Answer:
This solvent is not suitable because ΔS°vap > 0.
Explanation:
Let's consider a system at a higher temperature T1, and its surroundings at a lower temperature T2. Let's call q the heat that goes irreversible from the system to the surroundings:
ΔSsystem = -q/T1 (it's losing heat, so q must be negative)
ΔSsurroundings= q/T2
ΔSprocess = ΔSsystem + ΔSsurroundings
ΔSprocess = -q/T1 + q/T2
ΔSprocess = q*(1/T2 - 1/T1)
ΔSprocess = q*[(T1 - T2)/(T1*T2)]
As pointed above, T1> T2, so ΔSprocess > 0 and because of that, the reaction is spontaneous. It means that if ΔS°vap > 0, the solvent will vaporize. So, as we can notice, the solvent given is not suitable.
Final answer:
The overall entropy change for vaporization helps determine if a solvent is suitable for an experiment. A high value of entropy indicates increased disorder, making vaporization likely. In this case, with a high entropy of 112.9 J/(K⋅mol), the solvent's boiling at 75 ∘C may hinder its suitability for the experiment.
Explanation:
The overall entropy change associated with the vaporization reaction can be calculated considering the entropy of vaporization (ΔSvap) and the enthalpy of vaporization (ΔHvap) of the solvent. As stated in Trouton's rule, ΔSvap is approximately 10.5R, which is around 85-88 J/(mol K) for many liquids. In this case, with a ΔSvap of 112.9 J/(K⋅mol) and a ΔHvap of 38.0 kJ/mol, the high ΔSvap value indicates that the solvent has a highly disordered structure, making it likely to vaporize at the given temperature of 75 ∘C, hence making it unsuitable for the experiment.
Nylon 6,6 may be formed by means of a condensation polymerization reaction in which hexamethylene diamine [NH2-(CH2)6-NH2] and adipic acid react with one another with the formation of water as a byproduct. What masses of________(a) hexamethylene diamine and(b) adipic acid are necessary to yield 26 kg of completely linear nylon 6,6? This polymerization reaction occurs according to the following equation:
Answer:
Mass of hexamethylenediamine= 13,470.62 g
Mass of adipic acid = 16,940.13 g
Explanation:
Nylon -6,6 is synthesized by poly-condensation of hexamethylenediamine [NH₂-(CH₂)₆-NH₂] and adipic acid (C₆H₁₀O₄).
Both of these are taken in equal equivalents and reacted to form nylon 6,6 and by-product water is also formed.
Molecular mass of hexamethylenediamine [NH₂-(CH₂)₆-NH₂] =116.21 g/mol
Molecular mass of adipic acid (C₆H₁₀O₄) = 146.1412 g/mol
Molecular mass of Nylon 6,6 (C₁₂H₂₀N₂O₂) = 224.3 g/mol
Using unitary method :
Mass of hexamethylenediamine required for 26,000 g of Nylon 6,6 if 116.21 g/mol is required for 224.3 g/mol, is:
= [tex]\frac{26,000 * 116.21}{224.3}[/tex]
= 13,470.62 g
Mass of adipic acid required for 26,000 g of Nylon 6,6 if 146.1412 g/mol is required for 224.3 g/mol, is:
= [tex]\frac{26,000 * 146.1412}{224.3}[/tex]
= 16,940.13 g
A) Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6, and an unknown amount of propane, C3H8) were added to the same 10.0- L container. At 23.0 C, the total pressure in the container is 4.60atm . Calculate the partial pressure of each gas in the container.
Express the pressure values numerically in atmospheres, separated by commas. Enter the partial pressure of methane first, then ethane, then propane.
B) A gaseous mixture of O2 and N2 contains 39.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 585mmHg ?
Answer:
A. PpCH4 = 1.21 atm, Pp C2H6 = 1.46 atm, Pp C3H6 = 1.93 atm
B. Pp O2 = 352.17 mmHg
Explanation:
A. First step: Get the total mols of mixture to know the mass of propane.
P . V = n . R . T
T° K = T°C + 273 → 23°C + 273 = 296K
4.60 atm . 10L = n . 0.082 L.atm/mol.K . 296K
(4.60 atm . 10L) / (0.082 mol.K/L.atm . 296K) = n
1.89 mol = n
Mass/Molar weight = mol
Mass CH4 = 16 g/m
Mol CH4: 8g / 16g/m = 0.5 mol
Mass C2H6 = 30 g/m
Mol C2H6 = 0.6 mol
Total mols - Mol CH4 - Mol C2H6 = Mol propane
1.89 mols - 0.5 mol - 0.6 mol = 0.79 mol propane
2nd step: We can use the molar fraction with the partial pressure.
Partial pressure gas / Total pressure = mols gas / total moles
Sum of molar fraction = 1
0.5 / 1.89 + 0.6 /1.89 + 0.79/1.89 = 1
Partial pressure CH4 → Pp CH4 /4.60 atm = 0.5 / 1.89
PpCH4 = (0.5 /1.89) . 4.60 atm = 1.21 atm
Partial pressure C2H6 → Pp C2H6 /4.60atm = 0.6 /1.89
Pp C2H6 = (0.6 /1.89) . 4.60atm = 1.46 atm
Partial pressure C3H6 → Pp C3H6 /4.60atm = 0.79/1.89
Pp C3H6 = (0.79/1.89) . 4.60atm = 1.93 atm
B. 39.8% means a percent of the molar fraction, so the fraction is 0.398
N2 mass /total mass = Partial pressure N2/ Total pressure
0.398 . 585 mmHg = Partial pressure N2
232.83 mmHg = Partial pressure N2
Total pressure - Pp N2 = Pp O2
585 mmHg - 232.82 mmHg = 352.17 mmHg
Cyanobacteria______.a.can fix nitrogen, thus causing many freshwater ecosystems to be limited by phosphorus availability.b.can mineralize phosphorus from wetland sediments, raising phosphorus levels in inland waters to the point of eutrophication.c.mineralize the phosphates in detergent from an organic corn into an inorganic form that can cause algal blooms and eutrophication.d.consume excess N and P in inland waters and so are essential tools for watershed restoration and management.e.occur when aquatic ecosystems are disrupted by invasive species, and can cause pathogenic infections of native species, further enhancing the invasive nature of introduced exotic invaders.
