Answer:
[tex]\triangle G= -6.7 KJ/mol[/tex]
Explanation:
From the question we are told that:
Chemical Reactions:
X=A⇌B,ΔG= 14.8 kJ/mol
Y=B⇌C,ΔG= -29.7 kJ/mol
Z=C⇌D,ΔG= 8.10 kJ/mol
Since
Hess Law
The law states that the total enthalpy change during the complete course of a chemical reaction is independent of the number of steps taken.
Therefore
Generally the equation for the Reaction is mathematically given by
[tex]T = +1 * X +1 * Y +1 *Z[/tex]
Therefore the free energy, ΔG is
[tex]\triangle G=1 * \triangle G*X +1 * \triangle G*Y +1 * \triangle G *Z[/tex]
[tex]\triangle G= +1 * (14.9) +1 * (-29.7) +1 * (8.10)[/tex]
[tex]\triangle G= -6.7 KJ/mol[/tex]
g A high altitude balloon is filled with 1.41 x 104 L of hydrogen gas (H2) at a temperature of 21oC and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km, where the temperature is -48oC and the pressure is 63.1 torr
Answer:
1.27 × 10⁵ L
Explanation:
Step 1: Given data
Initial pressure (P₁): 745 TorrInitial volume (V₁): 1.41 × 10⁴ LInital temperature (T₁): 21 °CFinal pressure (P₂): 63.1 TorrFinal volume (V₂): ?Final temperature (T₂): -48 °CStep 2: Convert the temperatures to the Kelvin scale
We will use the following expression.
K = °C + 273.15
K = 21 °C + 273.15 = 294 K
K = -48 °C + 273.15 = 225 K
Step 3: Calculate the final volume of the balloon
We will use the combined gas law.
P₁ × V₁ / T₁ = P₂ × V₂ / T₂
V₂ = P₁ × V₁ × T₂/ T₁ × P₂
V₂ = 745 Torr × 1.41 × 10⁴ L × 225 K/ 294 K × 63.1 Torr
V₂ = 1.27 × 10⁵ L
Using the periodic table, choose the more reactive nonmetal.
Br or As
CAN HF USED TO CLEAVE ETHERS EXPLAIN
Answer:
no
Explanation:
Fluoride is not nucleophilic (having the tendency to donate electrons) enough to allow for the use of HF to cleave ethers in protic media(protic solvents are polar liquid compounds that have dissociable hydrogen atoms). The rate of reaction is comparably low, so that heating of the reaction mixture is required.
Sodium peroxide (Na2O2) is often used in self-contained breathing devices, such as those used in fire emergencies, because it reacts with exhaled CO2 to form Na2CO3 and O2. How many liters of respired air can react with 96.7 g of Na2O2 if each liter of respired air contains 0.0755 g of CO2
Answer:
725.15 L
Explanation:
The balanced chemical equation for the reaction between Na₂O₂ and CO₂ is the following:
Na₂O₂ + CO₂ → Na₂CO₃ + 1/2 O₂
From the stoichiometry of the reaction, 1 mol of Na₂O₂ reacts with 1 mol CO₂. So, the stoichiometric ratio is 1 mol Na₂O₂/1 CO₂.
Now, we convert the mass of reactants to moles by using the molecular weight (Mw) of each compound:
Mw (Na₂O₂) = (23 g/mol x 2) + (16 g/mol x 2) = 78 g/mol
moles Na₂O₂ = 96.7 g/(78 g/mol) = 1.24 mol Na₂O₂
Mw (CO₂) = 12 g/mol + (16 g/mol x 2) = 44 g/mol
moles CO₂ = 0.0755 g/(44 g/mol) = 1.71 x 10⁻³ mol CO₂
Now, we calculate the number of moles of CO₂ we need to completely react with the mass of Na₂O₂ we have:
1.24 mol Na₂O₂ x 1 mol CO₂/1 mol Na₂O₂ = 1.24 mol CO₂
In 1 L of respired air we have 1.71 x 10⁻³ mol CO₂ (0.0755 g), so we need the following number of liters to have 1.24 mol of CO₂:
1.24 mol CO₂ x 1 L/1.71 x 10⁻³ mol CO₂ = 725.15 L
g A gas is compressed in a cylinder from a volume of 20.0 L to 2.0 L by a constant pressure of 10.0 atm. Calculate the amount of work done on the system in Joules. (1 L.atm
Answer:
The amount of work done on the system is 18234 J and the final positive sign means that this work corresponds to an increase in internal energy of the gas.
