Answer:
Here's what I get
Explanation:
When a solid is converted to a liquid, heat is absorbed. TRUE.
Temperature remains constant while all of a solid is converted to a liquid. TRUE.
When heat is applied to a solid, the molecular motion decreases as the temperature increases. False. Molecular motion increases as the temperature increases.
The average kinetic energy of the system changes while all of a solid is converted to a liquid. False. The temperature remains constant during a phase change, so the average kinetic energy of the system does not change.
The temperature increases while all of a liquid is converted to a gas. False.
The complex [Zn(NH3)2Cl2]2+ does not exhibit cis-trans isomerism. The geometry of this complex must be ________.
a. tetrahedral
b. trigonal bipyramidal
c. square planar
d. octahedral
e. either tetrahedral or square planar
Answer:
A. tetrahedral
Explanation:
The only complex ion geometries that can exhibit cis-trans isomerim are the octahedral and square planar.With the above information in mind, the correct answer would then have to be either a) or b), however the complex [Zn(NH₃)₂Cl₂]²⁺ has 4 ligands. This means that the correct answer is a), because the trigonal bipyramidal geometry has 5 ligands.
Gaseous cyclobutene undergoes a first-order reaction to form gaseous butadiene. At a particular temperature, the partial pressure of cyclobutene in the reaction vessel drops to one-eighth its original value in 124 seconds. What is the half-life for this reaction at this temperature?
Answer:
41.3 minutes
Explanation:
Since the reaction is a first order reaction, therefore, half life is independent of the initial concentration, or in this case, pressure.
[tex]t_{1/2}= \frac{0.693}{K}[/tex]
So, fraction of original pressure = [tex]\frac{1}{2}^2[/tex]
n here is number of half life
therefore, [tex]\frac{1}{8}= \frac{1}{2}^3[/tex]
⇒ n= 3
it took 124 minutes to drop pressure to 1/8 of original value, half life = 124/3= 41.3 minutes.
Answer : The half-life of this reaction at this temperature is, 41.5 seconds.
Explanation :
First we have to calculate the rate constant.
Expression for rate law for first order kinetics is given by:
[tex]k=\frac{2.303}{t}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = ?
t = time passed by the sample = 124 s
a = let initial amount of the reactant = X
a - x = amount left after decay process = [tex]\frac{1}{8}\times (X)=\frac{X}{8}[/tex]
Now put all the given values in above equation, we get
[tex]k=\frac{2.303}{124s}\log\frac{X}{(\frac{X}{8})}}[/tex]
[tex]k=0.0167s^{-1}[/tex]
Now we have to calculate the half-life.
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]t_{1/2}=\frac{0.693}{0.0167s^{-1}}[/tex]
[tex]t_{1/2}=41.5s[/tex]
Therefore, the half-life of this reaction at this temperature is, 41.5 seconds.
The room temperature electrical conductivity of a semiconductor specimen is 2.8 x 104 (?-m)-1. The electron concentration is known to be 2.9 x 1022 m-3. Given that the electron and hole mobilities are 0.14 and 0.023 m2/V-s, respectively, calculate the hole concentration (in m-3)
Answer:
7.43 × 10²⁴ m⁻³
Explanation:
Data provided in the question:
Conductivity of a semiconductor specimen, σ = 2.8 × 10⁴ (Ω-m)⁻¹
Electron concentration, n = 2.9 × 10²² m⁻³
Electron mobility, [tex]\mu_n[/tex] = 0.14 m²/V-s
Hole mobility, [tex]\mu_p[/tex]= 0.023 m²/V-s
Now,
σ = [tex] nq\mu_n+pq\mu_p[/tex]
or
σ = [tex] q(n\mu_n+p\mu_p)[/tex]
here,
q is the charge on electron = 1.6 × 10⁻¹⁹ C
p is the hole density
thus,
2.8 × 10⁴ = 1.6 × 10⁻¹⁹( 2.9 × 10²² × 0.14 + p × 0.023 )
or
1.75 × 10²³ = 0.406 × 10²² + 0.023p
or
17.094 × 10²² = 0.023p
or
p = 743.217 × 10²²
or
p = 7.43 × 10²⁴ m⁻³
A flask is charged with 3.00 atm of dinitrogen tetroxide gas and 2.00 atm of nitrogen dioxide gas at 25ºC and allowed to reach equilibrium.
N2O4(g)-->2 NO2(g) Kp = 0.316
Which of the following best describes the system within the flask once equilibrium has been established?
The rate of the decomposition of N2O4(g) is equal to the rate of formation of NO2(g).
The partial pressure of N2O4(g) is equal to the partial pressure of NO2(g).
The rate of the decomposition of N2O4(g) is greater than the rate of formation of NO2(g).
The rate of the decomposition of N2O4(g) is less than the rate of formation of NO2(g).
Answer:
The correct answer is The rate of decomposition of N2O4 is equal to the rate of formation of NO2.
Explanation:
According to theory of reaction kinetics we all know that at equilibrium the amount of reactant is equal to the amount of product.
In other word it can be stated that the rate of the decomposition of N2O4 is equal to the rate of the formation of NO2.
Final answer:
At equilibrium, the partial pressure of N2O4 is equal to the partial pressure of NO2 in the flask.
Explanation:
Once equilibrium has been established, the partial pressure of N2O4(g) is equal to the partial pressure of NO2(g) in the flask. This is because the equilibrium constant (Kp) expression for the decomposition of N2O4 is (PNO2)^2 / PN2O4, which means the partial pressures of the two gases are directly related. Therefore, answer choice B, 'The partial pressure of N2O4(g) is equal to the partial pressure of NO2(g),' best describes the system within the flask at equilibrium.
Draw all of the constitutional isomeric monochlorination products resulting from the reaction of 2,3−dimethylbutane under radical substitution conditions.
(a) primary alkyl halide
(b) tertiary alkyl halide
Answer:
a) 2-chloro-2,3-dimethylbutane
b) 1-chloro-2,3-dimethylbutane
Explanation:
The monochlorination is a reaction in which we have to add only 1 Cl to the molecule. In this case, we will have to add a Cl to a primary carbon (a) and to a tertiary carbon (b).
In the monochlorination of the primary carbon, we can choose any methyl carbon. For the monochlorination of the terciary carbon we have to choose an CH carbon.
(See the figure)
Final answer:
When 2,3-dimethylbutane undergoes monochlorination under radical substitution conditions, multiple constitutional isomers can be formed due to the presence of primary and tertiary carbon atoms.