Answer:
Cyanophtya
Explanation:
Cyanobacteria also known as Cyanophyta are responsible for fixing nitogen and raising phosphorus level through mineralization they are also called blu-green-algae. They commonly obtain their energy from 'oxygenic' photosynthesis.
Which of these incorrectly pairs a hormone with its function? A. thyroxin—regulates metabolism B. insulin—enables the body to use glucose (blood sugar) for energy C. adrenaline (epinephrine)—prepares the body to "fight or flee" D. progesterone—controls muscle mass
Answer:
The correct answer is D progesterone control muscle mass.
Explanation:
Progesterone is a steroid hormone synthesized from cholesterol and secreted from corpus lutenum and placenta.progesterone regulate the luteal phase of the menstrual cycle, progesterone helps in the development of secondary sexual characteristics such as breast development,pubic hair development etc.
On the other hand muscle mass is controlled by the growth hormone released from the anterior lobe of the pituitary gland.
Hafnium, Hf, is also found in group 4B. Write the electron configuration for Hf. Express your answer in order of increasing orbital energy as a string without blank space between orbitals. For example, the electron configuration of Li could be entered as 1s^22s^1 or [He]2s^1.
The electron configuration for Hafnium (Hf) is [Xe]6s²4f¹⁴5d².
Explanation:The electron configuration for Hafnium (Hf) can be determined by looking at its position in the periodic table. Hafnium is in group 4B, which means it has 4 valence electrons. To write the electron configuration, we start with the noble gas that comes before Hafnium, which is Xenon (Xe).
The electron configuration for Hafnium, therefore, is [Xe]6s24f145d2.
Hafnium (Hf) has an electron configuration of 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d², or [Xe] 4f¹⁴ 5d² 6s².
It's part of the transition metals in group 4B.
The element hafnium (Hf) is part of the group 4B elements in the periodic table, which includes transition metals like titanium (Ti) and zirconium (Zr). Hafnium has an atomic number of 72. To determine the electron configuration of hafnium in order of increasing orbital energy, we start by filling the orbitals based on the Aufbau principle:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d².
Alternatively, in shorthand notation, this can be written as [Xe] 4f¹⁴ 5d² 6s². This configuration follows the orderly filling of orbitals, ensuring that lower energy orbitals are filled before higher energy orbitals.
The enthalpy of combustion of ethane gas, C2H6(g), is about -1.5*103 kJ/mol. When ethane reacts with O2(g), the products are carbon dioxide CO2(g) and water H2O(l). How much heat is released during the combustion of ethane gas when 14 mols of O2(g) are consumed. Express your answer in kJ.
Answer:
[tex]6.0\times 10^3 kJ[/tex] heat is released during the combustion of ethane gas when 14 moles of oxygen are consumed.
Explanation:
[tex]C_2H_6(g)+\frac{7}{2}O_2(g)\rightarrow 2CO_2(g)+3H_2O(l),\Delta H_{rxn}=-1.5\times 10^3 kJ/mol[/tex]
According to reaction, when 7/2 moles of oxygen is consumed by 1 mol of ethane [tex]1.5\times 10^3 kJ[/tex] energy is released.
Energy released when 1 mole of oxygen are consumed: [tex]\frac{1.5\times 10^3 kJ\times 2}{7}[/tex]
Then energy released when 14 moles of oxygen are consumed:
[tex]\frac{1.5\times 10^3 kJ\times 2}{7}\times 14=6.0\times 10^3 kJ[/tex]
[tex]6.0\times 10^3 kJ[/tex] heat is released during the combustion of ethane gas when 14 moles of oxygen are consumed.
Answer:
6.0×10^3 KJmol-1
Explanation:
Equation of the reaction
C2H6(g) + 7/2 O2(g) ------> 2CO2(g) + 3H2O(g)
From the balanced reaction equation:
1 mole of ethane reacts with 3.5 moles of oxygen
x moles of ethane will react with 14 moles of oxygen
x= 14/3.5 = 4 moles of ethane
If heat of combustion of ethane= -1.5*103 kJ/mol
Then for 4 moles of ethane= 4× -1.5*103 kJ/mol= 6.0×10^3 KJmol-1
A gas mixture used for anesthesia contains 2.83 mol oxygen (O2), and 8.41 mol of nitrous oxide (N2O). The total pressure of the mixture is 192 kPa. What is the partial pressure of N2O? Report your answer with three significant figures.
Answer:
The partial pressure of N2O is 143.6 kPa
Explanation:
Step 1: Data given
Moles O2 = 2.83 mol
Moles N2O = 8.41 mol
Total pressure = 192 kPa
Step 2: Calculate total number of moles
Total numberof moles = moles O2 + molesN2O = 2.83 + 8.41 = 11.24 mol
Step 3: Calculate mole fraction of N2O
Mole fraction N2O = mole N2O / Total moles
Mole fraction N2O = 8.41 / 11.24 = 0.748
Step 4: Calculate partial pressure of N2O
pN2O = 0.748 * 192 kPa
pN2O = 143.6 kPa
The partial pressure of N2O is 143.6 kPa
Answer: 144 kPa
Explanation:
To determine the partial pressure of N2O, first determine the mole fraction of N2O in the mixture, then multiply the mole fraction of N2O by the total pressure.
XN2O = nN2O/ ntotal = 8.41 mol / 2.83 mol + 8.41 mol = 0.7482
PN2O=XN2O×Ptotal = 0.7482 × 192 kPa = 143.7 kPa
Rounding the answer to three significant figures, the partial pressure of N2O is 144kPa.
Calcium chloride, CaCl2, is commonly used as an electrolyte in sports drinks and other beverages, including bottled water. A solution is made by adding 6.50 g of CaCl2 to 60.0 mL of water at 25∘C. The density of the solvent at that temperature is 0.997 g/mL. Calculate the mole percent of CaCl2 in the solution.