Explanation:
Thermodynamic work is called the transfer of energy between the system and the environment by methods that do not depend on the difference in temperatures between the two. When a system is compressed or expanded, a thermodynamic work is produced which is called pressure-volume work (p - v).
The pressure-volume work done by a system that compresses or expands at constant pressure is given by the expression:
W system= -p*∆V
Where:
W system: Work exchanged by the system with the environment. Its unit of measure in the International System is the joule (J) p: Pressure. Its unit of measurement in the International System is the pascal (Pa) ∆V: Volume variation (∆V = Vf - Vi). Its unit of measurement in the International System is cubic meter (m³)In this case:
p= 10 atm= 1.013*10⁶ Pa (being 1 atm= 101325 Pa)ΔV= 2 L- 20 L= -18 L= -0.018 m³ (being 1 L=0.001 m³)Replacing:
W system= -1.013*10⁶ Pa* (-0.018 m³)
Solving:
W system= 18234 J
The amount of work done on the system is 18234 J and the final positive sign means that this work corresponds to an increase in internal energy of the gas.
If 0.650 mL of benzaldehyde reacts with enough of the Grignard reagent, calculate the theoretical yield (in grams) of the alcohol product. Show calculation with units for full credit.
Answer:
Theoretical yield of 1-Phenyl-1-propanol in grams = 0.871 g
Explanation:
A Grignard reagent is any of the numerous organic derivatives of magnesium (Mg) which are commonly represented by the general formula RMgX (where R is a hydrocarbon radical e.g. methyl, ethyl, propyl, etc.; and X is a halogen atom, e.g. chlorine, bromine, or iodine)
A Grignard reaction converts an aldehyde to a secondary alcohol. In the grignard reaction involving benzaldehyde as in this experiment, the grignard reagent used is ethyl magnesium bromide, EtMgBr, and the resulting product is 1-Phenyl-1-propanol, a secondary alcohol. The reaction is shown in the figure attached below.
Mass of benzaldehyde in 0.650 mL = density × volume
Density of Benzaldehyde = 1.044 g/mL
Mass of benzaldehyde = 1.044 g/mL × 0.650 mL = 0.6786 g
Molar mass of benzaldehyde = 106 g/mol
Molar mass of 1-Phenyl-1-propanol = 136 g/mol
Mass of = mass of benzaldehyde × mole ratio of 1-Phenyl-1-propanol and benzaldehyde
Mass of 1-Phenyl-1-propanol = 0.6786 g × (136 g/mol)/(106 g/mol) = 0.871 g
Therefore, the theoretical yield of 1-Phenyl-1-propanol in grams = 0.871 g
why beta carbon hydrogen is easily replaceable but not alpha carbon hydrogen
Answer:
Four common types of reactions involving carbonyl reactions: 1) nucleophilic addition; 2) nucleophilic acyl substitution; 3) alpha substitution; 4) carbonyl condensations. The first two were previously discussed and the second two involve the properties of the carbon directly adjacent to the carbonyls, α carbons.
Alpha-substitution reactions results in the replacement of an H attached to the alpha carbon with an electrophile.
The nucleophile in these reactions are new and called enols and enolates.
Explanation:
The carbon in the carbonyl is the reference point and the alpha carbon is adjacent to the carbonyl carbon.
Hydrogen atoms attached the these carbons denoted with Greek letters will have the same designation, so an alpha hydrogen is attached to an alpha carbon.
Aldehyde hydrogens not given Greek leters.
α hydrogens display unusual acidity, due to the resonance stabilization of the carbanion conjugate base, called an enolate.
Tautomers are readily interconverted constitutional isomers, usually distinguished by a different location for an atom or a group, which is different than resonance.