Explanation:
When 2,3-dimethylbutane undergoes monochlorination under radical substitution conditions, multiple constitutional isomers can be formed. The monochlorination reaction involves the substitution of a hydrogen atom on the carbon chain with a chlorine atom. Since 2,3-dimethylbutane has two different types of carbon atoms (primary and tertiary), the resulting constitutional isomers will also differ.
(a) Primary alkyl halide: The primary carbon atom is bonded to one other carbon atom and one hydrogen atom. So, a constitutional isomer could be formed by substituting the hydrogen atom with a chlorine atom.
(b) Tertiary alkyl halide: The tertiary carbon atom is bonded to three other carbon atoms. So, a constitutional isomer could be formed by substituting one of the carbon atoms with a chlorine atom.
Calculate the standard potential for the following galvanic cell: Fe(s) | Fe2+(aq) || Ag+(aq) | Ag(s) which has the overall balanced equation: Fe(s)+2Ag+(aq)→Fe2+(aq)+2Ag(s) Express your answer to three significant figures and include the appropriate units
Answer:
The standard potential for this galvanic cell is 1.25 V
Explanation:
Step 1
Oxidation: Fe(s) → Fe^2+ (aq) + 2e- E° = 0.447 V
Reduction: Ag+(aq) + e- → Ag(s) E° = 0.7996 V
The electron number for the oxidation is 2; the electron number for the reduction is 1.
By multiplying the reduction reaction by 2, we make the electron number the same between the two half-equations
2 Ag+(aq) + 2e- → 2Ag(s) E° = 0.7996 V
Step 2: The overall balanced equation
Fe(s) + 2Ag+(aq) → Fe^2+(aq) + 2ag(s) E° = 0.447 V + 0.7996 V = 1.25V
The standard potential for this galvanic cell is 1.25 V
During the combustion of a peanut that weighed 0.341 g, the temperature of the 100 mL of water in the calorimeter rose from 23.4ºC to 37.9ºC. The peanut didn't burn completely. If the serving size is 28.0 grams of peanuts, what is the Cal from fat per serving (Cal/serving)?
Select one:
46.2 Cal/serving 1
19 Cal/serving
28.5 Cal/serving
138 Cal/serving
Answer:
119 kCal per serving.
Explanation:
The heat energy necessary to elevates water's temperature from 23.4°C to 37.9°C can be calculated by the equation below:
Q = mcΔT
Q: heat energy
m: mass in g
c: specific heat capacity in cal/g°C
ΔT = temperature variation in °C
m is the mass of water, considering the density of water to be 1g/mL, 100 mL of water weights 100g. Therefore:
Q = 100 g x 1.00 cal/g°C x (37.9 - 23.4)°C
Q = 1450 cal
1450 cal ____ 0.341 g peanuts
x ____ 28 g peanuts
x = 119061.58 cal
This means that the cal from fat per serving of peanuts is at least 119 kCal.
The temperature change of the water indicated 1.45 Calories for 0.341 grams of peanut. Therefore, a serving size of 28.0 grams contains option 2) 119 Calories from fat.
To determine the Calories from fat per serving, we must first calculate the energy released by the peanut during combustion and then scale it to a full serving size.
Determine the temperature change in the water:Thus, the Calories from fat per serving is option 2) 119 Cal/serving.
Give two examples (each) of strong electrolyte, weak electrolyte, and nonelectrolyte. 2. Predict the products for reactions below. Which of the following reaction(s) produce a precipitate? A) LiOH + Na2S B) (NH4)2SO4 + LiCl C) Sr(C2H3O2)2 + Na2SO4 D) KNO3 + NaOH E) None of the above solution pairs will produce a precipitate. 3. What are the spectator ions in the precipitation reaction you chose above? 4. Write the molecular, complete ionic, and net ionic equations for the reactions in Q2.
Answer:
1. Strong electrolytes: HCl and NaOH.
Weak electrolytes: CH₃COOH and NH₃.
Nonelectrolytes: (NH₂)₂CO (urea) and CH₃OH (methanol).
2. C) [tex]C) Sr(C₂H₃O₂)₂ (aq) + Na₂SO₄ (aq) → SrSO₄ (s) + 2NaC₂H₃O₂ (aq)[/tex]
3. The spectator ions are Na⁺ and C₂H₃O₂⁻.
4. Written in explanation section.
Explanation:
An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity.A characteristic of strong electrolytes is that the solute is assumed to be 100 percent dissociated into ions in solution, therefore they are great conductors of electricity. Eg: HCl and NaOH.
On the other hand, weak electrolytes are not completely dissociated in solution, therefore they are poor conductors of electricity. Eg: CH₃COOH and NH₃.
(By dissociation we mean the breaking up of the compound into cations and anions).
A nonelectrolyte does not conduct electricity when dissolved in water. Eg: (NH₂)₂CO (urea) and CH₃OH (methanol).
2. The products formed on each reaction are (always remember to balance the equations):
A) [tex]2 LiOH (aq) + Na₂S (aq) → Li₂S (aq) + 2NaOH (aq)[/tex]
B) [tex](NH₄)₂SO4 (aq) +2 LiCl (aq) → 2NH₄Cl (aq) + Li₂SO₄ (aq)[/tex]
C) [tex]C) Sr(C₂H₃O₂)₂ (aq) + Na₂SO₄ (aq) → SrSO₄ (s) + 2NaC₂H₃O₂ (aq)[/tex]
D) [tex]D) KNO₃ (aq) + NaOH (aq) → NaNO₃ (aq) + KOH (aq) [/tex]
The reaction C will produce SrSO₄, a white color precipitate.
3. When ionic compounds dissolve in water, they break apart into their component cations and anions. To be more realistic, the equations should show the dissociation of dissolved ionic compounds into ions. This is called a ionic equation which shows dissolved species as free ions. To see whether a precipitate might form from this solution, we first combine the cation and anion from different compounds, and refer to the solubility rules. The spectator ions are ions that are not involved in the overall reaction.
Therefore, for the equation chosen above:
[tex]Sr²⁺ + 2C₂H₃O₂⁻ + 2Na⁺ + SO₄²⁻ → SrSO₄ + 2Na⁺ + 2C₂H₃O₂⁻[/tex]
Because spectator ions appear on both sides of an equation, they can be eliminated from the ionic equation.
[tex]Sr²⁺ + SO₄²⁻ → SrSO₄[/tex]
Finally, we end up with the net ionic equation, which shows only the species that actually take part in the reaction.
In this reaction, the spectator ions are Na⁺ and C₂H₃O₂⁻.