Final answer:
The mole percent of CaCl2 in the solution would be 100% if we consider only the solute, but this does not account for the water's contribution. If water's contribution to total moles were considered, the mole percent for CaCl2 would be less than 100%.
Explanation:
To calculate the mole percent of CaCl2 in the solution, we must first determine the number of moles of CaCl2.
The molar mass of CaCl2 is approximately 110.98 g/mol (40.08 g/mol for calcium and 35.45 g/mol for each chlorine, with two chlorines). With 6.50 g of CaCl2, the number of moles of CaCl2 is:
6.50 g ÷ 110.98 g/mol = 0.05855 mol
To find the total mass of the solution, the mass of CaCl2 is added to the mass of water. With the water density given as 0.997 g/mL, the mass of 60.0 mL of water is:
60.0 mL × 0.997 g/mL = 59.82 g
The total mass of the solution is thus:
6.50 g (CaCl2) + 59.82 g (H2O) = 66.32 g
To find the mole fraction of CaCl2, we divide the moles of CaCl2 by the total moles (assuming the water in the solution contributes negligibly to the total moles):
Mole fraction of CaCl2 = 0.05855 mol ÷ 0.05855 mol = 1 (since there are no other solutes)
To express this as a mole percent, we multiply the mole fraction by 100%:
Mole percent of CaCl2 = 1 × 100% = 100%
However, this calculation does not account for the water's contribution to the mole fraction. To be precise, the mole fraction and mole percent should include the moles of water as well, but since the question seems to focus solely on CaCl2, the calculation reflects this perspective. If we were to include the moles of water, the mole percent would be significantly less than 100%.
An example of a synthesis is the Statue of Liberty turning green. Oxygen in the air combines with the copper statue to form copper oxide, which is green. Which of these is the correct chemical reaction for this?
A) Cu + O → CuO
B) Cu + O2 → CuO
C) CuO → Cu + O2
D) 2 Cu + O2 → 2CuO
Answer:
D. 2Cu + O2 = 2CuO
Explanation:
Then CuO, which is black, reacts with CO2 to form CuCO3 which is green.
Answer : The correct chemical reaction is, (D)
Explanation :
Synthesis reaction : It is defined as a chemical reaction where multiple substances or reactants combine to form a single product.
It is represented as,
[tex]X+Y\rightarrow XY[/tex]
When copper react with an oxygen then it react to give copper oxide as a product.
The balanced chemical reaction will be:
[tex]2Cu+O_2\rightarrow 2CuO[/tex]
Hence, the correct chemical reaction is, (D)
For a beaker containing the reaction below that is at equilibrium, using LeChatlier's Principle, predict which way the equilibrium conditions will shift (type in left, right, or no change) if silver chloride was added to a beaker.
AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)
Explanation:
LeChatlier's principle states that any change in pressure, temperature, or concentration of a reactant applied to a system in equilibrium, the equilibrium will shift so as to tend to counteract the effect of the change.
In our question,[tex]AgCl[/tex] is added.
[tex]AgCl[/tex] is a product.
When product concentration increases,the reaction goes to the left so as to decrease the product concentration.
Adding more silver chloride (AgCl) to the system in equilibrium causes the equilibrium to shift to the left, producing more reactants (AgNO3 and NaCl) and reducing the AgCl.
Explanation:According to Le Chatelier's Principle, when a change in concentration, pressure, or temperature is applied to a system in equilibrium, the system will adjust itself to counteract the effect of the change. The reaction equation provided is AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq).
If additional silver chloride (AgCl) is added to the beaker, the equilibrium will shift to the left (reactants) to reduce this excess. More reactants AgNO3 and NaCl will be produced, reducing the amount of AgCl. This is because the system strives to maintain equilibrium by utilising the 'extra' AgCl to form more reactants.
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Bromine (Br2) is produced by reacting HBr with O2, with water as a byproduct. The O2 is part of an air (21 mol % O2, 79 mol % N2) feed stream that is flowing sufficiently fast to provide 25% excess oxygen ("excess" has a precise meaning in process analysis: in this case there is 25% more oxygen than the amount needed to completely react with the limiting reactant). The fractional conversion of HBr is 78%.
a) Show the degree of freedom analysis. Be as specific as possible about labeling the unknowns and completely write out all of the independent equations.
b) Calculate the composition (mole fractions) of the product stream.
Answer:
The mole fractions:
[tex]x_{HBr}=\frac{100mol}{318.5}=0.314[/tex]
[tex]x_{Br_2}=\frac{78mol}{318.5}=0.245[/tex]
[tex]x_{H_2O}=\frac{78mol}{318.5}=0.245[/tex]
[tex]x_{O_2}=\frac{62.5mol}{318.5}=0.196[/tex]
Explanation:
The reaction described is:
[tex]2 HBr (g) + 1/2 O_2 (g) \longrightarrow Br_2 (g) + H_2O (g)[/tex]
The limiting reactant is the HBr (oxygen is in excess).
a) The mass (in moles) balance for this sistem:
[tex]n_{Br_2}=\frac{ 1 mol Br_2}{1 mol HBr} *n_{HBr}*0.78[/tex]
(the 0.78 is because of the fractional conversion)
[tex]n_{O_2}=\frac{ 0.5 mol O_2}{1 mol HBr} *n_{HBr}*1.25[/tex]
(the 1.25 is because of the oxygen excess)
[tex]n_{H_2O}=\frac{ 1 mol H_2O}{1 mol Br_2} *n_{Br_2}[/tex]
There is only one degree of freedom in this sistem, you can either deffine the moles of HBr you have or the moles of Br2 you want to produce. The other variables are all linked by the equations above.
b) Base of calculation 100 mol of HBr:
[tex]nn_{HBr}=100 mol HBr[/tex]
[tex]n_{Br_2}=\frac{ 1 mol Br_2}{1 mol HBr} *100mol HBr*0.78[/tex]
[tex]n_{Br_2}=78 mol Br2[/tex]
[tex]n_{O_2}=\frac{ 0.5 mol O_2}{1 mol HBr} *100 mol HBr*1.25[/tex]
[tex]n_{O_2}=62.5 mol O_2[/tex]
[tex]n_{H_2O}=n_{Br_2}= 78 mol[/tex]
[tex]n_{total}=(78+78+100+62.5)mol= 318.5mol[/tex]
The mole fractions:
[tex]x_{HBr}=\frac{100mol}{318.5}=0.314[/tex]
[tex]x_{Br_2}=\frac{78mol}{318.5}=0.245[/tex]
[tex]x_{H_2O}=\frac{78mol}{318.5}=0.245[/tex]
[tex]x_{O_2}=\frac{62.5mol}{318.5}=0.196[/tex]
Match the vocabulary terms to their definitions.
