Answer:
Kp=1.68×10⁴∆1.7×10⁴
Kc=1.06∆1.1
Explanation:
Value of Kp at 227°C is 2.86×10² and value of Kc at 1000 K is 1.56.
How are Kp and Kc related?Kp and Kc are related by the formula Kp=Kc(RT).For part A , Kp is calculated as,
Kp=6.9×10⁵×8.314×500=28.683×10² and for part B Kc is calculated as,
Kc=1.3×10[tex]^-2[/tex]/(8.314×1000)=1.56
Kc and Kp are equilibrium constants of a mixture of ideal gases. Kp is equilibrium constant when concentrations at equilibrium are in atmospheric pressure and Kc is equilibrium constant when concentrations are in molarity. The relation is only valid for gaseous mixtures. The relation between these two parameters is obtained through ideal gas equation.
Kc and Kp of reaction change with temperature of reaction but remain unaffected by change in concentration , pressure and presence of catalyst.
Learn more about equilibrium constants,here:
https://brainly.com/question/15118952
#SPJ2
Define the following terms - you may need to consult your lecture text or other suitable resource:
a. monomer,
b. repeating unit,
c. condensation polymerization,
d. cross-linked polymer
Answer:
a) Monomers: monomers are unit molecules, that can react together with other monomers, to form a long chain molecule called a polymer. Th polymer formed can also be in a three dimensional network. The process of this conversion of monomers to polymers is called polymerization.
b) Repeating unit: A repeating unit is a unit of the polymer formed, whose repetition would produce a long complete polymer chain. A polymer is made up of these repeating links of molecules that form a long chain of molecules.
c) Condensation polymerization: This is a form of condensation reaction, that involves the combination of molecules into polymers with the loss of small molecules such as water or methanol as by products.
d) Cross-linked polymer: This is a polymer formed from a type of bonding of molecules. The bonding is usually in the form of covalent bonds or ionic bonds and the polymers can be either synthetic polymers or natural polymers. The cross-links leads to an alteration in the physical properties of the polymer.
The definition of following terms are :
a) Monomers:
The monomers are unit atoms, that can respond in conjunction with other monomers, to create a long chain molecule called a polymer.
The polymer shaped can too be in a three dimensional arrange.
b) Repeating unit:
A rehashing unit may be a unit of the polymer shaped, whose reiteration would produce a long total polymer chain.
A polymer is made up of these rehashing joins of atoms that shape a long chain of molecules.
c) Condensation polymerization:
This is often a frame of condensation response, that includes the combination of particles into polymers with the misfortune of little particles such as water or methanol as by products.
d) Cross-linked polymer:
This can be a polymer shaped from a sort of holding of particles.
The cross-links leads to an modification within the physical properties.
DefinitionsDefinition is a rhetorical style that uses various techniques to impress upon the reader the meaning of a term, idea, or concept.
Definition may be used for an entire essay but is often used as a rhetorical style within an essay that may mix rhetorical styles.
Learn more about "Polymers":
https://brainly.com/question/17804565?referrer=searchResults
Which of the following happens to a molecule of an object when the object is heated? (1 point)
Answer:
They get more energy, so they vibrate!
Explanation:
The decomposition of ethylene oxide(CH₂)₂O(g) → CH₄(g) + CO(g)is a first order reaction with a half-life of 58.0 min at 652 K. The activation energy of the reaction is 218 kJ/mol. Calculate the half-life at 629 K.
Answer:
Half-life at 629K = 252.4min
Explanation:
Using Arrhenius equation:
[tex]ln\frac{K_1}{K_2} = \frac{Ea}{R} (\frac{1}{T_2} -\frac{1}{T_1})[/tex]
And as Half-life in a first order reaction is:
[tex]t_{1/2}=\frac{ln2}{K}[/tex]
We can convert the half-life of 58.0min to know K₁ adn replacing in Arrhenius equation find half-life at 629K:
[tex]58.0min=\frac{ln2}{K}[/tex]
K = 0.01195min⁻¹ = K₁
[tex]ln\frac{0.01195min^{-1}}{K_2} = \frac{218kJ/mol}{8.314x10^{-3}kJ/molK} (\frac{1}{629K} -\frac{1}{652K})[/tex]
[tex]ln\frac{0.01195min^{-1}}{K_2} =1.47[/tex]
[tex]\frac{0.01195min^{-1}}{K_2} =4.35[/tex]
K₂ = 2.75x10⁻³ min⁻¹
And, replacing again in Half-life expression:
[tex]t_{1/2}=\frac{ln2}{2.75x10^{-3}min^{-1}}[/tex]
Half-life at 629K = 252.4minThe half-life of the first-order reaction of ethylene oxide decomposition at 629 K is 251.1 min when the half-life at 652 K is 58.0 min and the activation energy is 218 kJ/mol.