4. Molecular equations:
A) [tex]2 LiOH (aq) + Na₂S (aq) → Li₂S (aq) + 2NaOH (aq)[/tex]
B) [tex](NH₄)₂SO4 (aq) +2 LiCl (aq) → 2NH₄Cl (aq) + Li₂SO₄ (aq)[/tex]
C) [tex]C) Sr(C₂H₃O₂)₂ (aq) + Na₂SO₄ (aq) → SrSO₄ (s) + 2NaC₂H₃O₂ (aq)[/tex]
D) [tex]D) KNO₃ (aq) + NaOH (aq) → NaNO₃ (aq) + KOH (aq) [/tex]
Complete ionic equations:
A) [tex]2Li⁺ + OH⁻ + 2Na⁺ + S²⁻ → 2Li⁺ + S²⁻ + 2Na⁺ +OH⁻[/tex]
B) [tex]2NH₄⁺ + SO₄²⁻ + 2Li⁺ + 2Cl⁻ → 2Li⁺ +S²⁻ +2NH₄⁺ +2Cl⁻ + 2Li⁺ + SO₄²⁻[/tex]
C) [tex]Sr²⁺ + 2C₂H₃O₂⁻ + 2Na⁺ + SO₄²⁻ → SrSO₄ + 2Na⁺ + 2C₂H₃O₂⁻[/tex]
D) [tex]K⁺ + NO₃⁻ + Na⁺ +OH⁻ → Na⁺ + NO₃⁻ + K⁺ + OH⁻[/tex]
If all products are aqueous, a net ionic equation cannot be written because all ions are canceled out as spectator ions. Therefore, no precipitation reaction occurs. The only net equation can be written for reaction C):
C) [tex]Sr²⁺ + SO₄²⁻ → SrSO₄[/tex]
You want to make 100 mL of a 2.5 M stock solution of calcium chloride (molecular weight 110.98 g/mol). How many grams of calcium chloride do you need to weigh out in order to make this solution? Round your answer to the nearest tenth of a gram.
Answer:
27.7 g
Explanation:
100 mL = 0.100 LTo solve this problem we use the definition of molarity:
M = mol / LFrom the problem we're given M = 2.5 M and volume = 0.100 L. We use the above formula and calculate the required moles of calcium chloride (CaCl₂):
2.5 M = moles / 0.100 Lmoles = 0.25 moles CaCl₂Finally we use the molecular weight of calcium chloride to calculate the required mass:
0.25 moles CaCl₂ * 110.98 g/mol = 27.745 g CaCl₂Rounding to the nearest tenth of a gram the answer is 27.7 g.
To make a 2.5 M stock solution of calcium chloride, you would need to weigh out approximately 276.2 grams of calcium chloride.
Explanation:To make a 2.5 M stock solution of calcium chloride, you need to calculate the number of grams of calcium chloride required. The molecular weight of calcium chloride is 110.98 g/mol. The formula to calculate the number of grams is:
Grams = Moles x Molecular Weight
Since you want to make 100 mL of a 2.5 M solution, you first need to convert mL to moles using the formula:
Moles = Volume (L) x Molarity (M)
Finally, you can substitute the calculated moles into the first formula to find the grams, and round to the nearest tenth of a gram.
Using these calculations, you would need to weigh out approximately 276.2 grams of calcium chloride to make the 2.5 M stock solution.
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Estimate ΔG°rxn for the following reaction at 387 K. HCN (g) + 2 H2 (g) → CH3NH2 (g) ΔH° = −158.0 kJ; ΔS° = −219.9
Answer:
ΔG°rxn = -72.9 kJ
Explanation:
Let's consider the following reaction.
HCN(g) + 2 H₂(g) → CH₃NH₂(g)
We can calculate the standard Gibbs free energy of the reaction (ΔG°rxn) using the following expression:
ΔG°rxn = ΔH° - T.ΔS°
where,
ΔH° is the standard enthalpy of the reaction
T is the absolute temperature
ΔS° is the standard entropy of the reaction
ΔG°rxn = -158.0 KJ - 387 K × (-219.9 × 10⁻³ J/K)
ΔG°rxn = -72.9 kJ
The estimated ΔG°rxn for the given reaction HCN(g) + 2 H2(g) → CH3NH2(g) at 387K, calculated using the equation ΔG = ΔH - TΔS and the given values ΔH° = −158.0 kJ and ΔS° = −219.9 J/K, is -72.86 kJ.
Explanation:To estimate ΔG ° rxn for the given reaction HCN(g) + 2 H2(g) → CH3NH2(g) at 387K, we need to use the equation ΔG = ΔH - TΔS. Given that ΔH° = −158.0 kJ and ΔS° = −219.9 J/K (remember to convert kJ to J, so ΔH is -158000 J), we substitute the values into the equation: ΔG = (-158000 J) - (387K * -219.9 J/K) = -158 kJ + 85.14 kJ = -72.86 kJ. So, the estimated ΔG°rxn for the reaction under these conditions is -72.86 kJ.
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Which of the following correctly represents the transmutation in which a curium-242 nucleus is bombarded with an alpha particle to produce a californium-245 nucleus?^242_96 Cm(^1_0 n, ^4_2 He)^245_98 Cf^242_96 Cm(^4_2 He, ^1_1 p)^245_98 Cf^242_96 Cm(^4_2 He, 2^1_1 p)^245_98 Cf^242_96 Cm(^4_2 He, ^1_0 n)^245_98 Cf^242_96 Cm(^4_2 He, ^1_-1 e)^245_98 Cf
Answer: The chemical equation is written below.
Explanation:
Transmutation is defined as the process in which one chemical isotope gets converted to another chemical isotope. The number of protons or neutrons in the isotope gets changed.
The chemical equation for the reaction of curium-242 nucleus with alpha particle (helium nucleus) follows:
[tex]_{96}^{242}\textrm{Cm}+_4^2\textrm{He}\rightarrow _{98}^{245}\textrm{Cf}+_0^1\textrm{n}[/tex]
The product formed in the nuclear reaction are californium-245 nucleus and a neutron particle.
A mixture of carbon dioxide and hydrogen gases contains carbon dioxide at a partial pressure of 401 mm Hg and hydrogen at a partial pressure of 224 mm Hg. What is the mole fraction of each gas in the mixture?
Answer:
[tex]X_{H_2}=\0.3584[/tex]
[tex]X_{CO_2}=\0.6416[/tex]
Explanation:
According to the Dalton's law of partial pressure, the total pressure of the gaseous mixture is equal to the sum of the pressure of the individual gases.