1. standard metric unit of length
2. a proposed explanation for a scientific problem
3. standard unit of volume
4. the curved top surface of a liquid column
5. a quantity in an experiment that remains unchanged or constant
6. amount of matter in an object
a.mass
b.hypothesis
c.liter
d.control
e.meter
f.meniscus
You may find bellow the association of therms with the definitions.
Explanation:
1. standard metric unit of length - e. meter
2. a proposed explanation for a scientific problem - b. hypothesis
3. standard unit of volume - c. liter
4. the curved top surface of a liquid column - f. meniscus
5. a quantity in an experiment that remains unchanged or constant - d. control
6. amount of matter in an object - a. mass
Learn more about:
standard units of measurement
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Write a balanced chemical equation, including physical state symbols, for the decomposition of solid calcium carbonate (CaCO₃) into solid calcium oxide and gaseous carbon dioxide. Suppose 23.0 L of carbon dioxide gas are produced by this reaction, at a temperature of 380.0 °C and pressure of exactly 1 atm. Calculate the mass of calcium carbonate that must have reacted. Be sure your answer has the correct number of significant digits.
Answer:
CaCO3 (s) → CaO (s) + CO2 (g)
The mass of carbonate that must have reacted was 43.03 grams
Explanation:
CaCO3 → CaO + CO2
Relation between reactant and product is 1:1
Let's apply the Ideal Gas Law to find out the moles of CO2 which were produced.
P . V = n . R . T
1 atm . 23 L = n . 0.082 L.atm/mol.K . 653K
(1atm . 23L) / (0.082 mol.K/L.atm . 653K) = n
0.43 moles = n
0.43 moles of CO2, were produced from 0.43 moles of CaCO3.
Molar weight of CaCO3 = 100.08 g/m
Mass = Molar weight . moles
Mass = 100.08 g/m 0.43 m = 43.03 g
The decomposition of calcium carbonate (CaCO₃) can be represented by the equation: CaCO₃(s) → CaO(s) + CO₂(g). Using the ideal gas law, we calculate that approximately 43.1 grams of CaCO₃ reacted to produce 23.0 liters of CO₂ at 380.0 °C and 1 atm pressure.
Decomposition of Calcium Carbonate:
The balanced chemical equation for the decomposition of solid calcium carbonate (CaCO₃) into solid calcium oxide (CaO) and gaseous carbon dioxide (CO₂) is:
CaCO₃(s) → CaO(s) + CO₂(g)
To calculate the mass of calcium carbonate that reacted to produce 23.0 L of CO₂ gas at 380.0 °C and 1 atm, we use the ideal gas law. The ideal gas law is expressed as:
PV = nRT
Where:
P = pressure (1 atm)V = volume (23.0 L)n = moles of gasR = ideal gas constant (0.0821 L·atm/mol·K)T = temperature in Kelvin (380.0 °C + 273.15 = 653.15 K)Rearranging the ideal gas law to solve for n (moles of CO₂) gives:
n = PV / RT
Substituting the known values:
n = (1 atm * 23.0 L) / (0.0821 L·atm/mol·K * 653.15 K) ≈ 0.431 moles of CO₂
From the balanced equation, 1 mole of CaCO₃ produces 1 mole of CO₂. Hence, the moles of CaCO₃ that reacted is also 0.431 mol.
The molar mass of CaCO₃ is approximately 100.09 g/mol.
The mass of CaCO₃ that reacted is:
mass = moles * molar mass
mass = 0.431 mol * 100.09 g/mol ≈ 43.1 g
Thus, about 43.1 grams of calcium carbonate must have reacted.
Acidic solution In acidic solution, the iodate ion can be used to react with a number of metal ions. One such reaction is IO3−(aq)+Sn2+(aq)→I−(aq)+Sn4+(aq) Since this reaction takes place in acidic solution, H2O(l) and H+(aq) will be involved in the reaction. Places for these species are indicated by the blanks in the following restatement of the equation: IO3−(aq)+Sn2+(aq)+ −−−→I−(aq)+Sn4+(aq)+ −−−
Answer:
[tex]\large \text{IO$_{3}$$^{-}$(aq) + 3Sn$^{2+}$(aq) + 6H$^{+}$(aq) $\longrightarrow \,$ I$^{-}$(aq) + 3Sn$^{4+}$(aq) + 3H$_{2}$O(l)}[/tex]
Explanation:
IO₃⁻ + Sn²⁺ ⟶ I⁻ + Sn⁴⁺
1: Separate the equation into two half-reactions.
IO₃⁻ ⟶ I⁻
Sn²⁺ ⟶ Sn⁴⁺
2: Balance all atoms other than H and O.
Done
3: Balance O.
IO₃⁻ ⟶ I⁻ + 3H₂O
Sn²⁺ ⟶ Sn⁴⁺
4: Balance H
IO₃⁻ + 6H ⟶ I⁻ + 3H₂O
Sn²⁺ ⟶ Sn⁴⁺
5: Balance charge.
IO₃⁻ + 6H⁺ + 6e⁻ ⟶ I⁻ + 3H₂O
Sn²⁺ ⟶ Sn⁴⁺ + 2e⁻
6: Equalize electrons transferred.
1 × [IO₃⁻ + 6H⁺ + 6e⁻ ⟶ I⁻ + 3H₂O]
3 × [Sn²⁺ ⟶ Sn⁴⁺ + 2e⁻]
7: Add the two half-reactions.