The activation energy of a reaction is related to its rate constant as follows:
[tex] k = Ae^{-\frac{E_{a}}{RT}} [/tex] (1)
Where:
k: is the rate constant A: is the pre-exponential factor[tex]E_{a}[/tex]: is the activation energy of the reaction = 218 kJ/mol R: is the gas constant = 8.314 J/(K*mol)T: is the temperature
We can find the rate constant of the first-order reaction at 652 K with the half-life as follows:
[tex]k_{652} = \frac{ln(2)}{t_{1/2}_{(652)}}[/tex] (2)
Where [tex]t_{1/2}_{(652)}[/tex] is the half-life at 652 K= 58.0 min
Hence, the rate constant at 652 K is:
[tex] k_{652} = \frac{ln(2)}{58.0 min} = 0.012 min^{-1} [/tex]
Now, from equation (1) we can find the pre-exponential factor (A):
[tex]A = \frac{k_{652}}{e^{(-\frac{E_{a}}{RT_{1}})}} = \frac{0.012 \:min{-1}}{e^{(-\frac{218\cdot 10^{3} \:J/mol}{8.314 \:J/(K*mol)*652 \:K})}} = 3.51 \cdot 10^{15} min^{-1}[/tex]
With the pre-exponential factor we can calculate the rate constant at 629 K (eq 1):
[tex]k_{629} = 3.51 \cdot 10^{15} min^{-1}*e^{(-\frac{218 \cdot 10^{3} J/mol}{8.314 J/(K*mol)*629 K})} = 2.76 \cdot 10^{-3} min^{-1}[/tex]
Finally, the half-life at 629 K is (eq 2):
[tex] t_{1/2}_{629} = \frac{ln(2)}{2.76\cdot 10^{-3} min^{-1}} = 251.1 min [/tex]
Therefore, the half-life at 629 K is 251.1 min.
Find more about activation energy here:
https://brainly.com/question/14725142?referrer=searchResultshttps://brainly.com/question/11334504?referrer=searchResultsI hope it helps you!
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.
What are some geographic features that could be found in the hydrosphere?
Lakes, oceans, glaciers, clouds, etc. It categorizes all forms of water on earth.
hydro = water
Answer:
Lakes, streams, ground water, polar ice caps, glaciers, water vapor, and rivers!
Explanation:
The hydrosphere is made up of all the water on Earth. So anything that is water, like oceans, can be found in the hydrosphere:)
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:
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/molwhat is the lewis structure for OP(N3)3
Explanation:
this is the ans
hope this helps
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!
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.
For more information on intermolecular forces, visit
https://brainly.com/subject/chemistry
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.
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:
what is ammonium nitrate
Answer:
Ammonium nitrate is a chemical compound with the chemical formula NH₄NO₃. It is a white crystalline solid consisting of ions of ammonium and nitrate.
There are 2.4g of calcium hydroxide reacted with nitric acid. Calculate the number of moles of calcium hydroxide used. Write your answer using proper significant digits and units. Show all your work.
Answer:
0.032 moles
Explanation:
no of moles =
[tex] \frac{mass \: in \: grams}{relative \: molecular \: mass} [/tex]
=
[tex] \frac{2.4}{40 + 32 + 2} [/tex]
= 0.032
Calcium hydroxide reacted with nitric acid the total number of moles will be 0.032 moles.
What is a mole?
A mole is Avogadro's number of particles, which is exactly 6.02214076×1023.