Partial pressure of carbon dioxide = 401 mmHg
Partial pressure of hydrogen gas = 224 mmHg
Total pressure,P = sum of the partial pressure of the gases = 401 + 224 mmHg = 625 mmHg
Also, the partial pressure of the gas is equal to the product of the mole fraction and total pressure.
So,
[tex]P_{CO_2}=X_{CO_2}\times P[/tex]
[tex]X_{CO_2}=\frac{P_{CO_2}}{P}[/tex]
[tex]X_{CO_2}=\frac{401\ mmHg}{625\ mmHg}[/tex]
[tex]X_{CO_2}=\0.6416[/tex]
Similarly,
[tex]X_{H_2}=\frac{P_{H_2}}{P}[/tex]
[tex]X_{H_2}=\frac{224\ mmHg}{625\ mmHg}[/tex]
[tex]X_{H_2}=\0.3584[/tex]
A phosphate buffer solution (25.00 mL sample) used for a growth medium was titrated with 0.1000 M hydrochloric acid. The components of the buffer were sodium monohydrogenphosphate and sodium dihydrogenphosphate. The first endpoint occurred at a volume of 10.32 mL, and the second occurred after an additional 18.62 mL was added, for a total volume of 28.94 mL. What was the total concentration of phosphate (in any form) in the buffer?
Answer:
0,07448M of phosphate buffer
Explanation:
sodium monohydrogenphosphate (Na₂HP) and sodium dihydrogenphosphate (NaH₂P) react with HCl thus:
Na₂HP + HCl ⇄ NaH₂P + NaCl (1)
NaH₂P + HCl ⇄ H₃P + NaCl (2)
The first endpoint is due the reaction (1), When all phosphate buffer is as NaH₂P form, begins the second reaction. That means that the second endpoint is due the total concentration of phosphate that is obtained thus:
0,01862L of HCl×[tex]\frac{0,1000mol}{L}[/tex]= 1,862x10⁻³moles of HCl ≡ moles of phosphate buffer.
The concentration is:
[tex]\frac{1,862x10^{-3}moles}{0,02500L}[/tex] = 0,07448M of phosphate buffer
I hope it helps!
WILL MARK BRAINLIEST
What pressure (in atm and in bars) is exerted by a column of methanol (CH3OH) 305 m high? The density of methanol is 0.787 g/cm3.
Answer
23.52 atm pressure is exerted by a column of methanol
Explanation:
formula for pressure
Pressure = force / area
And we know that Force F= mass x acceleration (ma)
And mass = volume x density
And volume = cross sectional area x height?
so that...
P = F/A
Putting value of F
P = (mg)/A
Putting value for mass
P = (ρ V) g / A
Putting value for volume
P = ρ (A h) g / A
Canceling A
P = ρgh
Now add the value
now watch the units...
P = (0.787g/cm³) x 9.80 m/s² x 305m... units= g x m² / (cm³ s²)
and if you recall for pressure unit Pressure ...
1 Pa = 1 N / m² = (1 kg m / s²) / m²
so know
[g x m² / (cm³ s²)] x (1 kg / 1000g) x (100 cm / m)³
converts the units to (1 kg m² / s²) / m
= (1 kg m / s²) / m²
Addition of all values
P = (0.787g/cm³) x 9.80 m/s² x 305m x (1 kg / 1000g) x (100 cm / m)³ x (1 atm / 101325 Pa)
Pressure= 23.52 atm
In Part B you calculated ΔG∘ for the following redox reaction:Cu2+(aq)+Co(s)→Cu(s)+Co2+(aq), ΔG∘=−1.18×105JBased on the value of ΔG∘ for the given redox reaction, identify the spontaneity of the reaction.a) nonspontaneousb) at equilibriumc) spontaneous
Answer:
c) spontaneous
Explanation:
According the equation of Gibb's free energy -
∆G = ∆H -T∆S
∆G = is the change in gibb's free energy
∆H = is the change in enthalpy
T = temperature
∆S = is the change in entropy .
And , the sign of the ΔG , determines whether the reaction is Spontaneous or non Spontaneous or at equilibrium ,
i.e. ,
if
ΔG < 0 , the reaction is Spontaneous ΔG > 0 , the reaction is non Spontaneous ΔG = 0 , the reaction is at equilibriumFrom the question ,
The value for ΔG is negative ,
Hence ,
ΔG < 0 , the reaction is Spontaneous
Which of the following statement(s) is/are correct? i) The mass defect is the difference in mass between that of a nucleus and the sum of the masses of its component nucleons. ii) The splitting of a heavier nucleus into two nuclei with smaller mass numbers is known as nuclear fission. iii) The first example of nuclear fission involved bombarding with nuclei. A. i) only B. ii) only C. iii) only D. i) and ii) only E. i) and iii) only F. ii) and iii) only G. i), ii) and iii) H. None of the choices are correct.
Answer:
Correct option -D
Explanation:
Mass defect:
The difference between the sum of the masses of individual nucleons that form an atomic nucleus and the mass of the nucleus.
Hence, i- is correct
Nuclear fission:
It is the process of splitting a nucleus into two nuclei with smaler masses.
Hence, ii- is correct
Nuclear fission involved bombarding with nuclei [tex]^{235}_{92}U[/tex] with [tex]^{4}_{2}He[/tex]
Nuclear fission of [tex]^{235}_{92}U[/tex] with neutrons forms two smaller mass nuclei.
[tex]^{1}_{0}n+^{235}_{92}U \rightarrow ^{91}_{36}Kr+^{142}_{56}B+3^{1}_{0}n+Energy[/tex]
Hence, iii- is incorrect
Therefore, option -D is correct
Statements i) and ii) are correct. Mass defect refers to the difference in mass between that of a nucleus and the sum of the masses of its component nucleons. Nuclear fission refers to the splitting of a heavier nucleus into two nuclei with smaller mass numbers.
Explanation:The correct statement(s) from the ones provided are: i) Mass defect is the difference in mass between that of a nucleus and the sum of the masses of its component nucleons. This is because the mass of a nucleus is not just the sum of the individual masses of the protons and neutrons. Some mass is converted into energy to provide the binding force that holds the nucleus together, so the mass of the nucleus has a slight deficit (or defect) compared to the combined masses of its components.
ii) Nuclear fission involves the splitting of a heavier nucleus into two nuclei with smaller mass numbers. This process usually does not occur naturally, but is induced by bombardment with neutrons. The first reported nuclear fission occurred in 1939 when uranium-235 atoms were bombarded with slow-moving neutrons that split the uranium nuclei into smaller fragments.