1 × [IO₃⁻ + 6H⁺ + 6e⁻ ⟶ I⁻ + 3H₂O]
3 × [Sn²⁺ ⟶ Sn⁴⁺ + 2e⁻]
IO₃⁻ + 3Sn²⁺ + 6H⁺ ⟶ I⁻ + 3Sn⁴⁺ + 3H₂O
8: Check mass balance.
On the left: 1 I, 3 O, 3 Sn, 6 H
On the right: 1 I, 3 O, 3 Sn, 6 H
Step 9: Check charge balance.
On the left: 1- + 12+ = 11+
On the right: 1- + 12+ = 11+
The equation is balanced.
[tex]\text{The balanced equation is }\\\large \textbf{IO$_{3}$$^{-}$(aq) + 3Sn$^{2+}$(aq) + 6H$^{+}$(aq) $\longrightarrow \,$ I$^{-}$(aq) + 3Sn$^{4+}$(aq) + 3H$_{2}$O(l)}[/tex]
The reaction in an acid solution involves IO3−(aq) reacting with Sn2+(aq) to form I−(aq) and Sn4+(aq). The complete balanced redox reaction is 2IO3−(aq) + 10H+(aq)+ 6Sn2+(aq) → I−(aq) + 4H2O(l) + 6Sn4+(aq). It is a redox reaction where iodate ions act as strong oxidizing agents.
Explanation:The reaction under consideration involves the iodate ion (IO3−) reacting with a tin(II) ion (Sn2+) to create iodide (I−) and a tin(IV) ion (Sn4+) in an acidic solution. In an acid solution, water (H2O) and hydrogen ions (H+) play crucial roles, hence they have places in the reaction. The complete balanced redox reaction is:
2IO3−(aq) + 10H+(aq)+ 6Sn2+(aq) → I−(aq) + 4H2O(l) + 6Sn4+(aq)
. To balance this reaction, we consider the change in oxidation state, the law of conservation of charge, and the law of conservation of mass. This reaction is a prominent example of a redox reaction where iodate ions act as strong oxidizing agents, oxidizing Sn2+ to Sn4+.
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What is the molarity of a potassium triiodide solution, KI3(aq), if 30.00 mL of the solution is required to completely react with 25.00 mL of a 0.300 M thiosulfate solution, K2S2O3(aq)? The chemical equation for the reaction is 2 S2O32-(aq) + I3-(aq) → S4O62-(aq) + 3 I-(aq).
Answer:
Molarity of triiodide solution is 0.125M
Explanation:
According to balanced equation, 2 moles of [tex]S_{2}O_{3}^{2-}[/tex] completely react with 1 mol of [tex]I_{3}^{-}[/tex].
Moles of [tex]S_{2}O_{3}^{2-}[/tex] in 25.00 mL of 0.300 M of [tex]K_{2}S_{2}O_{3}[/tex] solution = [tex](0.300\times 0.02500)moles=0.0075moles[/tex]
So, 0.0075 moles of [tex]S_{2}O_{3}^{2-}[/tex] completely reacts with [tex](\frac{1}{2}\times 0.0075)moles[/tex] of [tex]I_{3}^{-}[/tex] or 0.00375 moles of [tex]I_{3}^{-}[/tex]
If molarity of [tex]KI_{3}[/tex] solution is S (M) then-
[tex]S\times 0.0300=0.00375[/tex]
or, [tex]S=0.125[/tex]
So, molarity of triiodide solution is 0.125M
To determine the molarity of the KI3 solution, calculate the moles of S2O32- that reacted and use the stoichiometry of the reaction to find the moles of I3-. Then, divide the moles of I3- by the volume of KI3 solution in liters to find the molarity.
Explanation:The question asks for the molarity of a potassium triiodide (KI3) solution. To find the molarity of the KI3(aq) solution, we need to use the provided stoichiometry of the reaction between thiosulfate (S2O32-) and triiodide (I3-).
The reaction is given as: 2 S2O32-(aq) + I3-(aq) → S4O62-(aq) + 3 I-(aq). Since 25.00 mL of a 0.300 M thiosulfate solution is required to completely react with the KI3 solution, we first calculate the moles of S2O32- reacted.
Moles of S2O32- = 0.025 L * 0.300 mol/L = 0.0075 moles.
According to the balanced equation, 1 mole of I3- reacts with 2 moles of S2O32-. Therefore, we have:
Moles of I3- = 0.0075 moles S2O32- / 2 = 0.00375 moles.
Because 30.00 mL (or 0.03000 L) of KI3 solution contains 0.00375 moles I3-, the molarity of KI3 is calculated by:
Molarity of KI3 = 0.00375 moles / 0.03000 L = 0.125 M.
Which one of the following statements is not true?
a. Basalt magmas, in general, have higher temperatures than rhyolite magmas.
b. When magma reaches the surface, its dissolved gas content increases.
c. Melting temperatures of silicate rocks are lowered by small amounts of water.
d. Melting temperatures of silicate rocks increase with increased pressure.
Answer:
When magma reaches the surface, its dissolved gas content increases is true
Explanation:
The volcanic eruptions happen because of magma that is expelled on the earth’s surface. At the earth’s depth, all magma have gas dissolved in liquid. When the pressure has decreased the magma rises towards the earth’s surface creating a separate vapour phase.
As pressure reduces the volume of gas will expands and giving magma its 'explosive character'. Thus, as magma reaches the surface the dissolved gas content decreases and magma comes out of earth’s surface.
The untrue statement is b, as the dissolved gas content in magma decreases as it reaches the surface due to reduced pressure. Basalt magma has higher temperatures than rhyolite magma. Water reduces the melting temperature of silicate rocks, and increased pressure raises it.
Explanation:The subject here pertains to volcanic activity and magma characteristics. The statement that is not true is: b. When magma reaches the surface, its dissolved gas content increases. In fact, as magma rises to the surface, the pressure decreases which allows the dissolved gases to escape, thus its dissolved gas content actually decreases. Statements a, c and d are true - Basalt magmas do generally have higher temperatures than rhyolite magmas. Additionally, the presence of water does indeed lower the melting temperature of silicate rocks, and increased pressure results in a higher melting temperature for these rocks.