The tautomerization in this chapter focuses on the carbonyl group with alpha hydrogen, which undergo keto-enol tautomerism.
Keto refers to the tautomer containing the carbonyl while enol implies a double bond and a hydroxyl group present in the tautomer.
The keto-enol tautomerization equilibrium is dependent on stabilization factors of both the keto tautomer and the enol tautomer, though the keto form is typically favored for simple carbonyl compounds.
The 1,3 arrangement of two carbonyl groups can work synergistically to stabilize the enol tautomer, increasing the amount present at equilibrium.
The positioning of the carbonyl groups in the 1,3 arrangement allows for the formation of a stabilizing intramolecular hydrogen bond between the hydroxyl group of the enol and the carbonyl oxygen as well as the alkene group of the enol tautomer is also conjugated with the carbonyl double bond which provides additional stabilization.
Aromaticity can also stabilize the enol tautomer over the keto tautomer.
Under neutral conditions, the tautomerization is slow, but both acid and base catalysts can be utilized to speed the reaction up.
Biological enol forming reactions use isomerase enzymes to catalyze the shifting of a carbonyl group in sugar molecules, often converting between a ketose and an aldose in a process called carbonyl isomerization.
A solution contains only sucrose and water. If the mole fraction of sucrose is 0.0558, determine the molality of sucrose.
Answer:
The correct solution is "3.28 m".
Explanation:
According to the question,
Mol fraction of solvent,
= 0.0558
Molar mass of water,
= 18 g/mol
Mol of H₂O in 1000 g water,
= 55.55 mol
Now,
Let the mol of solute will be "x mol".
Total mol in solution will be "55.55 + x".
As we know,
⇒ The mol fraction of solvent = [tex]\frac{x}{55.55+x}[/tex]
[tex]0.0558=\frac{x}{55.55+x}[/tex]
[tex]x=0.0558[55.55+x][/tex]
[tex]x=3.09969+0.0558x[/tex]
[tex]x-0.0558x=3.09969[/tex]
[tex]x=\frac{3.09969}{0.9442}[/tex]
[tex]=3.38 \ m[/tex]
Draw the structure of the neutral product formed in the reaction of dimethyl malonate and methyl vinyl ketone.
Answer:
Explanation:
The reaction between dimethyl malonate which is an active methylene group with an (∝, β-unsaturated carbonyl compound) i.e methyl vinyl ketone is known as a Micheal Addition reaction. The reaction mechanism starts with the base attack on the β-carbon to remove the acidic ∝-hydrogens and form a carbanion. The carbanion formed(enolate ion) attacks the methyl vinyl ketone(i.e. a nucleophilic attack at the β-carbon) to give a Micheal addition product, this is followed by the protonation to give the neutral product.
How much of a 24-gram sample of Radium-226 will remain unchanged at the end of three half-life periods?
Answer:
The right answer is "3 g".
Explanation:
Given:
Initial mass substance,
[tex]M_0=24 \ g[/tex]
By using the relation between half lives and amount of substances will be:
⇒ [tex]M=\frac{M_0}{2^n}[/tex]
[tex]=\frac{24}{2^3}[/tex]
[tex]=3 \ g[/tex]
Thus, the above is the correct answer.
A 0.245-L flask contains 0.467 mol co2 at 159 °c. Calculate the pressure using the ideal gas law.
Answer:
Pressure, P = 67.57 atm
Explanation:
Given the following data;
Volume = 0.245 LNumber of moles = 0.467 molesTemperature = 159°CIdeal gas constant, R = 0.08206 L·atm/mol·KConversion:
We would convert the value of the temperature in Celsius to Kelvin.
T = 273 + °C
T = 273 + 159
T = 432 Kelvin
To find the pressure of the gas, we would use the ideal gas law;
PV = nRT
Where;
P is the pressure.V is the volume.n is the number of moles of substance.R is the ideal gas constant.T is the temperature.Making P the subject of formula, we have;
[tex] P = \frac {nRT}{V} [/tex]
Substituting into the formula, we have;
[tex] P = \frac {0.467*0.08206*432}{0.245} [/tex]
[tex] P = \frac {16.5551}{0.245} [/tex]
Pressure, P = 67.57 atm
A compound contains only carbon, hydrogen, and oxygen. Combustion of 65.76 g of the compound yields 96.38 g of CO2 and 39.46 g of H2O.