Thus, the correct option is D. i) and ii) only.
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The standard cell potential (E°cell) for the reaction below is +1.10V. The cell potential for this reaction is ________ V when the concentration of [Cu2+]=1.0⋅10−5M and [Zn2+]=2.5M. Zn (s) + Cu2+ (aq) → Cu (s) + Zn2+ (aq) The standard cell potential () for the reaction below is . The cell potential for this reaction is ________ when the concentration of and (s) + (aq) (s) + (aq) 0.78 1.10 0.94 1.26 1.42
Answer: 0.94 V
Explanation:
For the given chemical reaction :
[tex]Zn(s)+Cu^{2+}(aq)\rightarrow Cu(s)+Zn^{2+}[/tex]
Using Nernst equation :
[tex]E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Cu^{2+}]}[/tex]
where,
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature = [tex]298K[/tex]
n = number of electrons in oxidation-reduction reaction = 2
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +1.10 V
[tex]E_{cell}[/tex] = emf of the cell = ?
Now put all the given values in the above equation, we get:
[tex]E_{cell}=+1.10-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{2.5}{1.0\times 10^{-5}}[/tex]
[tex]E_{cell}=+1.10-0.16V=0.94V[/tex]
The cell potential for this reaction is 0.94 V
The cell potential for the given reaction depends on the concentration of the reactants. We can use the Nernst equation to calculate the cell potential when the concentration of [Cu2+]=1.0⋅10−5M and [Zn2+]=2.5M.
Explanation:The standard cell potential (E°cell) for the given reaction is +1.10V. To find the cell potential for this reaction when the concentrations of [Cu2+]=1.0⋅10−5M and [Zn2+] = 2.5M, we need to use the Nernst equation:
Ecell = E°cell - (0.0592/n) x log(Q)
Here, n is the number of electrons involved in the reaction, and Q is the reaction quotient, which is equal to [Zn2+]/[Cu2+]. In this case, n = 2. Plugging in the values, we get:
Ecell = 1.10V - (0.0592/2) x log((2.5M)/ (1.0⋅10−5M))
Solving this equation will give us the cell potential for the given concentrations.
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What is/are the product(s) of a neutralization reaction of a carboxylic acid? View Available Hint(s) What is/are the product(s) of a neutralization reaction of a carboxylic acid? a neutral compound and water a carboxylate salt and water a carboxylate salt a base and water g
Answer:
A carboxylate salt and water
Explanation:
A carboxylic acid is an organic compound that has general formula RCOOH, where R is a carbon chain. Because it's an acid, the neutralization will happen when it reacts with a base, such as NaOH.
When this reaction occurs, the base will dissociate in Na⁺ and OH⁻, and the acid will ionize in RCOO⁻ and H⁺, so the products will be RCOO⁻Na⁺ (a carboxylate salt) and H₂O (water).
When 15.5 g of an organic compound known to be 70.58 % C , 5.9 % H , and 23.50 % O by mass is dissolved in 718.1 g of cyclohexane, the freezing point is 3.29 ∘ C . The normal freezing point of cyclohexane is 6.59 ∘ C . What is the molecular formula for the organic compound?Assume that the organic compound is a molecular solid and does not ionize in water. Kf values for various solvents are given here.
Answer:
The molecular formula for the organic compound is C₈H₈O₂
Explanation:
70.58 % C , 5.9 % H , and 23.50 % O
This is a percent by mass, so:
in 100 g of compound I have 70.58 g of C, 5.9 g of H and 23.5 g of O.
We need to apply the formula for freezing point depression, the colligative property.
ΔT = Kf . m
Where ΔT means T°F (pure solvent) − T°F (solution)
Kf, the cryoscopic constant and m is molality (mol of solute in 1kg of solvent); Kf for cyclohexane is 20.8 °C/m
6.59°C - 3.29°C = 20.8 °C/m . m
3.3°C = 20.8 °C/m . m
3.3°C /20.8 m/°C = molal
0.158 m/kg
In 1kg of solvent I have 0.158 moles, but I have 718.1 g of cyclohexane
1000 g _____ 0.158 moles of solute
718.1 g _____ 0.114 moles of solute
To find out the molar mass of my compound I have to divide
mass/mol → 15.5 g /0.114 m = 136.04 g/m
Now I have to work with the percent and think this rules of three:
100 g of compound has ____ 70.58 g of C __5.9 g of H __23.5 g of O
136.04 g of compound has ___ 96.02 g of C __8.02 g of H __ 32 g of O
Molar mass of C = 12 g/m
Molar mass of H = 1 g/m
Molar mass of O = 16 g/m
g / molar mass = moles
C: 96 g / 12 g/m = 8
H: 8 g / 1 g/m =8
O: 32 g/ 16 g/m = 2
The molecular formula for the organic compound is C₈H₈O₂
To find the molecular formula of the organic compound, calculate the empirical formula first, which is CH₃O. Then calculate the molar mass of the compound, which is 29g/mol. Divide the molar mass of the compound by the molar mass of the empirical formula to find the molecular formula ratio. The molecular formula is CH₃O.
Explanation:To determine the molecular formula of the organic compound, we need to calculate the empirical formula first.
The percentages of carbon, hydrogen, and oxygen in the compound represent the mass ratios of these elements. Assume we have 100g of the compound, which means we have 70.58g of carbon, 5.9g of hydrogen, and 23.50g of oxygen. Now we convert these masses to moles using the molar masses: 12g/mol for carbon, 1g/mol for hydrogen, and 16g/mol for oxygen.
Next, we divide the moles of each element by the smallest number of moles to get the simplest mole ratio. The empirical formula of the compound is CH3O.
To find the molecular formula, we need to know the molar mass of the compound. This can be calculated by adding the atomic masses of the elements in the empirical formula: 12g/mol for carbon, 1g/mol for hydrogen, 16g/mol for oxygen. The sum is 29g/mol.
Since the molar mass of the compound is 29g/mol and 15.5g of the compound is dissolved in 718.1g of cyclohexane, we can calculate the moles of the compound: 15.5g / 29g/mol = 0.534moles.
The molecular formula is found by multiplying the empirical formula by an integer value to match the molar mass. To do this, divide the molar mass of the compound by the molar mass of the empirical formula: 29g/mol / 29g/mol = 1.
Therefore, the molecular formula of the organic compound is CH3O.
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Zinc reacts with hydrochloric acid according to the reaction equation Zn ( s ) + 2 HCl ( aq ) ⟶ ZnCl 2 ( aq ) + H 2 ( g ) How many milliliters of 5.50 M HCl (aq) are required to react with 6.25 g Zn (s) ?
