Answer:
W = - 500 KJ
∴ the work is done on the system
Explanation:
isothermal system:
∴ ΔU = 0; ⇒ Q = W
∴ W = P1V1 -P2V2
⇒ W = ((100KPa)*(25m³)) - ((300KPa)*(10m³))
⇒ W = 2500KPa.m³ - 3000KPa,m³
⇒ W = - 500 KPa.m³ = - 500 KJ
∴ W (-) the work is done on the system
A pellet of Zn of mass 10.0g is dropped into a flaskcontaining
dilute H2SO4 at a pressure of P=1.00 bar and
temperature of 298K. What is the reaction thatoccures? Calculate w
for the process.
Answer: Work done for the process is -390 J
Explanation:
The chemical equation for the reaction of zinc metal with sulfuric acid follows:
[tex]Zn+H_2SO_4\rightarrow ZnSO_4+H_2[/tex]
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of zinc = 10.0 g
Molar mass of zinc = 65.38 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of zinc}=\frac{10.0g}{65.38g/mol}=0.153mol[/tex]
The equation given by ideal gas follows:
[tex]P\Delta V=nRT[/tex]
where, P = pressure of the gas
[tex]\Delta V[/tex] = Change in volume of the gas
T = Temperature of the gas = 298 K
R = Gas constant = 8.314 J/mol.K
n = number of moles of gas = 0.153 mol
Putting values in above equation, we get:
[tex]P\Delta V=0.153mol\times 8.314J/mol.K\times 298K\\\\P\Delta V=397J[/tex]
To calculate the work done, we use the equation:
[tex]\text{Work done}=-P\Delta V\\\\W=-390J[/tex]
Hence, work done for the process is -390 J
Assume that the NO, concentration in a house with a gas stove is 150 pg/m°. Calculate the equivalent concentration in ppm at STP.
Explanation:
It is known that for [tex]NO_{2}[/tex], ppm present in 1 [tex]mg/m^{3}[/tex] are as follows.
1 [tex]\frac{mg}{m^{3}}[/tex] = 0.494 ppm
So, 150 [tex]pg/m^{3}[/tex] = [tex]\frac{150}{1000} mg/m^{3}[/tex]
= 0.15 [tex]mg/m^{3}[/tex]
Therefore, calculate the equivalent concentration in ppm as follows.
[tex]0.15 \times 0.494 ppm[/tex]
= 0.074 ppm
Thus, we can conclude that the equivalent concentration in ppm at STP is 0.074 ppm.
Calculate the freezing temperature of the following solution of 0.50 M glucose (a covalent compound). Assume that the molality of the solution is 0.50 m. (The molar and molal concentrations of dilute aqueous solutions are often identical to two significant figures.) Enter your answer in the provided box. 0.50 m glucose (a covalent compound) °C
Answer:
-0.93 °C
Explanation:
Hello,
The freezing-point depression is given by:
[tex]T_f-T_f^*=-iK_{solvent}m_{solute}[/tex]
Whereas [tex]T_f[/tex] is the freezing temperature of the solution, [tex]T_f^*[/tex] is the freezing temperature of the pure solvent (0 °C since it is water), [tex]i[/tex] the Van't Hoff factor (1 since the solute is covalent), [tex]K_{f,solvent}[/tex] the solvent's freezing point depression point constant (in this case [tex]1.86 C\frac{kg}{mol}[/tex]) and [tex]m_{solute}[/tex] the molality of the glucose.
As long as the unknown is [tex]T_f[/tex], solving for it:
[tex]T_f=T_f^*-iK_fm\\T_f=0C-1*1.86C\frac{kg}{mol}*0.5\frac{mol}{kg} \\T_f=-0.93C[/tex]
Best regards.
The specific reaction rate of a reaction at 492k is 2.46 second inverse and at 528k 47.5 second inverse.find activation energy and ko.
Answer:
Ea = 177x10³ J/mol
ko = [tex]1.52x10^{19}[/tex] J/mol
Explanation:
The specific reaction rate can be calculated by Arrhenius equation:
[tex]k = koxe^{-Ea/RT}[/tex]
Where k0 is a constant, Ea is the activation energy, R is the gas constant, and T the temperature in Kelvin.