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CAM plants function using crassulacean acid metabolism. Like C4 plants, CAM plants provide preparatory step for the Calvin cycle. CAM plants are found in hot, dry environments; to prevent desiccation, they keep their stomata closed during the day. They take in carbon dioxide at night while stomata are open. To increase the concentration of carbon dioxide available to the enzyme rubisco and minimize the degree of photorespiration, the CAM plants carboxylate ________.
Answer: ORGANIC ACIDS
Explanation:
CAM PLANTS CARBOXYLATE ORGANICS ACIDS through the addition of CO2 to PEP Carboxylase( a phosphoenolpyruvate carboxylase enzyme present in the mesophyll cells of the cytoplasm in a green plant) to produce Oxaloacetate (organic compound).
CO2 + PEP ⇒ C4H4O5 (oxaloacetate)
Oxaloacetate is then converted to a similar molecule, Malate (C4H6O5, another form of organic compound) that can be transported in to the bundle-sheath cells. Malate enters the plasmodesmata and releases the CO2. The CO2 then fixed by rubisco and made into sugars via the Calvin cycle.
Bromine has two isotopes 79Br and 81Br, whose masses (78.9183 and 80.9163 amu) and abundances (50.69% and 49.31%, respectively) were determined in earlier experiments. Calculate the average atomic mass of bromine based on these experiments.
Answer:
Average atomic mass = 79.9034 amu
Explanation:
The formula for the calculation of the average atomic mass is:
[tex]Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})[/tex]
Given that:
For first isotope:
% = 50.69 %
Mass = 78.9183 amu
For second isotope:
% = 49.31 %
Mass = 80.9163 amu
Thus,
[tex]Average\ atomic\ mass=\frac{50.69}{100}\times {78.9183}+\frac{49.31}{100}\times {80.9163}\ amu[/tex]
[tex]Average\ atomic\ mass=40.0036+39.8998\ amu[/tex]
Average atomic mass = 79.9034 amu
Final answer:
The average atomic mass of bromine is calculated by taking the sum of the products of each isotope's mass and its abundance in decimal form. The result is an average atomic mass of 79.898 amu for bromine.
Explanation:
To calculate the average atomic mass of bromine based on the given experimental results, we need to consider the masses and relative abundances of its isotopes. By multiplying the mass of each isotope by its abundance (converted to a decimal form) and adding these weighted masses together, we can determine the average atomic mass.
Calculation:
Adding the weighted masses of both isotopes gives us: 39.994 amu + 39.904 amu = 79.898 amu. Thus, the average atomic mass of bromine based on these experiments is 79.898 amu.
Checking for Accuracy:
It is important to make sure that this number makes sense and correlates with the given abundances, ensuring that no mathematical errors have been made.
In the laboratory, a student uses a coffee cup calorimeter to determine the specific heat of a metal. She heats 18.1 grams of lead to 99.50°C and then drops it into a cup containing 81.7 grams of water at 23.00°C. She measures the final temperature to be 23.67°C. Assuming that all of the heat is transferred to the water, she calculates the specific heat of lead to be _____ J/g°C.
Answer:
[tex]specificheat=0.177J/g^{0}C[/tex]
Explanation:
The heat evolved by lead will be absorbed by water.
The heat evolved by lead will be:
[tex]Q1=massXspecificheatXchangeintemperature[/tex]
[tex]Q1=18.1X(Specificheat)X(99.50-23.67)[/tex]
Heat absorbed by water will be:
[tex]Q2=massXspecificheatX(changeintemperature)[/tex]
[tex]Q2=81.7X4.184(23.67-23)=229.03J[/tex]
We know that
Q1=Q2
[tex]18.1X(Specificheat)X(99.50-23.67)=229.03[/tex]
[tex]1372.523Xspecificheat=229.03[/tex]
[tex]specificheat=0.177J/g^{0}C[/tex]
B5H9(l) is a colorless liquid that will explode when exposed to oxygen. How much heat is released when 0.211 mol of B5H9 reacts with excess oxygen where the products are B2O3(s) and H2O(l). The standard enthalpy of formation of B5H9(l) is 73.2 kJ/mol, the standard enthalpy of formation of B2O3(s) is -1272 kJ/mol and that of H2O(l) is -285.4 kJ/mol. Express your answer in kJ.
Answer:
Heat realesed is - 2192.7 kJ
Explanation:
First lets have the chemical equation balanced, and then solve the question based on the fact that the enthalpy change for a reaction is the sum of the enthalpies of formation of products minus reactants.
B₅H₉ (l) + O₂ (g) ⇒ B₂O₃ (s) + H₂O(l)
B atoms we have 5 reactants and 2 products, so the common mltiple is 10 and we have
2 B₅H₉ (l) + O₂ (g) ⇒ 5 B₂O₃ (s) + H₂O(l)
Now balance H by multiplying by 9 the H₂O
2 B₅H₉ (l) + O₂ (g) ⇒ 5 B₂O₃ (s) + 9 H₂O(l)
Finally, balance the O since we have 24 in products by multiplying by 12 the
O₂ ,
2 B₅H₉ (l) + 12 O₂ (g) ⇒ 5 B₂O₃ (s) + 9 H₂O
ΔHrxn = 5 x ΔHºf B₂O₃ + 9 x ΔHºf H₂O - ( 2 x ΔHºf B₅H₉ + 12 x ΔHºf O₂ )
We have all the ΔHºf s except oxygen but remember the enthalpy of formation of a pure element in its standard estate is cero.
ΔHrxn = 5 mol x ( -1272 kJ/mol )+ 9 mol x ( -285.4 kJ/mol ) - ( 2mol x 732 kJ/mol )
= -8928.6 kJ - 1464 kJ = -10,392 kJ
Now this enthalpy change was based in 2 mol reacted according to the balanced equation, so for 0.211 mol of B₅H₉ we will have:
-10,392 kJ/ 2 mol B₅H₉ x 0.211 mol B₅H₉ = - 2192.7 kJ
Answer:
ΔH=-957.41 kJ
Explanation:
The chemical equation is:
B₅H₉ (l) + O₂ (g) ⇒ B₂O₃ (s) + H₂O(l)
The law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.