The mole is widely used in chemistry as a convenient way to express amounts of reactants and products of chemical reactions. For example, the chemical equation 2H2 + O2 → 2H2O can be interpreted to mean that for each 2 mol dihydrogen (H2) and 1 mol dioxygen (O2) that react 2 mol of water (H2O) form.
Number of moles = Mass of substance / Mass of one mole Number of moles
mass of substance = 2.4g
molar mass of calcium hydroxide is (1 ×40.078g/mol Ca) +(2 × 15.999g/mol O) + (2 × 1.008g/mol H) = 74.092 g/mol Ca (OH)2
substituting the value,
number of moles = 2.4 / 74.029
= 0.032 moles
Therefore, moles of calcium hydroxide will be 0.032 moles
Learn more about moles, here :
https://brainly.com/question/15209553
#SPJ2
2NH3 → N2 + 3H2 If 2.22 moles of ammonia (NH3) decomposes according to the reaction shown, how many moles of hydrogen (H2) are formed? A) 2.22 moles of H2 B) 1.11 moles of H2 C) 3.33 moles of H2 D) 6.66 moles of H2
Answer:
C
Explanation:
According to the mole ratio, using 2NH3 will give you 3H2. Which means in order to find the moles of H2 you would only need to divide 2 and multiply 3 to get the amount of moles of H2 produced.
Answer:
I think it's C
Explanation:
Please, tell me if I'm incorrect.
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.
If 1 mol of a pure triglyceride is hydrolyzed to give 2 mol of RCOOH, 1 mol of R'COOH, and 1 mol of glycerol, which of the following compounds might be the triglyceride?
CHOC(O)R
A. CHOC(O)R
CHOC(O)R
CH,OC(O)R
B. CHOC(O)R
CH2OC(O)R
CHOC(O)R
C. CHOC(O)R
CHOC(O)R
CHOC(O)R
D. CHOC(O)R
CHOC(O)R
Answer:
The correct option is C.
Note the full question and structure of the moleculesis found in the attachment below.
Explanation:
Triglycerides or triacylglycerols are non-polar, hydrophobic lipid molecules composed of three fatty acids linked by ester bonds to a molecule of glycerol.
The fatty acids linked to the glycerol molecule are denoted by R and may be of the same kind or different. when the R group is the same, the R is attached in all the three positions for ester bonding in the glycerol molecule but when they are different are denoted by R, R' and R'' respectively.
During the hydrolysis of triglycerides, the three fatty acids molecules are obtained as well as a glycerol molecule.
From the question, when 1 mole of the triglyceride is hydrolysed, 2 moles of RCOOH, 1 mole of R'COOH and 1 mole of glycerol is obtained. The triglyceride must then be composed of two fatty acids which are the same denoted by R, and a different fatty acid molecule denoted by R'.
The correct option therefore, is C
Beginning with Na, record the number of energy levels, number of protons, and atomic radius for each element in period 3.
Answer:
Sodium, magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon are the elements of third period.
Explanation:
There are three energy levels in sodium atom. It has 11 electrons revolving around the nucleus. the atomic radius of sodium atom is 227 ppm. Magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon has also three energy levels like sodium because all these elements belongs to third period. There are 12 electrons present in magnesium, 13 in aluminium, 14 in silicon, 15 in phosphorus, 16 in sulfur, 17 in chlorine, and 18 electrons in argon. The atomic radius of magnesium atom is 173 ppm. The atomic radius of aluminium atom is 143 ppm. The atomic radius of silicon atom is 111 ppm. The atomic radius of phosphorus atom is 98 ppm. The atomic radius of sulfur atom is 87 ppm. The atomic radius of chlorine atom is 79 ppm and the atomic radius of argon atom is 71 ppm.
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.
The gas with an initial volume of 24.0 L at a pressure of 565 torr is compressed until the volume is 16.0 L. What is the final pressure of the gas, assuming the temperature and amount of gas does not change
Answer:
848 torr
Explanation:
The only variables are the pressure and the volume, so we can use Boyle's Law.
p₁V₁ = p₂V₂
Data:
p₁ = 565 torr; V₁ = 24.0 L
p₂ = ?; V₂ = 16.0 L
Calculations:
[tex]\begin{array}{rcl}p_{1}V_{1} & = & p_{2}V_{2}\\\text{565 torr} \times \text{24.0 L} & = & p_{2} \times \text{16.0 L}\\\text{13 560 torr} & = & 16.0p_{2}\\p_{2} & = & \dfrac{\text{13 560 torr}}{16.0}\\\\& = &\textbf{848 torr}\\\end{array}\\\text{The final pressure of the gas is $\large \boxed{\textbf{848 torr}}$}[/tex]
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.