The molar mass of the compound is 90.078 g/mol.
1. Calculate the grams of carbon (C) in 65.76 g of the compound:
2. Calculate the grams of hydrogen (H) in 65.76 g of the compound.
3. Calculate the grams of oxygen (O) in 65.76 g of the compound.
Answer:
1. 26.30 g C.
2. 4.42 g H.
3. 35.04 g O.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the required as follows:
1. Here, the only source of carbon is in CO2, and thus, we calculate the grams of carbon from the produced grams of this substance:
[tex]m_C=96.38gCO_2*\frac{1molCO_2}{44.01gCO_2} *\frac{1molC}{1molO_2} *\frac{12.01gC}{1molC} =26.30g[/tex]
2. Here, the only source of hydrogen is in H2O, and thus, we calculate the grams of hydrogen from the produced grams of this substance:
[tex]m_H=39.46gH_2O*\frac{1molH_2O}{18.02gH_2O} *\frac{2molH}{1molH_2O} *\frac{1.01gH}{1molH} =4.42gH[/tex]
3. Here, we subtract the mass of H and C from the mass of the sample, to obtain the mass of oxygen:
[tex]m_O=65.76g-26.30g-4.42g\\\\m_O=35.04g[/tex]
Regards!
How to solve this problem step by step
Answer:
[tex]V_2= 736mL[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to solve this problem by using the combined gas law:
[tex]\frac{P_2V_2}{T_2}= \frac{P_1V_1}{T_1}[/tex]
Thus, we solve for the final volume by solving for V2 as follows:
[tex]V_2= \frac{P_1V_1T_2}{T_1P_2}[/tex]
Now, we plug in the variables to obtain the result in milliliters and making sure we have both temperatures in Kelvins:
[tex]V_2= \frac{1.20atm*735mL*279K}{(112+273)K*660/760atm}\\\\V_2= \frac{1.20atm*735mL*279K}{(112+273)K*660/760atm}=736mL[/tex]
Regards!
Chemical reactions can exhibit different rate constants at differing: Select the correct answer below: initial concentrations volumes of container temperatures none of the above
Explanation:
Chemical reactions can exhibit different rate constants at differing:
i)initial concentrations
ii)volumes of container
iii) temperatures
iv)none of the above.
The rate constant of a reaction is constant at a particular temperature.
It is not depending on the initial concentration of the reactants. It varies with temperature.
Thus, among the given options the correct answer is Temperatures.
Chemical reactions can exhibit different rate constant at different temperatures. Hence, the correct option is temperature.
What is a rate constant?
The rate constant is the proportionality constant in the equation that expresses the relationship between the rate of a chemical reaction and the concentrations of the reacting substances.
Chemical reactions proceed at vastly different speeds depending on the nature of the reacting substances, the type of chemical transformation, the temperature, etc.
For a given reaction, the speed of the reaction will vary with the temperature, the pressure, and the amounts of reactants present.
The rate constant goes on increasing as the temperature goes up, but the rate of increase falls off quite rapidly at higher temperatures.
On the other hand, the volume of the container, initial concentration does not affect the rate constant.
Therefore, Chemical reactions can exhibit different rate constant at different temperatures. Hence, the correct option is temperature.
To learn more about rate constant, click here:
https://brainly.com/question/24658842
what is irrigation?Describe two methods of irrigation which conserve water
The process of watering the crops is called irrigation.
Any two methods of irrigation are:
(i) Sprinkler system:This system is used on the uneven land where less water is available. The perpendicular pipes, having rotating nozzles on top, are joined to the main pipeline at regular intervals. Water is allowed to flow through main pipe under pressure, which escapes from the rotating nozzles. In this way water gets sprinkled on the crop.