k depends on the temperature, so, we can divide the k of two different temperatures:
[tex]\frac{k1}{k2} = \frac{koxe^{-Ea/RT1}}{koxe^{-Ea/RT2}}[/tex]
[tex]\frac{k1}{k2} = e^{-Ea/RT1 + Ea/RT2}[/tex]
Applying natural logathim in both sides of the equations:
ln(k1/k2) = Ea/RT2 - Ea/RT1
ln(k1/k2) = (Ea/R)x(1/T2 - 1/T1)
R = 8.314 J/mol.K
ln(2.46/47.5) = (Ea/8.314)x(1/528 - 1/492)
ln(0.052) = (Ea/8.314)x(-1.38x[tex]10^{-4}[/tex]
-1.67x[tex]10^{-5}[/tex]xEa = -2.95
Ea = 177x10³ J/mol
To find ko, we just need to substitute Ea in one of the specific reaction rate equation:
[tex]k1 = koxe^{-Ea/RT1}[/tex]
[tex]2.46 = koxe^{-177x10^3/8.314x492}[/tex]
[tex]1.61x10^{-19}ko = 2.46[/tex]
ko = [tex]1.52x10^{19}[/tex] J/mol
Define "Triose", "Tetrose", "Pentose", and "Hexose" and identify which of these is/are most abundant
Answer:
Monosaccharides are the simplest form of carbohydrates that cannot be hydrolyzed to smaller compounds. Monosaccharides are the basic units of carbohydrates and are also known as simple sugars.
The monosaccharides are classified on the basis of number of carbon atoms present.
Triose is a type of monosaccharide molecule, which is composed of 3 carbon atoms.
Tetrose is a type of monosaccharide molecule, which is composed of 4 carbon atoms.
Pentose is a type of monosaccharide molecule, which is composed of 5 carbon atoms.
Hexose is a type of monosaccharide molecule, which is composed of 6 carbon atoms.
D-glucose is a hexose sugar and it is the most abundant monosaccharide in the nature.
If atom X has an atomic mass of 12, and atom Y has an atomic mass of 1, what would be the mass of 4.18 moles of the compound X7Y11, in grams? Please report your answer to the nearest whole gram.
Answer:
397 g
Explanation:
From the information given in the question ,
the atomic mass of X = 12 ,
and , the atomic mass of Y = 1 .
For the molecular formula , X₇Y₁₁ , the molecular mass is given as -
X₇Y₁₁ = 7 * atomic mass of X + 11 * atomic mass of Y
= 7 * 12 + 11 * 1
= 84 + 11 = 95 g / mol
Hence , molecular mass of X₇Y₁₁ = 95 g / mol .
Moles is denoted by given mass divided by the molecular mass ,
Hence ,
n = w / m
n = moles ,
w = given mass ,
m = molecular mass .
From question ,
moles n = 4.18 mol
and ,
m = molecular mass of X₇Y₁₁ = 95 g / mol .
To find the mass of the compound X₇Y₁₁ , the above formula can be used , and putting the respective values ,
n = w / m
4.18 = w / 95
w = 4.18 * 95 = 397 g
Dinitrogen monoxide gas is collected at-3.0 °C in an evacuated flask with a measured volume of 5.0 L. When all the gas has been collected, the press the flask is measured to be 0.100 atm . Calculate the mass and number of moles of dinitrogen monoxide gas that were collected. Round your answer to 2 significant digits. mass: mole: nol Explanation Check 2019 McGraw-Hill Education All Rights Reserved
Answer:
0.971 grams
Explanation:
Given:
Temperature = 3.0° C = 3 + 273 = 276 K
Volume, V = 5.0 L
Pressure, P = 0.100 atm
Now, from the relation
PV = nRT
where,
n is the number of moles,
R is the ideal gas constant = 0.082057 L atm/mol.K
thus,
0.1 × 5 = n × 0.082057 × 276
or
n = 0.022 moles
Also,
Molar mass of the Dinitrogen monoxide gas (N₂O)
= 2 × Molar mass of nitrogen + 1 × Molar mass of oxygen
= 2 × 14 + 16 = 44 grams/mol
Therefore, Mass of 0.022 moles of N₂O = 0.022 × 44 = 0.971 grams
AG' for the isomerization reaction glucose-1-phosphate (GIP) $ glucose-6-phosphate (G6P) is -7.1 kJ/mol. Calculate the equilibrium ratio of [G1P] to (G6P) at 25°C. Read this carefully to make sure you solve for the correct ratio! SHOW WORK! [G1P][G6P] =
Answer:
The ratio [G1P]/[G6P] = 5.7 . 10⁻².
Explanation:
Let us consider the reaction G1P ⇄ G6P, with ΔG° = -7.1 kJ/mol. According to Hess's Law, we can write the inverse reaction, and Gibbs free energy would have an opposite sign.
G6P ⇄ G1P ΔG° = 7.1 kJ/mol
This is the reaction for which we want to find the equilibrium constant (the equilibrium ratio of [G1P] to [G6P]):
[tex]Kc=\frac{[G1P]}{[G6P]}[/tex]
The equilibrium constant and Gibbs free energy are related by the following expression:
[tex]Kc=e^{-\Delta G\si{\textdegree}/R.T } } =e^{-7.1kJ/mol/8.314.10^{-3}kJ/mol.K.298K} } }=5.7.10^{-2}[/tex]
where,
R is the ideal gas constant (8.314 . 10⁻3 kJ/mol.K)
T is the absolute temperature (in kelvins)
Final answer:
To calculate the equilibrium ratio of [G1P] to [G6P], the Gibbs free energy equation is used with ΔG', universal gas constant R, and the temperature T substituted. The result is that the concentration of G6P is approximately 1.331 times that of G1P at equilibrium and at 25°C.