Answer:
34.7mL
Explanation:
First we have to convert our grams of Zinc to moles of zinc so we can relate that number to our chemical equation.
So: 6.25g Zn x (1 mol / 65.39 g) = 0.0956 mol Zn
All that was done above was multiplying the grams of zinc by the reciprocal of zincs molar mass so our units would cancel and leave us with moles of zinc.
So now we need to go to HCl!
To do that we multiply by the molar coefficients in the chemical equation:
[tex]\frac{0.0956g Zn}{1 } (\frac{2 mol HCl}{1molZn})[/tex]
This leaves us with 2(0.0956) = 0.1912 mol HCl
Now we use the relationship M= moles / volume , to calculate our volume
Rearranging we get that V = moles / M
Now we plug in: V = 0.1912 mol HCl / 5.50 M HCl
V= 0.0347 L
To change this to milliliters we multiply by 1000 so:
34.7 mL
When aqueous solutions of potassium fluoride and hydrochloric acid are mixed, an aqueous solution of potassium chloride and hydrofluoric acid results. Write the net ionic equation for the reaction. (Use the solubility rules provided in the OWL Preparation Page to determine the solubility of compounds.)
Answer:
F⁻(aq) + H⁺(aq) ⇄ HF(aq)
Explanation:
When aqueous solutions of potassium fluoride and hydrochloric acid are mixed, an aqueous solution of potassium chloride and hydrofluoric acid results. The corresponding molecular equation is:
KF(aq) + HCl(aq) ⇄ KCl(aq) + HF(aq)
The full ionic equation includes all the ions and the molecular species. HF is a weak acid so it exists mainly in the molecular form.
K⁺(aq) + F⁻(aq) + H⁺(aq) + Cl⁻(aq) ⇄ K⁺(aq) + Cl⁻(aq) + HF(aq)
The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the molecular species.
F⁻(aq) + H⁺(aq) ⇄ HF(aq)
The decomposition reaction of N2O5 in carbon tetrachloride is 2N2O5−→−4NO2+O2. The rate law is first order in N2O5. At 64 °C the rate constant is 4.82×10−3s−1. (a) Write the rate law for the reaction. (b) What is the rate of reaction when [N2O5]=0.0240M? (c) What happens to the rate when the concentration of N2O5 is doubled to 0.0480 M? (d) What happens to the rate when the concentration of N2O5 is halved to 0.0120 M?
Answer:
(a) rate = 4.82 x 10⁻³s⁻¹ [N2O5]
(b) rate = 1.16 x 10⁻⁴ M/s
(c) rate = 2.32 x 10⁻⁴ M/s
(d) rate = 5.80 x 10⁻⁵ M/s
Explanation:
We are told the rate law is first order in N₂O₅, and its rate constant is 4.82 x 10⁻³s⁻¹ . This means the rate is proportional to the molar concentration of N₂O₅, so
(a) rate = k [N2O5] = 4.82 x 10⁻³s⁻¹ x [N2O5]
(b) rate = 4.82×10⁻³s⁻¹ x 0.0240 M = 1.16 x 10⁻⁴ M/s
(c) Since the reaction is first order if the concentration of N₂O₅ is double the rate will double too: 2 x 1.16 x 10⁻⁴ M/s = 2.32 x 10⁻⁴ M/s
(d) Again since the reaction is halved to 0.0120 M, the rate will be halved to
1.16 x 10⁻⁴ M/s / 2 = 5.80 x 10⁻⁵ M/s
Answer:
a) r = 4.82x10⁻³*[N2O5]
b) 1.16x10⁻⁴ M/s
c) The rate is doubled too (2.32x10⁻⁴ M/s)
d) The rate is halved too (5.78x10⁻⁴ M/s)
Explanation:
a) The rate law of a generic reaction (A + B → C + D) can be expressed by:
r = k*[A]ᵃ*[B]ᵇ
Where k is the rate constant, [X] is the concentration of the compound X, and a and b are the coefficients of the reaction (which can be different from the ones of the chemical equation).
In this case, there is only one reactant, and the reaction is first order, which means that a = 1. So, the rate law is:
r = k*[N2O5]
r = 4.82x10⁻³*[N2O5]
b) Substituing the value of the concentration in the rate law:
r = 4.82x10⁻³*0.0240
r = 1.16x10⁻⁴ M/s
c) When [N2O5] = 0.0480 M,
r = 4.82x10⁻³*0.0480
r = 2.32x10⁻⁴ M/s
So, the rate is doubled too.
d) When [N2O5] = 0.0120 M,
r = 4.82x10⁻³*0.0120
r = 5.78x10⁻⁴ M/s
So, the rate is halved too.
Write complete ionic and net ionic equations for the reaction between sulfuric acid (H2SO4) and calcium carbonate (CaCO3).
H2SO4(aq) + CaCO3(s) —> H2O(I) + CO2(g) + CaSO4(aq)
Answer:
Complete ionic equation:
2H²⁺(aq) + SO₄²⁻(aq) + CaCO₃(s) → H₂O(l) + CO₂(g) + Ca²⁺(aq) + SO₄²⁻(aq)
Net ionic equation:
2H²⁺(aq) + CaCO₃(s) → H₂O(l) + CO₂(g) + Ca²⁺(aq)
Explanation:
Chemical equation:
H₂SO₄(aq) + CaCO₃(s) → H₂O(l) + CO₂(g) + CaSO₄(aq)
Balanced chemical equation:
H₂SO₄(aq) + CaCO₃(s) → H₂O(l) + CO₂(g) + CaSO₄(aq)
Complete ionic equation:
2H²⁺(aq) + SO₄²⁻(aq) + CaCO₃(s) → H₂O(l) + CO₂(g) + Ca²⁺(aq) + SO₄²⁻(aq)
Net ionic equation:
2H²⁺(aq) + CaCO₃(s) → H₂O(l) + CO₂(g) + Ca²⁺(aq)
Answer:
Complete ionic equation: 2 H⁺(aq) + SO₄²⁻(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + Ca²⁺(aq) + SO₄²⁻(aq)
Net ionic equation: 2 H⁺(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + Ca²⁺(aq)
Explanation:
The molecular equation includes all the species in the molecular form.
H₂SO₄(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + CaSO₄(aq)
The complete ionic equation includes all the ions and the molecular species.
2 H⁺(aq) + SO₄²⁻(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + Ca²⁺(aq) + SO₄²⁻(aq)
The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the molecular species.