(ii) Drip irrigation:This system is used to save water as it allows the water to flow drop by drop at the roots of the plants. It is the best technique for watering fruit plants, gardens and trees. Water is not wasted at all.
An electron moved from shell n = 2 to shell n = 1. What most likely happened during the transition?
When electron goes from a higher shell to lower shell then it loses energy .
So, when an electron moved from shell n = 2 to shell n = 1 then a photon of energy is released.
Postlab Questions (2 pts ea; 8 pts total be specific and answer all in complete sentences): 1. How would you distinguish benzaldehyde and acetophenone by the results of their qualitative tests
Answer:
Using the Iodoform test, we can differentiate both compounds.
Explanation:
Benzaldehyde (C6H5CHO -an aldehyde) and Acetophenone (C6H5COCH3 - a methyl ketone) can be differentiated by reacting both compounds with iodine in a basic (NaOH) solutions.
The methyl ketone (acetophenone) gives a pale yellow precipitate of triiodomethane (iodoform) while the aldehyde (benzaldehyde) would not react.
This is known as the IODOFORM test and is indicative for methyl ketones
-300g de acido comercial se disuelve en agua destilada contenidos en un cono, cuyo radio es de 0.005Km y 300cm de altura, si la densidades de 1.2g/m3 ¿Cuál es la concentración expresada en %m/m?
Answer:
[tex]\%m/m=76.1\%[/tex]
Explanation:
¡Hola!
En este caso, considerando la información dada, entendemos que se haría primero necesario calcular el volumen del cono en metros cúbicos, teniendo en cuenta que 0.005 km son 5 m y 300 cm son 3 m:
[tex]V=\frac{1}{3} \pi r^2h\\\\V=\frac{1}{3} \pi *(5m)^2(3m)=78.5m^3\\\\[/tex]
Ahora, convertimos esta cantidad a gramos por medio de la densidad para conocer la masa de la solución:
[tex]m_{sol}=78.5m^3*\frac{1.2g}{1m^3} =94.2g[/tex]
Finalmente, aplicamos la definición de %m/m para obtener:
[tex]\%m/m=\frac{300g}{300g+94.2g}*100\%\\\\ \%m/m=76.1\%[/tex]
¡Saludos!
Consider the following equation for the combustion of acetone (C3H6O), the main ingredient in nail polish remover.
C3H6O(l) + 4O2(g) → 3CO2(g) + 3H2O(g) ΔHrxn = −1790kJ
If a bottle of nail polish remover contains 143 g of acetone, how much heat would be released by its complete combustion? Express your answer to three significant figures.
Molar mass of Acetone
C3H6O3(12)+6+1658g/molNow
1 mol releases -1790KJ heat .Moles of Acetone:-
143/58=2.5molAmount of heat:-
2.5(-1790)=-4475kJWrite the functional isomers of C2H6O?
Answer:
See explanation
Explanation:
Isomers are molecules which have the same molecular formula but different structural formulas. Sometimes, isomers may even contain different functional groups.
The formula C2H6O may refer to an ether or an alcohol. The compound could be CH3CH2OH(ethanol) or CH3OCH3(methoxymethane).
Hence, the functional isomers of the formula C2H6O are ethanol and methoxymethane.
Using the balanced equation below
Answer:
27.99 g of CsF
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
CsF + XeF₆ —> CsXeF₇
Next, we shall determine the mass of CsF that reacted and the mass of CsXeF₇ produced from the balanced equation. This can be obtained as follow:
Molar mass of CsF = 133 + 19
= 152 g/mol
Mass of CsF from the balanced equation = 1 × 152 = 152 g
Molar mass of CsXeF₇ = 133 + 131 + (19× 7)
= 133 + 131 + 133
= 397 g/mol
Mass of CsXeF₇ from the balanced equation = 1 × 397 = 397 g
SUMMARY:
From the balanced equation above,
152 g of CsF reacted to produce 397 g of CsXeF₇.
Finally, we shall determine the mass of CsF required to produce 73.1 g of CsXeF₇. This can be obtained as follow:
From the balanced equation above,
152 g of CsF reacted to produce 397 g of CsXeF₇.