Explanation:
The student is asking about the equilibrium ratio of concentrations of glucose-1-phosphate ([G1P]) to glucose-6-phosphate ([G6P]) at 25°C when the standard free energy change (ΔG') for the isomerization reaction is given as -7.1 kJ/mol. To calculate this ratio, we can use the Gibbs free energy equation for the equilibrium constant (Keq):
ΔG' = -RT ln(Keq)
Where ΔG' is the standard free energy change, R is the universal gas constant (8.314 J/mol K), T is the temperature in Kelvin (25°C + 273.15 = 298.15 K), and Keq is the equilibrium constant which for this reaction is [G6P]/[G1P].
Substituting the values into the equation we get:
-7100 J/mol = -(8.314 J/mol K)(298.15 K) ln([G6P]/[G1P])
Now, we solve for ln([G6P]/[G1P]):
ln([G6P]/[G1P]) = ΔG' / (-R * T)
ln([G6P]/[G1P]) = -7100 J/mol / (-(8.314 J/mol K)(298.15 K))
ln([G6P]/[G1P]) = 0.286
Exponentiating both sides to remove the natural logarithm, we get:
[G6P]/[G1P] = e0.286 = 1.331
Therefore, at equilibrium and at 25°C, the concentration of G6P is approximately 1.331 times that of G1P.
The equilibrium reaction below has the Kc = 3.93. If the volume of the system at equilibrium is increased from 2.00 liters to 8.00 liters, how and for what reason will the equilibrium shift? Be sure to calculate the value of the reaction quotient, Q, and use this to confirm your answer.
co (g) + 3H2(g) -CH4(g) + H2O (g)
Answer:
the equilibium changes since Q > Kc, by increasing the volume, therefore, the reaction will try to use some of the excess product and favor the reverse reaction to reach equilibrium.
Explanation:
CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)
∴ Kc = [ H2O ] * [ CH4 ] / [ H2 ]³ * [ CO ] = 3.93.....equilibrium, V = 2.00L
PV = nRT; assuming P,T a standart conditions ( 1 atm, 298 K )
⇒ n / V = P / RT
∴ V = 8.00L
∴ R = 0.082 atm.L/K.mol
⇒ mol CO(g) = 0.327 mol = mol H2O = mol CH4
⇒ mol H2(g) = 0.327 mol CO * ( 3mol H2 / mol CO) = 0.981 mol
⇒ [ CO ] = 0.041 = [ CH4 ] = [ H2O]
⇒ [ H2 ] = 0.123 M
∴ Q = [ H2O ] * [ CH4 ] / [ H2 ]³ * [ CO ]
⇒ Q = ( 0.041² ) / (( 0.123³ ) * ( 0.041 ))
⇒ Q = 22.03
Q > Kc⇒ we have more product present than we would have in the equilibrium.
Answer:
Q = 62.9
Q > Kc, so the reaction will proceed to the left so that Q takes the value of Kc.
Explanation:
Let's consider the following reaction.
CO(g) + 3 H₂(g) ⇄ CH₄(g) + H₂O(g)
The equilibrium constant (Kc) is:
[tex]Kc= 3.93 =\frac{[CH_{4}].[H_{2}O]}{[CO].[H_{2}]^{3} }[/tex]
If the volume is multiplied by 4 (2.00 L → 8.00 L), the concentrations will be divided by 4. The reaction quotient (Q) is:
[tex]Q=\frac{(0.25[CH_{4}]).(0.25[H_{2}O])}{(0.25[CO]).(0.25[H_{2}])^{3} }=16\frac{[CH_{4}].[H_{2}O]}{[CO].[H_{2}]^{3} }=16Kc=16 \times 3.93 = 62.9[/tex]
Q > Kc, so the reaction will proceed to the left so that Q takes the value of Kc.
Show that 1 kJ/kg = 1000 m2/S2
Answer:
1000m2 / s2
Explanation:
Hello! In order to verify this, we have to do unit conversion. We also have to know that J (Joule) = kg * m2 / s2
Then we can start with the test.
1kJ / kg * (1000J / 1kJ) = 1000J / kg
1000J / kg = 1000kg * m2 / kg * s2
In this step we can simplify "kg".
So the result is
1000m2 / s2
Final answer:
To demonstrate that 1 kJ/kg equals 1000 m²/s², we recognize that the joule (J) is defined as kg-m²/s², and by converting kJ to J and canceling out the kg units, we affirm the equality.
Explanation:
To show that 1 kJ/kg is equal to 1000 m²/s², we start by recognizing that the unit of energy, the joule (J), is defined as 1 kilogram-meter²/second² (kg-m²/s²). Therefore, when we talk about energy per unit mass, we are effectively dividing energy by mass, leaving us with units of m²/s².