Then, you must balance the chemical equation. For that, you must first look at the subscripts next to each atom to find the number of atoms in the equation. If the same atom appears in more than one molecule, you must add its amounts
The coefficients located in front of each molecule indicate the amount of each molecule for the reaction. This coefficient can be modified to balance the equation, just as you should never alter the subscripts.
By multiplying the coefficient mentioned by the subscript, you get the amount of each element present in the reaction.
So in first place you balance B
2 B₅H₉ (l) + O₂ (g) ⇒ 5 B₂O₃ (s) + H₂O(l)
Then you balance H
2 B₅H₉ (l) + O₂ (g) ⇒ 5 B₂O₃ (s) + 9 H₂O(l)
Finally you balance O
2 B₅H₉ (l) + 12 O₂ (g) ⇒ 5 B₂O₃ (s) + 9 H₂O(l)
Then
Left side: 2*5=10 boron (B), 2*9=18 hydrogen. and 12*2=24 oxygen
Right side: 5*2=10 boron (B), 9*2=18 hydrogen. and 5*3 + 9*1=24 oxygen
Since you have the same amount of elements on each side of the equation, the equation is balanced.
You want to calculate the ∆H (heat of reaction) of the combustion reaction, that is, the heat that accompanies the entire reaction. For that you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (quantity of molecules of each compound that participates in the reaction) and finally subtract them:
Enthalpy of combustion = ΔH = ∑Hproducts - ∑Hreactants
Knowing that:
Heat of formation of B₅H₉ = 73.2 kJ/mol Heat of formation of water = -285.4 kJ/mol Heat of formation of B₂O₃ = -1,272 kJ/molFor the formation of one mole of a pure element the heat of formation is 0, in this case we have as a pure compound the oxygen O₂.
Then:
ΔH= 9 mol*(-285.4 kJ/mol) + 5 mol* (-1,272 kJ/mol) - [2 mol* 73.2 kJ/mol + 12 mol* 0 kJ/mol]
ΔH= -9,075 kJ
If you observe the previous balanced reaction, you can see that 2 moles of B₅H₉ (l) are necessary. And the calculation of the heat of reaction previously carried out is based on this reaction. This ultimately means that the energy that would result in the reaction of 2 moles of B₅H₉ (l) is -9,075 kJ.
To determine the heat that is released when 0.211 mol of B₅H₉ (l) react with excess oxygen, a rule of three is applied as follows: if 2 moles of B₅H₉ (l) produces a heat ΔH of -9,075 kJ, when reacting 0.211 mol of B₅H l (l) how much heat ΔH is released?
ΔH=[tex]\frac{0.211 moles*(-9075)kJ}{2 moles}[/tex]
ΔH=-957.41 kJ
To practice Problem-Solving Strategy 11.1 Energy efficiency problems. Suppose that you have left a 200-mL cup of coffee sitting until it has cooled to 30∘C , which you find totally unacceptable. Your microwave oven draws 1100 W of electrical power when it is running. If it takes 45 s for this microwave oven to raise the temperature of the coffee to 60∘C , what is the efficiency of heating with this oven?
Answer:
efficiency of heating with this oven is 51 %
Explanation:
to raise the temp of 200 ml of coffee from 30°C to 60°C the energy input to microwave oven is:
1100 J/s x 45 = 49,500 J
AT 100% efficiency
For 1°C the energy required to raise the temperature of 1 ml = 4.2 J
So for 30 C°, 1°C the energy required to raise the temperature of 200 ml =
Q = (4.2) (200)(30) = 25,200 J
efficiency = 25,200/49,500 = 0.51 = 51%
Answer:
51%
Explanation:
To practice Problem-Solving Strategy 11.1 Energy efficiency problems. Suppose that you have left a 200-mL cup of coffee sitting until it has cooled to 30∘C , which you find totally unacceptable. Your microwave oven draws 1100 W of electrical power when it is running. If it takes 45 s for this microwave oven to raise the temperature of the coffee to 60∘C , what is the efficiency of heating with this oven?
Using the formula Q=mCdT
Q=Energy (J)
m=mass
C=specific heat capacity
dT=temperature change
but Q=power xtime
to raise the coffee from 30∘C to 60∘C requires
1100 J/s x 45 = 49,500 J
Energy to raise 200ml coffee to 30∘C is at 4.2j/gC
Q = (4.2)(200)(30) = 25,200 J
Efficiency=output/input
efficiency = 25,200/49,500*100% = 0.51 = 51%
The substance that dissolves to make a solution is called
Answer: Solute or solvent
Explanation: A solute is the substance that dissolves to make a solution. The solvent is the solution that does the dissolving.
________ is an indication of how near the air is to saturation rather than the actual quantity of water vapor in the air.
Answer: Relative humidity
Explanation:
Relative humidity (RH) is the ratio of the partial pressure of water vapor to the equilibrium vapor pressure of water at a given temperature. Relative humidity depends on temperature and the pressure of the system of interest. The same amount of water vapor has higher relative humidity in cool air than in warm air. A related parameter to that regard is that of dew point.
The Goodyear blimp we usually see hovering above our heads at a football game contains 5600 m3 of helium gas. If there are 1070 kg of gas at an absolute pressure of 1.10 105 Pa in the blimp, what is the temperature of the gas? The atomic mass of helium 4.00260 g/mol.
The questioned tackled determining the temperature of gas in a Goodyear blimp using the provided specifications and the ideal gas law: PV=nRT.
Explanation:
The question is about determining the temperature of the helium gas in a Goodyear blimp given certain parameters. Using the ideal gas law: PV=nRT we can calculate the temperature (T). In this scenario, the volume (V) is 5600 m3, the pressure (P) is 1.10x105 Pa and the number of moles (n) can be calculated by dividing the total mass of helium gas by its molar mass. By rearranging the gas law formula, the temperature could be calculated as T=(PV)/(nR), where R is the gas constant.