Therefore, Xg of CsF will react to produce 73.1 g of CsXeF₇ i.e
Xg of CsF = (152 × 73.1)/397
Xg of CsF = 27.99 g
Thus, 27.99 g of CsF is required for the reaction.
Biphenyl, C12H10, is a nonvolatile, nonionizing solute that is soluble in benzene, C6H6.
At 25 ∘C, the vapor pressure of pure benzene is 100.84 Torr. What is the vapor pressure of a solution made from dissolving 10.9 g of biphenyl in 25.3 g of benzene?
Answer:
The correct answer is - 83. torr
Explanation:
We know that:
p = X solvent * P pure solvent,
where, X solvent = number of moles of solvent / total number of moles.
in this solution, solute is 10.9 g of C12 H10 and the solvent is 25.3 g C6H6.
molar mass of C6H6 = 6 * 12 g/mol + 6 * 1g/mol = 78 g/mol ( => moles of solvent = mass in grams / molar mass)
moles of solvent = 25.3 g / 78 g/mol = 0.3243 mol
molar mass of C12H10 = 12 * 12g/mol + 10*1g/mol = 154 g/mol
moles of solute = 10.9 g / 154 g/mol = 0.07077 mol
=> X solvent = 0.3243 / (0.3243 + 0.07077) = 0.8208
=> p = 0.8208 * 100.84 torr = 82.77 torr ≈ 83.0 torr
Answer: 83.0 torr
3. After 7.9 grams of sodium are dropped into a bathtub full of water, how many grams of hydrogen gas are released?
4. Excess oxygen gas is added to 34.5 grams of aluminum and heated under pressure. How many grams of aluminum oxide are produced?
Please explain as well if possible!
Answer:
3) About 0.35 grams of hydrogen gas.
4) About 65.2 grams of aluminum oxide.
Explanation:
Question 3)
We are given that 7.9 grams of sodium is dropped into a bathtub of water, and we want to determine how many grams of hydrogen gas is released.
Since sodium is higher than hydrogen on the activity series, sodium will replace hydrogen in a single-replacement reaction for sodium oxide. Hence, our equation is:
[tex]\displaystyle \text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}[/tex]
To balance it, we can simply add another sodium atom on the left. Hence:
[tex]\displaystyle 2\text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}[/tex]
To convert from grams of sodium to grams of hydrogen gas, we can convert from sodium to moles of sodium, use the mole ratios to find moles in hydrogen gas, and then use hydrogen's molar mass to find its amount in grams.
The molar mass of sodium is 22.990 g/mol. Hence:
[tex]\displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}[/tex]
From the chemical equation, we can see that two moles of sodium produce one mole of hydrogen gas. Hence:
[tex]\displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}[/tex]
And the molar mass of hydrogen gas is 2.016 g/mol. Hence:
[tex]\displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}[/tex]
Given the initial value and the above ratios, this yields:
[tex]\displaystyle 7.9\text{ g Na}\cdot \displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}\cdot \displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}[/tex]
Cancel like units:
[tex]=\displaystyle 7.9\cdot \displaystyle \frac{1}{22.990}\cdot \displaystyle \frac{1}{2}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1}[/tex]
Multiply. Hence:
[tex]=0.3463...\text{ g H$_2$}[/tex]
Since we should have two significant values:
[tex]=0.35\text{ g H$_2$}[/tex]
So, about 0.35 grams of hydrogen gas will be released.
Question 4)
Excess oxygen gas is added to 34.5 grams of aluminum and produces aluminum oxide. Hence, our chemical equation is:
[tex]\displaystyle \text{O$_2$} + \text{Al} \rightarrow \text{Al$_2$O$_3$}[/tex]
To balance this, we can place a three in front of the oxygen, four in front of aluminum, and two in front of aluminum oxide. Hence:
[tex]\displaystyle3\text{O$_2$} + 4\text{Al} \rightarrow 2\text{Al$_2$O$_3$}[/tex]
To convert from grams of aluminum to grams of aluminum oxide, we can convert aluminum to moles, use the mole ratios to find the moles of aluminum oxide, and then use its molar mass to determine the amount of grams.