Given that 1 joule is 1 kg·m²/s², 1 kJ is 1000 joules (since the prefix 'kilo' means 1000). So when we have 1 kJ/kg, it's the same as saying 1000 J/kg. When we divide each term (kg·m²/s²) by kg, the kilograms cancel out, leaving us with m²/s². Thus, 1 kJ/kg is indeed equivalent to 1000 m²/s².
If the molality of a NaBr(aq) solution is 2.50 m, what is the weight percent of NaBr? The molar mass of NaBr is 1029 g/mol 20.5% 25.0% 25.7% 34.6% 65.4% ent Navigator J K L
Answer:
The weight percent of NaBr is 25,7%
Explanation:
Molality is a way to express the concentration of a chemical in terms of moles of substances per kg of solution. Weight percent is other way to express concentration in terms of mass of substance per mass of solution per 100
Thus, to obtain weight percent you must convert moles of NaBr to grams with molar mass and these grams to kilograms, thus:
2,50 mol / kg solution × (102,9 g / 1mol) × (1 kg / 1000 g) =
[tex]\frac{0,257 kg NaBr}{kg solution }[/tex] × 100 = 25,7 % (w/w)
I hope it helps!
For the following reaction, 101 grams of magnesium nitride are allowed to react with 144 grams of water. Mg3N2 (5) + 6 H20 (1) — 3 Mg(OH)2 (aq) + 2 NH2 (aq) What is the FORMULA for the limiting reagent? What is the maximum amount of magnesium hydroxide that can be formed? grams What amount of the excess reagent remains after the reaction is complete? grams Submit Answer Retry Entire Group 8 more group attempts remaining
Answer : The formula of limiting reagent is, [tex]Mg_3N_2[/tex].
The mass of [tex]Mg(OH)_2[/tex] is, 174 grams.
The mass of excess reactant is, 36 grams.
Solution : Given,
Mass of [tex]Mg_3N_2[/tex] = 101 g
Mass of [tex]H_2O[/tex] = 144 g
Molar mass of [tex]Mg_3N_2[/tex] = 101 g/mole
Molar mass of [tex]H_2O[/tex] = 18 g/mole
Molar mass of [tex]Mg(OH)_2[/tex] = 58 g/mole
First we have to calculate the moles of [tex]Mg_3N_2[/tex] and [tex]H_2O[/tex].
[tex]\text{ Moles of }Mg_3N_2=\frac{\text{ Mass of }Mg_3N_2}{\text{ Molar mass of }Mg_3N_2}=\frac{101g}{101g/mole}=1moles[/tex]
[tex]\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{144g}{18g/mole}=8moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]Mg_3N_2+6H_2O\rightarrow 3Mg(OH)_2+2NH_3[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]Mg_3N_2[/tex] react with 6 mole of [tex]H_2O[/tex]
So, given 1 moles of [tex]Mg_3N_2[/tex] react with 6 moles of [tex]H_2O[/tex]
From this we conclude that, [tex]H_2O[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]Mg_3N_2[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]Mg(OH)_2[/tex]
From the reaction, we conclude that
As, 1 mole of [tex]Mg_3N_2[/tex] react to give 3 mole of [tex]Mg(OH)_2[/tex]
So, given 1 mole of [tex]Mg_3N_2[/tex] react to give 3 moles of [tex]Mg(OH)_2[/tex]
Now we have to calculate the mass of [tex]Mg(OH)_2[/tex]
[tex]\text{ Mass of }Mg(OH)_2=\text{ Moles of }Mg(OH)_2\times \text{ Molar mass of }Mg(OH)_2[/tex]
[tex]\text{ Mass of }Mg(OH)_2=(3moles)\times (58g/mole)=174g[/tex]
The mass of [tex]Mg(OH)_2[/tex] is, 174 grams.
Now we have to calculate the moles of excess reactant [tex](H_2O)[/tex].
Moles of excess reactant = 8 - 6 = 2 moles
Now we have to calculate the mass of excess reactant.
[tex]\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O[/tex]
[tex]\text{ Mass of }H_2O=(2moles)\times (18g/mole)=36g[/tex]
The mass of excess reactant is, 36 grams.
Based on sulubility rules, which of the
followingwill occur when solutions of ZnSO4(aq)
andMgCl2(aq) are mixed? Write the NIE if aprecipitation
is considered likely.
A. ZnCl2 wil
precipitate;Mg2andSO42_ will be
spectatior ions.
B. ZnSO4 will precipitate;Mg2+ andCl-
will be spectator ions.
C. MgSO4 will precipitate; Zn2+and Cl-
will be spectator ions.
D. MgCl2 will precipitate; Zn2+
andSO42- will be spectator ions.
E. No precipitate will form.
Answer:
The correct option is: E. No precipitate will form.