Does the amount of methanol increase, decrease, or remain the same when an equilibrium mixture of reactants and products is subjected to the following changes?
a. the catalyst is removed
b. the temp is increased
c. the volume is decreased
d. helium is added
e. CO is added
Answer:
a. Methanol remains the same
b. Methanol decreases
c. Methanol increases
d. Methanol remains the same
e. Methanol increases
Explanation:
Methanol is produced by the reaction of carbon monoxide and hydrogen in the presence of a catalyst as follows; 2H2+CO→CH3OH.
a) The presence or absence of a catalyst makes no difference on the equilibrium position of the system hence the methanol remains constant.
b) The amount of methanol decreases because the equilibrium position shifts towards the left and more reactants are formed since the reaction is exothermic.
c) If the volume is decreased, there will be more methanol in the system because the equilibrium position will shift towards the right hand side.
d) Addition of helium gas has no effect on the equilibrium position since it does not participate in the reaction system.
e) if more CO is added the amount of methanol increases since the equilibrium position will shift towards the right hand side.
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
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
If you want additional help with chemistry or another subject for FREE, check out growthinyouth.org.
Experiment:
Part I: Voltaic Cell
Assume that you are provided with the following materials:
Strips of metallic zinc, metallic copper, metallic iron
1M aqueous solutions of ZnSO4, CuSO4, FeSO4, and aqueous iodine (I2)
Other required materials to create Voltaic cells such as beakers, porous containers, graphite rods, a voltmeter, and a few wires with alligator clips.
In this modified version of the lab, after thoroughly studying the lab hand out and watching the videos, identify 4 different combinations of Voltaic cells that are possible to be created with the above materials.
For each cell created, include the following details.
Which electrode was the anode and which was the Cathode?
The anode and cathode half reactions.
Balanced equation for each cell you propose to construct.
Calculated Eocell
Short hand notation (line notation) for each cell (be sure to include the inactive electrode if needed)
Answer:
Here are four possible voltaic cells.
Explanation:
1. Standard reduction potentials
E°/V
I₂(s) + 2e⁻ ⟶ 2I⁻(aq); 0.54
Cu²⁺(aq) + 2e⁻ ⟶ Cu(s); 0.34
Fe²⁺(aq) + 2e⁻ ⟶ Fe(s); -0.41
Zn²⁺(aq) + 2e⁻ ⟶ Zn(s); -0.76
2. Possible Voltaic cells
(a) Zn/I₂
E°/V
Anode: Zn(s) ⟶ Zn²⁺(aq) + 2e⁻; 0.76
Cathode: I₂(s) + 2e⁻ ⟶ 2I⁻(aq); 0.54
Cell: Zn(s) + I₂(s) ⟶ Zn²⁺(aq) + 2I⁻(aq); 1.30
Zn(s)|Zn²⁺(aq)∥I⁻(aq)|I₂(s)|C(s, graphite)
Zn is the anode; graphite is the cathode.
(b) Zn/Cu²⁺
E°/V
Anode: Zn(s) ⟶ Zn²⁺(aq) + 2e⁻; 0.76
Cathode: Cu²⁺(aq) + 2e⁻ ⟶ Cu(s); 0.34
Cell: Zn(s) + Cu²⁺(s) ⟶ Zn²⁺(aq) + Cu(s); 1.10
Zn(s)|Zn²⁺(aq)∥Cu²⁺(aq)|Cu(s)
Zn is the anode; Cu is the cathode.
(c) Zn/Fe²⁺
E°/V
Anode: Zn(s) ⟶ Zn²⁺(aq) + 2e⁻; 0.76
Cathode: Fe²⁺(aq) + 2e⁻ ⟶ Fe(s); -0.41
Cell: Zn(s) + Fe²⁺(s) ⟶ Zn²⁺(aq) + Fe(s); 0.35
Zn(s)|Zn²⁺(aq)∥Fe²⁺(aq)|Fe(s)
Zn is the anode; Fe is the cathode.