The molar mass of aluminum is 26.982 g/mol. Thus:
[tex]\displaystyle \frac{1\text{ mol Al}}{26.982 \text{ g Al}}[/tex]
According to the equation, four moles of aluminum produces two moles of aluminum oxide. Hence:
[tex]\displaystyle \frac{2\text{ mol Al$_2$O$_3$}}{4\text{ mol Al}}[/tex]
And the molar mass of aluminum oxide is 101.961 g/mol. Hence: [tex]\displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1\text{ mol Al$_2$O$_3$}}[/tex]
Using the given value and the above ratios, we acquire:
[tex]\displaystyle 34.5\text{ g Al}\cdot \displaystyle \frac{1\text{ mol Al}}{26.982 \text{ g Al}}\cdot \displaystyle \frac{2\text{ mol Al$_2$O$_3$}}{4\text{ mol Al}}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1\text{ mol Al$_2$O$_3$}}[/tex]
Cancel like units:
[tex]\displaystyle= \displaystyle 34.5\cdot \displaystyle \frac{1}{26.982}\cdot \displaystyle \frac{2}{4}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1}[/tex]
Multiply:
[tex]\displaystyle = 65.1852... \text{ g Al$_2$O$_3$}[/tex]
Since the resulting value should have three significant figures:
[tex]\displaystyle = 65.2 \text{ g Al$_2$O$_3$}[/tex]
So, approximately 65.2 grams of aluminum oxide is produced.
Answer:
Solution given:
3.
[tex]2Na+H_2O→Na _2O+H_2[/tex]
2Na=2*23g.
2O=18g.
[tex]Na_2O[/tex]=62g
[tex]H_2[/tex]=2 g
we have
2*23g of Na produce 2g of [tex]H_2[/tex]
Now
7.9 g of Na produce 2*7.9/(2*23)
=0.34g of [tex]H_2[/tex]
:. 0.34g of [tex]H_2[/tex] is produced.4.
we have
[tex]3O_2+4Al→2Al_2O_3[/tex]
[tex]3O_2[/tex]=3*16g*2g
4Al=4*27g
[tex]2Al_2O[/tex]= 2*27*2g+2*16*3g
4*27g of Al produces 204g of [tex]Al_2O_3[/tex]
34.5g of Al produces 204g*34.5/(4*27)
=65.17g of [tex]Al_2O_3[/tex] is producedwhat are the major specials presents in a solution of a strong acid like HCl
Answer:
hydrogen ions (H⁺) and chloride ions (Cl⁻)
Explanation:
Hydrochloric acid (HCl) is a strong acid. That means that the compound dissociates completely into ions when is dissolved in water, as follows:
HCl → H⁺ + Cl⁻
The equilibrium is completely shifted to the right side (products). Thus, it is considered that the concentration of the non-dissociated compound (HCl) is negligible, and the major specials present in the solution are the hydrogen ions (H⁺) and chloride ions (Cl⁻).
4.If 15.00 mL of 3.00 M potassium iodide is needed to reach the equivalence point with 10.00 mL of lead (Il) nitrate, determine the molarity of the lead (Il) nitrate solution
Answer:
2.25 M
Explanation:
The reaction that takes place is:
2KI + Pb(NO₂)₃ → PbI₂ + 2KNO₃First we calculate how many potassium iodide moles reacted, using the given volume and concentration:
15.00 mL * 3.00 M = 45 mmol KIThen we convert 45 mmoles of KI into mmoles of Pb(NO₂)₃, using the stoichiometric coefficients of the balanced reaction:
45 mmol KI * [tex]\frac{1mmolPb(NO_3)_2}{2mmolKI}[/tex] = 22.5 mmol Pb(NO₂)₃Finally we calculate the molarity of the Pb(NO₂)₃ solution, using the calculated number of moles and given volume:
22.5 mmol Pb(NO₂)₃ / 10.00 mL = 2.25 MIdentify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction.
half-reaction identification
Cu+(aq)--->Cu2+(aq) + e- _________
I2(s) + 2e--->2I-(aq) _________
Answer:
Cu+(aq)--->Cu2+(aq) + e- : oxidation
reason: there is loss of electrons.