Explanation:
A solubility chart refers to the list of solubility of various ionic compounds. It shows the solubility of the various compounds in water at room temperature and 1 atm pressure.
Also, according to the solubility rules, the salts of chlorides, bromides and iodides are generally soluble and mostly all salts of sulfate are soluble.
Since, all the compounds formed in this double replacement reaction are soluble in water. Therefore, no precipitate will be formed.
ZnSO₄ (aq) + MgCl₂ (aq) → ZnCl₂ (aq) + MgSO₄ (aq)
Enter your answer in the provided box. Calculate the number of g of CO2 produced from the combustion of 5.24 mol of CzHg. The balanced equation is: C3H2(g) + 502(g) → 3CO2(g) + 4H2O(g). g CO2
Answer:
691.84g
Explanation:
I'm assuming that by CzHg, you mean C3H2
First, use the mole ratio in the equation to find the moles of CO2
n (CO2)= n ( C3H2) × 3
= 5.24 × 3
= 15.72
To find the mass of CO2 produced in grams, complete the following calculation
m= n × MM
where
m = mass
n= moles
MM= molecular mass
m= 15.72 × (12.01 +( 16×2))
m =691.8372
m= 691.84g
Glucose is a carbohydrate that contains carbon, hydrogen, and oxygen. The empirical formula of glucose is CH2O and its molar mass is 180.12 g/mol. Find the molecular formula of glucose.
Answer: The molecular formula of glucose is [tex]C_6H_{12}O_6[/tex]
Explanation:
We are given:
Empirical formula of the compound = [tex]CH_2O[/tex]
Empirical mass of the compound = [tex][(1\times 12)+(2\times 1)+(1\times 16)]=30g/mol[/tex]
For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.
The equation used to calculate the valency is:
[tex]n=\frac{\text{Molecular mass}}{\text{Empirical mass}}[/tex]
We are given:
Mass of molecular formula = 180.12 g/mol
Mass of empirical formula = 30 g/mol
Putting values in above equation, we get:
[tex]n=\frac{180.12g/mol}{30g/mol}=6[/tex]
Multiplying this valency by the subscript of every element of empirical formula, we get:
[tex]C_{(1\times 6)}H_{(2\times 6)}O_{(1\times 6)}=C_6H_{12}O_6[/tex]
Hence, the molecular formula of glucose is [tex]C_6H_{12}O_6[/tex]
4. Each time that you prepare a diluted bleach solution you will want it to have a [OH-] = 0.02. Calculate how much 1.00 M NaOH you need to add to d.1. water to get every 20 mL of this concentration of OH- ion. Express your answer in drops of 1.00 M NaOH. (Assume that there are 20 drops in 1 m.)
Answer:
8 drops of 1.00 M NaOH will be needed.
Explanation:
Concentration of [tex][OH^-][/tex] in bleach solution = 0.02 M
[tex]NaOH\rightarrow OH^-+Na^+[/tex]
[tex]NaOH=[OH^-]=0.02M[/tex]
Concentration of bleach solution we want ,[tex]M_1[/tex] = 0.02 M
Volume of the bleach solution,[tex]V_1[/tex] = 20 ml
Concentration of NaOH solution,[tex]M_2[/tex] = 1.00 M
Volume of the NaOH solution required ,[tex]V_2[/tex] = ?
[tex]M_1V_1=M_2V_2[/tex]
[tex]0.02 M\times 20 mL=1.00 M\times V_2[/tex]
[tex]V_2=\frac{0.02 M\times 20 mL}{1.00 M}=0.4 mL[/tex]
1 mL = 20 drops
0.4 mL = 0.4 × 20 drops = 8 drops
8 drops of 1.00 M NaOH will be needed.
Steve holds the world record for the 100-meter dash. At his fastest, Bolt ran 100 meters in 9.58 seconds. Convert his speed into miles per hour. Report your answer using 3 significant figures.
Conversion factors: 1 mile = 1609 meters 60 seconds = 1 minute 60 minutes = 1 hour
Use appropriate Sig Figs in answer!
Answer:
23.4 mph
Explanation:
The conversion factors are multiplied so that the units cancel out:
(100 meters / 9.58 sec) x (1 mile / 1609 meters) x (60 sec / min) x (60 min /1 hr) = 23.4 mph
Which form of bond between two carbons is capable of rotation without breaking the bond? * single bond double bond triple bond
Answer:
single bond
Explanation:
Single bond is the chemical bond in which 2 valence electrons are involved. These are the sigma bond which are formed formed by the head-on overlapping between the orbitals of the two atoms involving in the bond. The bond lies on the inter nuclear axis and thus , the bond can be rotated without the breaking of bond.
On the other hand, double and triple bonds contains one sigma bond and also bonds which are formed by the sideways overlapping of the orbitals and thus, these do not lie on inter nuclear axis and breaks on rotation.