Following these steps, we can substitute the known values into the equation to solve for the temperature inside the Goodyear blimp. This problem involves concepts of physics, specifically gas laws, and would be covered typically in a college-level physics course.
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The Lewis structure for CO2 has a central The Lewis structure for C O 2 has a central blank atom attached to blank atoms.
1. atom attached to The Lewis structure for C O 2 has a central blank atom attached to blank atoms. atoms.
2. These atoms are held together by These atoms are held together by blank bonds. bonds.
3. Carbon dioxide has a Carbon dioxide has a blank electron geometry. electron geometry.
4. The carbon atom is The carbon atom is blank hybridized. hybridized.
5. Carbon dioxide has two Carbon dioxide has two blankπ bonds and two blankσ bonds.π bonds and two Carbon dioxide has two blankπ bonds and two blankσ bonds.σ bonds.
Answer:
See the explanation
Explanation:
1) The Lewis structure for [tex]CO_2[/tex] has a central Carbon atom attached to Oxygen atoms.
In the [tex]CO_2[/tex] we will have a structure: O=C=O the central atom "carbon" we will have 2 sigma bonds and 2 pi bonds, therefore, we have an Sp hybridization. For O we have 1 pi and 1 sigma bond, therefore, we have an Sp2 hybridization.
2) These atoms are held together by double bonds.
Again in the structure of [tex]CO_2[/tex]: O=C=O we only have double bonds.
3. Carbon dioxide has a Carbon dioxide has a Linear electron geometry.
Due to the double bonds we have to have a linear structure because in this geometry the atoms will be further apart from each other.
4. The carbon atom is Sp hybridized.
We will have for carbon 2 pi bonds, so we will have an Sp hybridization.
5. Carbon dioxide has two Carbon dioxide has two C(p) - O(p) π bonds and two C(sp) - O(Sp2) σ bonds.
(See figures)
Figure 1: Carbon hybridization
Figure 2: Oxygen hybridization
The Lewis structure of CO₂ has a central carbon atom with double bonds to two oxygen atoms, resulting in a linear geometry and sp hybridization. Each double bond contains one pi bond and one sigma bond.
Explanation:The Lewis structure for CO₂ features a central carbon atom attached to two oxygen atoms. These atoms are held together by double bonds. The electron geometry of carbon dioxide is linear, which is determined by the Valence Shell Electron Pair Repulsion (VSEPR) theory.
The carbon atom in CO₂ is sp hybridized. In a molecule of carbon dioxide, there are two π (pi) bonds and two σ (sigma) bonds. Each double bond consists of one σ bond and one π bond, contributing to the overall structure and stability of the molecule.
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The natural tendence of an object to resist change in its state of motion is:
friction
force
inertia
gravity
Answer:
inertia
Explanation:
The principal of inertia explains why objects do not voluntarily change their speed or direction without the influence of an external force. Inertia is not unaffected by other forces in planetary systems such as Earth and is, as such, not really observable on planetary surfaces. This is due to the resistance forces of gravity and friction from both the air and the ground. The drag slows objects down and makes it appear as if a constant force were necessary to keep them in motion.
Answer:
c
Explanation:
Glycerin (C3H8O3) is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 170 g of glycerin to 337 mL of H2O at 40.0°C?
The vapor pressure of pure water at 40.0°C is 55.32 torr and its density is 0.992 g/cm3.
Answer:
mass water = 347 x 0.992 = 344.2 g
moles water = 344.2 g / 18.02 g/mol= 19.1
moles C3H8O3 = 164 g / 92.097 g/mol= 1.78
mole fraction water = 19.1 / 19.1 + 1.78 = 0.915
p = 0.915 x 54.74 = 50.07 torr
Explanation:
Methane burns in oxygen to yield carbon dioxide and water. The chemical formula for this reaction is CH4+2O2→CO2+2H2O. What is the product, or what are the products, of this reaction?
a. water and methane
b. carbon dioxide and water
c. oxygen and methane
d. carbon dioxide and methane
Answer: the product of this reaction is option B which is carbondioxide and water.
Explanation:A chemical reaction involves the process in which one or more substances, the reactants, are converted to one or more different substances, the products. In the equation above, methane and oxygen gas are the reactants while carbondioxide and water are d products.
Answer:B
Explanation:
co 2 is the same as carbon dioxide and H 20 stand for water.
Liquid hydrogen peroxide, an oxidizing agent in many rocket fuel mixtures, releases oxygen gas on decomposition.
2 H2O2(l) 2 H2O(l) + O2(g) Hrxn = -196.1 kJ
How much heat is released when 765 kg H2O2 decomposes?
Answer:
4.3 x 10⁶ kJ are released
Explanation:
From the balanced chemical equation we know that when 2 moles of hydrogen peroxide decompose, 191 kJ of heat are released. So what we need to calculate is the number of moles the 765 kg of H2O2 represent and calculate the heat released.
Molar Mass H2O2 = 34.01 g/mol
Mass H2O2 = 765 Kg x 1000 g/Kg = 765000 g
moles H2O2 = 1 Mol / 34.01 g x 765000 grams = 22.5 Mol
(-196.1 kJ/ 2 mol H2O2 ) x 22.5 mol H2O2 = -4.3 x 10⁶ kJ
A chemistry student needs 65.0g of diethylamine for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student discovers that the density of diethylamine is ·0.706gcm−3. Calculate the volume of diethylamine the student should pour out.
Final answer:
To find the volume of diethylamine needed for an experiment, dividing the required mass (65.0 g) by the density of diethylamine (0.706 g/cm3) calculates a volume of 92.1 mL that the student should measure out.
Explanation:
To calculate the volume of diethylamine that a student needs for an experiment, use the formula that relates mass, density, and volume: volume = mass / density. Here, the mass required is 65.0 g, and the density of diethylamine is given as 0.706 g/cm3.
Substitute the given values into the formula:
Volume = 65.0 g / 0.706 g/cm3 = 92.1 cm3 or 92.1 mL, since 1 cm3 = 1 mL.
Therefore, the student should pour out 92.1 mL of diethylamine for their experiment.