I2(s) + 2e--->2I-(aq) : reduction
reason: There is reduction of electrons.
Carbon monoxide and chlorine gas react to produce phosgene (COCl2) gas according to the following reaction at 100.0C CO(g) + Cl2 (g) <-> COCl2 (g) Kp = 1.49x10^8
In an equilibrium mixture of the three gases, PCO = PCl2 = 2.55 x 10^-4 what is the equilibrium partial pressure (in atm) of Phosgene?
Express your answer below in decimal form (you will not be able to use scientific notation) to THREE significant figures. If your answer is negative include the sign.
Answer:
0.000000000000000969 atm
Explanation:
Based on the reaction:
CO(g) + Cl2(g) ⇄ COCl2(g)
Where equilibrium constant, Kp, is defined as:
Kp = 1.49x10⁻⁸ = PCOCl2 / PCO * PCl2
As PCO = PCl2 = 2.55x10⁻⁴atm:
1.49x10⁻⁸ = PCOCl2 / PCO * PCl2
1.49x10⁻⁸ = PCOCl2 / 2.55x10⁻⁴atm * 2.55x10⁻⁴atm
9,69x10⁻¹⁶atm = PCOCl2
In decimal form:
0.000000000000000969 atmhelp plsssssssssssss
Answer:
The rate of the forward reaction equals the rate of the reverse reaction.
Explanation:
To know which option is correct, it important that we know what dynamic equilibrium is all about.
We'll begin by defining chemical equilibrium.
A chemical system is said to be in chemical equilibrium when there is no observable change in the properperties of the chemical system.
Dynamic equilibrium on the other hand can be defined as a state in which the forward and backward reaction is occurring at the same time. Thus, in dynamic equilibrium, the rate of the forward reaction is equal to the rate of the backward (i.e reverse) reaction.
In aqueous solution the Hg2+ ion forms a complex with four iodide anions. Write the formation constant expression for the equilibrium between the hydrated metal ion and the aqueous complex. Under that, write the balanced chemical equation for the first step in the formation of the complex.
Answer:
Following are the solution to the given question:
Explanation:
In the aqueous solution of [tex]Hg^{2+}[/tex] ion is used to represent a [tex][Hg(H_2O)_4]^{2+}[/tex].
It is used to form a complex with iodide ions as [tex][Hgl_4]^{2-}[/tex].
Conversion reaction:
[tex][Hg(H_2O)_4]^{2+} +4I^{-} \to [HgI_4]^{2-}+4H_20[/tex]
Following are the formation constants:
[tex]k_f=\frac{[HgI_4]^{2-}}{[Hg(H_2O)_4]^{2+} [I^{-}]^4}[/tex]
Following are the first step to the conversion:
[tex][Hg(H_2O)_4]^{2+} +I^{-} \to [HgI(H_2O)_3]^{+}+H_20[/tex]
What will be the equilibrium temperature when a 245 g block of lead at 300oC is placed in 150-g aluminum container containing 820 g of water at 12.0oC?
Answer:
The correct approach is "12.25°C".
Explanation:
Given:
Mass of lead,
mc = 245 g
Initial temperature,
tc = 300°C
Mass of Aluminum,
ma = 150 g
Initial temperature,
ta = 12.0°C
Mass of water,
mw = 820 g
Initial temperature,
tw = 12.0°C
Now,
The heat received in equivalent to heat given by copper.
The quantity of heat = [tex]m\times s\times t \ J[/tex]
then,
⇒ [tex]245\times .013\times (300-T) = 150\times .9\times (T-12.0) + 820\times 4.2\times (T-12.0)[/tex]
⇒ [tex]3.185(300-T) = 135(T-12.0) + 3444(T-12.0)[/tex]
⇒ [tex]955.5-3.185T=135T-1620+3444T-41328[/tex]
⇒ [tex]43903.5 = 3582.185 T[/tex]
⇒ [tex]T = 12.25^{\circ} C[/tex]