Two moles of ideal He gas are contained at a pressure of 1 atm and a temperature of 300 K. 34166 J of heat are transferred to the gas, as a result of which the gas expands and does 1216 J of work against its surroundings. The process is reversible. (Note: C = 1.5R) Please calculate the final temperature of the gas
Explanation:
The given data is as follows.
n = 2 mol, P = 1 atm, T = 300 K
Q = +34166 J, W= -1216 J (work done against surrounding)
[tex]C_{v}[/tex] = [tex]\frac{3R}{2}[/tex]
Relation between internal energy, work and heat is as follows.
Change in internal energy ([tex]\Delta U[/tex]) = Q + W
= [34166 + (-1216)] J
= 32950 J
Also, [tex]\Delta U = n \times C_{v} \times \Delta T[/tex]
= [tex]3R \times (T_{2} - T_{1})[/tex]
32950 J = [tex]3 \times 8.314 J/mol K \times (T_{2} - 300 K)[/tex]
[tex]\frac{32950}{24.942} = T_{2} - 300 K[/tex]
1321.06 K + 300 K = [tex]T_{2}[/tex]
[tex]T_{2}[/tex] = 1621.06 K
Thus, we can conclude that the final temperature of the gas is 1621.06 K.
How many significant figures are represented in each of the following numbers? a) 7.1 x 10^-5 b) 0.00677 c) 750
Answer:
a) 7.1 x 10⁻⁵ : 2 significant figures
b) 0.00677 : 3 significant figures
c) 750 : 2 significant figure
Explanation:
The significant digits or figures refers to the digits of a given number that carry meaning and also contributes to precision of the given number.
a) 7.1 x 10⁻⁵ = 0.000071 : 2 significant figures, leading zeros are not significant.
b) 0.00677 : 3 significant figures, leading zeros are not significant.
c) 750 : 2 significant figure, trailing zeros are not significant.
The diffusion flux is defined as: a) Mass, per unit surface area, per unit time b) Surface area, per unit time c) Time, per unit surface area. d) The product of time and surface area, per unit mass
Answer:
The correct answer is option a.
Explanation:
Diffusion flux is defined as movement of mas of atoms diffusing through the unit area in per unit time.It measured in ([tex]kg/m^2 s[/tex]).
[tex]J=\frac{1}{A}\frac{dM}{dt}[/tex]
J = diffusion flux
A = Unit area A through which atoms moves.
M = mass of atoms passes in t interval of time.
Diffusion flux is the mass of gas passing through a unit surface area per unit time, influenced by the concentration gradient, surface area, and particle travel distance.
Explanation:The diffusion flux can be defined as the amount of gas passing through a given area over a unit of time. Therefore, the correct answer is a) Mass, per unit surface area, per unit time. The rate of diffusion is dictated by several factors such as the concentration gradient, the surface area available for diffusion, and the distance the gas particles must travel. In addition, it's important to note that the time required for diffusion is inversely proportional to the diffusion rate.
Learn more about Diffusion Flux here:https://brainly.com/question/33792345
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Why is it important to have an energy balance on a chemical process facility
Final answer:
The importance of an energy balance in a chemical process facility lies in ensuring efficient energy use, maintaining safety, optimizing chemical processes for increased yield, and reducing environmental impact.
Explanation:
An energy balance is crucial in a chemical process facility because it ensures the efficient use of energy, minimizes waste, and helps maintain the safety of the operation. Since energy is neither created nor destroyed (law of conservation of energy), it's important to know where energy is being consumed and produced within chemical processes. This understanding allows engineers to optimize the process, increase the yield of the desired product, and reduce environmental impact by minimizing wasted energy and reducing unwanted by-products.
An energy balance aids in maintaining the operation within the desired range of conditions, preventing unsafe levels of energy which could lead to accidents. Moreover, certain reactions require specific amounts of energy to produce raw materials or to synthesize products, and tracking energy use is essential for economic and environmental reasons.
Which one of the following conditions is met at the equivalence point of the titration of a monoprotic weak acid with a strong base?A. The volume of strong base added from the buret must equal the volume of weak acid.B. The moles of strong base added must equal the moles of weak acid.C. pH
Answer:
At the equivalence point of the titration of a monoprotic weak acid with a strong base: B. the moles of strong base added must equal the moles of weak acid.
Explanation:
In every titration, the equivalence point is defined as the point where the moles of the titrant and analyte are equal. For every acid-base titration, the equivalence point is defined as the point where the moles of the base is equal to the moles of the acid.
If the solutions of the acid and base are at a different concentration the volume added from the buret will not be the same as the volume of the analyte.