(d) Fe/I₂
E°/V
Anode: Fe(s) ⟶ Fe²⁺(aq) + 2e⁻; 0.41
Cathode: I₂(s) + 2e⁻ ⟶ 2I⁻(aq); 0.54
Cell: Zn(s) + I₂(s) ⟶ Zn²⁺(aq) + 2I⁻(aq); 0.95
Fe(s)|Fe²⁺(aq)∥I⁻(aq)|I₂(s)|C(s, graphite)
Fe is the anode; graphite is the cathode.
Sulfur dioxide reacts with oxygen to form sulfur trioxide. What change in hybridization of the sulfur occurs in this reaction ? g
Answer:
PLEASE LOOK INN TO THE FILE YOU WILL GET ANSWER AND ALSO SUMMARY THANKS FOR ASKING QUESTION.
Explanation:
Atomic mass is calculated by _____. subtracting protons from neutrons averaging the mass of isotopes adding protons and neutrons subtracting neutrons from protons
Answer:
Atomic mass is calculated by adding protons and neutrons.
Explanation:
Atomic mass is the sum of protons and neutrons in an atomic nucleus. For example, the element Oxygen has 8 protons (derived from the atomic number) and 8 neutrons (derived from subtracting the amount of protons from the atomic mass).
We can craft an equation to show the relationship between these variables.
M - N = P, where M = Mass, N = Neutrons, and P = Protons
This equation can be rearranged to show the relationship between the neutrons and protons leading to the atomic mass. Simply add N to both sides of the equation.
M = N + P
This shows that atomic mass is equivalent to the sum of protons and neutrons in an atom's nucleus.
Calculate the amount of heat needed to melt of ice () and bring it to a temperature of . Round your answer to significant digits. Also, be sure your answer contains a unit symbol.
Answer:
your answer is given below
Explanation:
Take it stepwise, and you are going to have to look up the various heat values.
You have 36.5grams of ice, presumably at 0C. You will need to add heat to take the ice at 0 C to water at 0 C. (Latent heat of melting) Then you add in a different heat value to take the water at 0 C to water at 82.3 C. (Specific heat of water)
Add the two heat amounts together.
According to specific heat capacity, to calculate the amount of heat needed to melt ice given mass of ices is multiplied by specific heat of ice and the temperature change which it undergoes.
What is specific heat capacity?Specific heat capacity is defined as the amount of energy required to raise the temperature of one gram of substance by one degree Celsius. It has units of calories or joules per gram per degree Celsius.
It varies with temperature and is different for each state of matter. Water in the liquid form has the highest specific heat capacity among all common substances .Specific heat capacity of a substance is infinite as it undergoes phase transition ,it is highest for gases and can rise if the gas is allowed to expand.
It is given by the formula ,
Q=mcΔT
Learn more about specific heat capacity,here:
https://brainly.com/question/28941910
#SPJ5
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.
Please mark Brainliest...
At standard temperature and pressure conditions, the volume of an ideal gas contained in a jar is 55.3 L. How many molecules are in the jar. This question is to be answered in scientific notation.(eg. 1.5 e5)
Answer:
1.49e24
Explanation:
Standars temperature and pressure are 273.15K and 1atm, respectively.
Using ideal gas law, we can find moles of an ideal gas if we know its pressure, temperature and volume as follows:
PV = nRT
PV / RT = n
Where P is pressure (1atm), V is volume (55.3L), R is gas constant (0.082atmL/molK), T is temperature (273.15K) and n moles of the ideal gas.
Replacing:
PV / RT = n
1atm*55.3L / 0.082atmL/molK*273.15K = n
2.47 moles = n
Now, the question is about the number of molecules in the jar. By definition, 1 mole = 6.022x10²³ molecules.
As we have 2.47 moles:
2.47 mol × (6.022x10²³ molecules / 1 mole) =
1.49x10²⁴ molecules that are in the jar
In scientific notation:
1.49e24