2 H⁺(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + Ca²⁺(aq)
What is the mass of 3 L of water vapor at 120◦C and 388 torr?
Answer:
Explanation:
Use gas equation to calculate mol of H2O gas:
P = pressure = 389/760 = 0.512atm
V = volume = 4L
n=???
R = 0.082057
T = 121+273 = 394K
0.512 * 4 = n*0.082057*394
n = (0.512*4) / (0.082057*394)
n = 2.048 / 32.33
n = 0.063 mol H2O
Molar5 mass H2O = 18g/mol
Mass of water = 0.063*18
Mass of water valour = 1.14g
What is the percentage composition of each element in hydrogen peroxide, H2O2?
A)7.01% H and 92.99% O
B)7.22% H and 92.78% O
C)6.32% H and 93.68% O
D)5.88% H and 94.12% O
Answer:
The percentage composition of each element in H2O2 is 5.88% H and 94.12% O (Option D).
Explanation:
Step 1: Data given
Molar mass of H = 1.0 g/mol
Molar mass of O = 16 g/mol
Molar mass of H2O2 = 2*1.0 + 2*16 = 34.0 g/mol
Step 2: Calculate % hydrogen
% Hydrogen = ((2*1.0) / 34.0) * 100 %
% hydrogen = 5.88 %
Step 3: Calculate % oxygen
% Oxygen = ((2*16)/34)
% oxygen = 94.12 %
We can control this by the following equation
100 % - 5.88 % = 94.12 %
The percentage composition of each element in H2O2 is 5.88% H and 94.12% O (Option D).
The percentage composition of hydrogen peroxide (H₂O₂) is D) 5.88% H and 94.12% O.
The calculation can be represented as follows:
1) Molar masses:
H = 1.008 g/mol O = 15.999 g/mol2) Molar mass of [tex]H_2O_2:[/tex]
[tex]$\ce{H_2O_2} = (2 \times 1.008) + (2 \times 15.999) = 34.014$ g/mol[/tex]3) Percentage of hydrogen:
[tex]\[ \% \ce{H} = \frac{\text{mass of H}}{\text{total mass}} \times 100\% = \frac{2 \times 1.008}{34.014} \times 100\% \approx 5.88\% \][/tex]4) Percentage of oxygen:
[tex]\[ \% \ce{O} = 100\% - \% \ce{H} = 100\% - 5.88\% = 94.12\% \][/tex]Therefore, the percentage composition of [tex]$\ce{H_2O_2}$[/tex] is approximately 5.88% H and 94.12% O.
Comparing with the given options, we can see that the closest match is D) 5.88% H and 94.12% O
The overall reaction in a commercial heat pack can be representedas shown below.
4 Fe(s) + 3 O2(g) 2Fe2O3(s) ΔH = -1652 kJ
(a) How much heat (kJ) is released when4.48 mol iron is reacted with excessO2?
(b) How much heat is released when 1.76 mol Fe2O3 isproduced?
(c) How much heat is released when 2.66 g iron is reacted with excessO2?
(d) How much heat is released when 12.8 g Fe and 1.49 gO2 are reacted?
Explanation:
[tex]4 Fe(s) + 3O_2(g)\rightarrow 2Fe_2O_3(s) ,\Delta H = -1652 kJ[/tex]
a) Heat released when 4.48 moles of iron is reacted with excessive oxygen:
According to reaction, when 4 moles of an iron reacts with 3 moles of oxygen it gives 1625 kilo Joules of heat.
Then 4.48 moles of iron will give:
[tex]\frac{1652 kJ}{4}\times 4.48=1850.24 kJ[/tex]
1850.24 kJ of heat is released when 4.48 moles of an iron is reacted with excess oxygen.
b) Heat released when 1.76 mol [tex]Fe_2O_3[/tex] is produced.
According to reaction, when 2 moles of an [tex]Fe_2O_3[/tex] are produced 1625 kilo Joules of heat is released
Then heat released on production of 1.76 mol [tex]Fe_2O_3[/tex] :
[tex]\frac{1652 kJ}{2}\times 1.76=1453.76 kJ[/tex]
1453.76 kJ heat is released when 1.76 mol [tex]Fe_2O_3[/tex] is produced.
c) Heat released when 2.66 grams of iron is reacted with excessive oxygen:
Moles of iron = [tex]\frac{2.66 g}{56 g/mol}=0.0475 mol[/tex]
According to reaction, when 4 moles of an iron reacts with 3 moles of oxygen it gives 1625 kilo Joules of heat.
Then 0.0475 moles of iron will give:
[tex]\frac{1652 kJ}{4}\times 0.0475=19.62 kJ[/tex]
19.62 kJ of heat is released when 2.66 grams of iron is reacted with excessive oxygen.
d) Heat released when 12.8 g Fe and 1.49 g oxygen gas are reacted:
Moles of iron = [tex]\frac{12.8 g}{56 g/mol}=0.2286 mol[/tex]
According to reaction, 4 moles of iron reacts with 3 moles of oxygen.Then 0.2286 mol will react with:
[tex]\frac{3}{4}\times 0.2286 mol=0.17145 mol[/tex] of oxygen
Moles of oxygen = [tex]\frac{1.49 g}{56 g/mol}=0.04656 mol[/tex]
According to reaction, 3 moles of oxygen gas reacts with 4 moles of iron .Then 0.04656 mol will react with:
[tex]\frac{4}{3}\times 0.04656 mol=0.06208 mol[/tex] of iron.
From this we can conclude that oxygen is in limiting amount. So, the amount of energy release will depend upon moles of oxygen gas.
According to reaction, when 4 moles of an iron reacts with 3 moles of oxygen it gives 1625 kilo Joules of heat.
Then 0.04656 moles of oxygen gas will give:
[tex]\frac{1652 kJ}{3}\times 0.04656 =25.7011 kJ[/tex]
25.7011 kJ of Heat is released when 12.8 g Fe and 1.49 g oxygen gas are reacted
The heat released in an exothermic reaction can be calculated using stoichiometry based on the ΔH value. The released heat varies with the amount of reactant or product involved. By calculating the molar ratios, the heat released is computed for each sub-question accordingly.
Explanation:The reaction you provided loosens heat, hence it is an exothermic reaction. The heat released can be calculated using stoichiometry if we know the amount of reactants or products. ΔH (-1652 kJ) is the amount of heat released for the reaction of 4 moles of Fe with 3 moles of O2 to form 2 moles of Fe2O3.