Solve for t In(A/B) -kt where B-1.65 x 102 M A 1.00 x 10 M. k-4.80 x 104s* and t=
Answer:
[tex]t=5.84\times 10^{-5}s[/tex]
Explanation:
First rearrange the whole equation by keeping t in one side and rest parameters in another side of equationThen plug-in all the given values for each parameters and get the solution[tex]ln(\frac{A}{B})=-kt[/tex]
or, [tex]t=\frac{1}{-k}\times ln(\frac{A}{B})[/tex]
Now plug-in all given values:
[tex]t=\frac{1}{-4.80\times 10^{4}s^{-1}}\times ln(\frac{1.00\times 10M}{1.65\times 10^{2}M})[/tex]
So, [tex]t=5.84\times 10^{-5}s[/tex]
A 1.24g sample of a hydrocarbon, when completely burned in an excess of O2 yields 4.04g Co2 and 1.24g H20. Draw plausible structure for the hydrocarbon molecule
Answer:
Plausible structure has been given below
Explanation:
Molar mass of [tex]CO_{2}[/tex] is 44 g/mol and molar mass of [tex]H_{2}O[/tex] is 18 g/molNumber of mole = (mass/molar mass)4.04 g of [tex]CO_{2}[/tex] = [tex]\frac{4.04}{44}moles[/tex] [tex]CO_{2}[/tex] = 0.0918 moles of [tex]CO_{2}[/tex]
1 mol of [tex]CO_{2}[/tex] contains 1 mol of C atom
So, 0.0918 moles of [tex]CO_{2}[/tex] contains 0.0918 moles of C atom
1.24 g of [tex]H_{2}O[/tex] = [tex]\frac{1.24}{18}moles[/tex] [tex]H_{2}O[/tex] = 0.0689 moles of [tex]H_{2}O[/tex]
1 mol of [tex]H_{2}O[/tex] contain 2 moles of H atom
So, 0.0689 moles of [tex]H_{2}O[/tex] contain [tex](2\times 0.0689)moles[/tex] of [tex]H_{2}O[/tex] or 0.138 moles of [tex]H_{2}O[/tex]
Moles of C : moles of H = 0.0918 : 0.138 = 2 : 3
Empirical formula of hydrocarbon is [tex]C_{2}H_{3}[/tex]
So, molecular formula of one of it's analog is [tex]C_{4}H_{6}[/tex]
Plausible structure of [tex]C_{4}H_{6}[/tex] has been given below.
A sample of nitrogen gas, stored in a 2.96-L container at 32.0°C, exerts a pressure of 4.27 atm. Calculate the number of moles of nitrogen gas in the container. Enter your answer in the box provided. mol
Answer: The number of moles of nitrogen gas is 0.505 moles.
Explanation:
To calculate the number of moles of nitrogen gas, we use ideal gas equation, which is:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 4.27 atm
V = Volume of the gas = 2.96 L
T = Temperature of the gas = [tex]32.0^oC=[32.0+273]K=305K[/tex]
R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
n = number of moles of gas = ?
Putting values in above equation, we get:
[tex]4.27atm\times 2.96L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 305K\\n=0.505mol[/tex]
Hence, the number of moles of nitrogen gas is 0.505 moles.
Explanation:
The given data is as follows.
Volume = 2.96 L, Temperature = [tex]32.0^{o}C[/tex] = (32 + 273) K = 305 K,
Pressure = 4.27 atm, n = ?
And, according to ideal gas equation, PV = nRT
[tex]4.27atm \times 2.96 L= n \times 0.0821 Latm/mol K \times 305K[/tex]
n = [tex]\frac{12.64 atm L}{25.04 Latm/mol}[/tex]
= 0.505 mol
Thus, we can conclude that the number of moles of nitrogen gas in the container is 0.505 mol.
Consider the titration of 100 mL of 0.200 M HCHO, with 1.00 M NaOH. The pK, of HCHO2 is 3.75. a) What is the pH before ANY NaOH is added? b) What is the pH after 5.00 mL of NaOH are added? c) After 10 mL of NaOH are added? d) What is the pH when 20 mL of NaOH have been added? What is this point in the titration called?
Answer:
a) pH = 2,23
b) pH = 3,26
c) pH = 3,74
d) pH = 7,98. Here we have the equivalence point of the titration
Explanation:
In a titration of a strong base (NaOH) with a weak acid (HCOOH) the reaction is:
HCOOH + NaOH → HCOONa + H₂O
a) Here you have just HCOOH, thus:
HCOOH ⇄ HCOO⁻ + H⁺ where ka =1,8x10⁻⁴ and pka = 3,74
When this reaction is in equilibrium:
[HCOOH] = 0,200 -x
[HCOO⁻] = x
[H⁺] = x
Thus, equilibrium equation is:
1,8x10⁻⁴ = [tex]\frac{[x][x] }{[0,200-x]}[/tex]
The equation you will obtain is:
x² + 1,8x10⁻⁴x - 3,6x10⁻⁵ = 0
Solving:
x = -0,006090675 ⇒ No physical sense. There are not negative concentrations
x = 0,005910674
As x = [H⁺] and pH = - log [H⁺]
pH = 2,23
b) Here, it is possible to use:
HCOOH + NaOH → HCOONa + H₂O
With adition of 5,00 mL of 1,00M NaOH solution the initial moles are:
HCOOH = [tex]0,100 L.\frac{0,200 mol}{L} =[/tex] = 2,0x10⁻² mol
NaOH = [tex]0,005 L.\frac{1,00 mol}{L} =[/tex] = 5,0x10⁻³ mol
HCOO⁻ = 0.
