED IS 7836.99 L
Explanation:
Using Boyle's law,
P1V1 = P2V2
P1 = 150 atm
V1 = 50 L
P2 = 735 Torr
V2 = unknown
We must first convert the pressures into the same SI unit for easy calculation
1torr = 1/760 atm
So converting 735 torr to atm; we have:
1 torr = 1/ 760 atm
735 torr = 735 * 1 / 760 atm
= 0.967 atm
In other words, P2 = 0.957 atm
So rearranging the formula by making V2 the subject of the equation, we have:
V2 = P1 V1 / P2
V2 = 150 * 50 / 0.957
V2 = 7836.99 L
The volume of the oxygen cylinder after deflation to a final pressure of 735 torr or 0.967 atm pressure is 7836.99 L.