When 4.48 moles of iron (Fe) reacts with an excess of O2, the heat released would be in direct proportion. So, heat released = (4.48 / 4) * -1652 = -1848 kJ.When 1.76 mole of Fe2O3 is produced, heat released = (1.76 / 2) * -1652 = -1456 kJ.When 2.66 g of iron is reacted, convert it to moles (2.66g / 55.85g/mol = 0.048 moles). Hence, the heat released in this case = (0.048 / 4) * -1652 = -19.92 kJ.When 12.8 g of Fe and 1.49 g O2 are reacted, convert both into moles and use the reactant which is in lesser amount to calculate the heat. In this case, it's O2 (1.49g / 32g/mol = 0.0466 moles). The heat released = (0.0466 /(3/2)) * -1652 = -51.40 kJ.Learn more about Heat Release in Exothermic Reaction here:https://brainly.com/question/1047086
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Identify the precipitate or lack thereof for the following:CaCl2(aq) + K2CO3(aq) --> CaCO3 + 2KClA) CaCO3B) KClC) No precipitateFeCl2(aq) + (NH4)2SO4(aq) --> FeSO4 + 2NH4ClA) FeSO4B) NH4ClC) No precipitate
Answer:
A) CaCO₃
C) No precipitate
Explanation:
To answer these questions we need to consider the solubility rules.
Identify the precipitate or lack thereof for the following:
CaCl₂(aq) + K₂CO₃(aq) ⇒ CaCO₃(s) + 2 KCl(aq)
Group II carbonates are insoluble. Thus, CaCO₃ is insoluble.Salts containing Group I cations are soluble. Thus, KCl is soluble.FeCl₂(aq) + (NH₄)₂SO₄(aq) ⇒ FeSO₄(aq) + 2 NH₄Cl(aq)
Most sulfates are soluble. Thus, FeSO₄ is soluble.Salts containing the ammonium ion are soluble. Thus, NH₄Cl is soluble.A chemist prepares a solution of mercury(II) iodide HgI2 by measuring out 0.0067μmol of mercury(II) iodide into a 350.mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in /μmolL of the chemist's mercury(II) iodide solution. Round your answer to 2 significant digits.
Final answer:
The concentration of the chemist's mercury(II) iodide solution is 0.019 μmol/L after rounding to two significant digits.
Explanation:
The concentration of the chemist's mercury(II) iodide solution can be calculated using the formula for molarity, which is the number of moles of solute divided by the volume of the solution in liters.
In this case, we are given 0.0067μmol of HgI2 and a volume of 0.350 L:
Molarity (μM) = amount of solute (in micromoles) / volume of solution (in liters)Molarity (μM) = 0.0067μmol / 0.350 L = 0.01914μmol/L
After rounding to two significant digits, the concentration of the solution is 0.019 μM.
Glucose (C6H12O6) is a key nutrient for generating chemical potential energy in biological systems. We were provided 16.55 g of glucose. Please calculate:
a) The mass percent of carbon in glucose.
b) The mass of CO2 produced by the combustion of 16.55 g glucose with sufficient oxygen gas.
c) How many oxygen molecules needed for the completely combustion of 16.55 g glucose?
Answer:
a) 40 %
b) [tex]4.04~g~CO_2[/tex]
c) [tex]5.53x10^2^3~molecules~of~O_2[/tex]
Explanation:
For a) we will have to calculate the molar mass of [tex]C_6H_1_2O_6[/tex], so the first step is to find the atomic mass of each atom and multiply by the amount of atoms in the molecule.
C => 12*(6) = 72
H => 1*(12) = 12
O => 6*(16) = 96
Molar mass = 180 g/mol
Then we can calculate the percentage by mass:
[tex]Percentage~=~\frac{72}{180}*100=40[/tex]
For b) we have to start with the reaction of glucose:
[tex]C_6H_1_2O_6~+~6O_2~->~6CO_2~+~6H_2O[/tex]
Then we have to convert the grams of glucose to moles, the moles of glucose to moles of carbon dioxide and finally the moles of carbon dioxide to grams. To do this we have to take into account the following conversion ratios:
-) 180 g of glucose = 1 mol glucose
-) 1 mol glucose = 6 mol carbon dioxide
-) 1 mol carbon dioxide = 44 g carbon dioxide
[tex]16.55~g~C_6H_1_2O_6\frac{1~mol~C_6H_1_2O_6}{180~g~C_6H_1_2O_6}\frac{6~mol~CO_2}{1~mol~C_6H_1_2O_6}\frac{44~g~CO_2}{1~mol~CO_2}=4.04~g~CO_2[/tex]
For C, we have to start with the conversion from grams of glucose to moles, the moles of glucose to moles of oxygen and finally the moles of oxygen to molecules. To do this we have to take into account the following conversion ratios:
-) 180 g of glucose = 1 mol glucose
-) 1 mol glucose = 6 mol oxygen
-) 1 mol oxygen = 6.023x10^23 molecules of O2
[tex]16.55~g~C_6H_1_2O_6\frac{1~mol~C_6H_1_2O_6}{180~g~C_6H_1_2O_6}\frac{6~mol~O_2}{1~mol~C_6H_1_2O_6}\frac{6.023x10^2^3~molecules~O_2}{1~mol~O_2}=~5.53x10^2^3~molecules~of~O_2[/tex]
This detailed answer explains the mass percent of carbon in glucose, the mass of CO2 produced by its combustion, and the number of oxygen molecules needed for complete combustion.
a) Mass Percent of Carbon in Glucose: Glucose has a molar mass of 180.16 g/mol. The molar mass of the carbon in one mole of glucose is 72.06 g (6 carbons in a molecule of glucose). Therefore, the mass percent of carbon in glucose is 40%.
b) Mass of CO2 Produced: The combustion of 16.55 g of glucose will produce 44.01 g of CO2.
c) Oxygen Molecules Needed: For the complete combustion of 16.55 g of glucose, 23 molecules of O2 are required.
Metals with ________ electron configurations characteristically form diamagnetic, square planar complexes.
a. d6
b. d8
c. d10
d. d0
e. d9
Answer:
B
Explanation:
Diamagnetism , paramagnetism and ferromagnetism are terms which are used to describe the magnetic activity of transition metals. These terms helps to know the response the metal will have when placed in a magnetic field. These activities can be discerned from the d-block electronic configuration. If the number of electrons are even, this means they all form a pair and the metal is diamagnetic. If otherwise, there is an unpaired electron which causes the paramagnetic activity of the metal in magnetic field.
To the question, options A-D are diamagnetic but configurations with d8 are the ones that form square planar complexes