In equilibrium:
HCOOH = 2,0x10⁻² mol - 5,0x10⁻³ mol = 1,5x10⁻² mol
NaOH = 0 mol
HCOO⁻ = 5,0x10⁻³ mol
Now, you can use Henderson–Hasselbalch equation:
pH = 3,74 + log [tex]\frac{5,0x10^{-3} }{1,5x10^{-2} }[/tex]
pH = 3,26
c) With adition of 10 mL of 1,00M NaOH solution the initial moles are:
HCOOH = [tex]0,100 L.\frac{0,200 mol}{L} =[/tex] = 2,0x10⁻² mol
NaOH = [tex]0,010 L.\frac{1,00 mol}{L} =[/tex] = 1,0x10⁻² mol
HCOO⁻ = 0.
In equilibrium:
HCOOH = 2,0x10⁻² mol - 1,0x10⁻² mol = 1,0x10⁻² mol
NaOH = 0 mol
HCOO⁻ = 1,0x10⁻² mol
Now, you can use Henderson–Hasselbalch equation:
pH = 3,74 + log [tex]\frac{1,0x10^{-2} }{1,0x10^{-2} }[/tex]
pH = 3,74
d) With adition of 20 mL of 1,00M NaOH solution the initial moles are:
HCOOH = [tex]0,100 L.\frac{0,200 mol}{L} =[/tex] = 2,0x10⁻² mol
NaOH = [tex]0,020 L.\frac{1,00 mol}{L} =[/tex] = 2,0x10⁻² mol
HCOO⁻ = 0.
Here we have the equivalence point of the titration, thus, the equilibrium is:
HCOO⁻ + H₂O ⇄ HCOOH + OH⁻ kb = kw/ka where kw is equilibrium constant of water = 1,0x10⁻¹⁴; kb = 5,56x10⁻¹¹
Concentrations is equilibrium are:
[HCOOH] = x
[HCOO⁻] = 0,1667-x
[OH⁻] = x
Thus, equilibrium equation is:
5,56x10⁻¹¹ = [tex]\frac{[x][x] }{[0,01667-x]}[/tex]
The equation you will obtain is:
x² + 5,56x10⁻¹¹x - 9,27x10⁻¹³ = 0
Solving:
x = -9,628361x10⁻⁷⇒ No physical sense. There are not negative concentrations
x = 9,627806x10⁻⁷
As x = [OH⁻] and pOH = - log [OH⁻]; pH = 14 - pOH
pOH = 6,02
pH = 7,98
I hope it helps!
Write Huckel's rule below and determine how many electrons are required to make an aromatic ring with n = 0, 1, and 2.
Answer: The number of electrons for n = 0, 1 and 2 are 2, 6 and 10 respectively.
Explanation:
Huckel's rule is used to determine the aromaticity in a compound. The number of delocalized [tex]\pi-[/tex] electrons are calculated by using the equation:
[tex]\text{Number of delocalized }\pi-\text{ electrons}=4n+2[/tex]
where,
n = 0 or any whole number
Calculating the value of electrons for n = 0Putting values in above equation, we get:
[tex]\text{Number of delocalized }\pi-\text{ electrons}=4(0)+2=2[/tex]
Calculating the value of electrons for n = 1Putting values in above equation, we get:
[tex]\text{Number of delocalized }\pi-\text{ electrons}=4(1)+2=6[/tex]
Calculating the value of electrons for n = 2Putting values in above equation, we get:
[tex]\text{Number of delocalized }\pi-\text{ electrons}=4(2)+2=10[/tex]
Hence, the number of electrons for n = 0, 1 and 2 are 2, 6 and 10 respectively.
How much CrCl3 • 6H20 is needed to prepare 1 L of solution containing 20.0g Cr3+ per L?
Answer:
salt mass=102.5gCrCl3.6H2O
Explanation:
Hello !
To know how much CrCl3.6H2O we need, we follow the steps below
First we have to know what the mass of the compound is
Cr = 52g / mol
Cl = 35.5g / mol
H = 1g / mol
O = 16g / mol
We calculate the mass
52+ (35.5 * 3) + 6 * (2 * 1 + 16) = 266.5g
We have 20g Cr, so we calculate the amount of salt:
20gCr * (266.5g salt / 52gCr) = 102.5gCrCl3.6H2O
salt mass=102.5gCrCl3.6H2O
Compute 4.62 x 4.48697. Round the answer appropriately. Express your answer numerically using the proper number of sian
Answer:
20.7
Explanation:
20.7298014 rounded off to 3 sig fig =20.7