Entropy is INCREASING when a snowman melts, a document goes through paper shredder, silver tarnishes, while it is DECREASING when dissolved sugar precipitates, water vapor forms droplets and water cools down.
Entropy can be defined as the degree of randomness or disorder of a particular system.
Entropy is equal to zero (0) for a perfectly ordered system.
Heat increases the entropy of the system because more energy excites the molecules and it increases the amount of random activity.
Moreover, the cooling decreases the entropy of the system because molecules are more ordered and it decreases the amount of random activity.
In conclusion, entropy is INCREASING when a snowman melts, a document goes through paper shredder, silver tarnishes, while it is DECREASING when dissolved sugar precipitates, water vapor forms water droplets and the water cools down.
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When electrons are filling energy levels, the lowest energy sublevels are occupied first. This is ____________.
When electrons are filling energy levels, the lowest energy sublevels are occupied first. This is Hund's rule.
Hund's rules state that:
Every orbital in a sublevel has to be singularly occupied before any other orbital is able to be doubly occupied.
All of the electrons in single occupied orbitals have to have the same spin to maximize the total spin.
Determine whether each of the following salts will form a solution that is acidic, basic, or pH-neutral. Drag the appropriate items to their respective bins.
Al(NO3)3
C2H5NH3NO3
NaClO
RbI
CH3NH3CN
Answer:
Al(NO₃)₃: Acidic.
C₂H₅NH₃NO₃: Acidic.
NaClO: Basic
RbI: pH-neutral
CH₃NH₃CN: Solution basic
Explanation:
The general rules to determine if a solution is acidic, basic or neutral are:
If it is a salt of a strong acid and base, the solution will be pH-neutral. If it is a salt of a strong acid and a weak base, the solution will be acidic due to the hydrolysis of the weak base component (cation). If it is a salt of a strong base and a weak acid, the solution will be basic due to the hydrolysis of the weak acid component (anion).For the salts:
Al(NO₃)₃. The repective acid is HNO₃ (Strong acid) and the base is Al(OH)₃ (Weak base). As the salt comes from strong acid and weak base. SOLUTION ACIDIC
C₂H₅NH₃NO₃. The acid is HNO₃ (Strong acid) and the base C₂H₅NH₃OH (Weak base). SOLUTION ACIDIC.
NaClO. Tha acid is HClO (weak acid), and the base NaOH (Strong base). SOLUTION BASIC.
RbI: The acid is HI (Strong acid) and the base RbOH (Strong base). pH-NEUTRAL
CH₃NH₃CN. The acid is HCN (weak acid; pKb = 4.79) and the base CH₃NH₃OH (weak base; pKa = 10.64). Both weak acid and base will produce each hydrolisis. The lower pK will predominate. That is the weak acid. SOLUTION BASIC
Solution of Al(NO₃)₃ and C₂H₅NH₃NO₃ salts is acidic, NaClO is basic and of RbI & CH₃NH₃cyanide is neutral in nature.
What is pH?pH of any solution tells about the acidity or basicity of the solution, pH of any solution ranges from 0 to 14 and from acidity to basicity.
Al(NO₃)₃ is a salt which is formed by the mixing of strong acid HNO₃ (Nitric acid) and weak base Al(OH)₃, so the resultant solution of the salt is acidic in nature.C₂H₅NH₃NO₃ salt is formed by the mixing of strong acid HNO₃ (Nitric acid) and weak base C₂H₅NH₃OH, so the resultant solution of the salt is acidic in nature.NaClO is a salt of weak acid is HClO and strong base NaOH, so the resultant solution of the salt is basic in nature.RbI salt is formed by the combination of strong acid HI and strong base RbOH, so the resultant solution of the salt is neutral in nature.CH₃NH₃Cyanide is a salt of weak acid hydrogen cyanide and weak base CH₃NH₃OH, so the resultant solution of the salt is neutral in nature.Hence, appropriate differentiation was done above.
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Which of the following is an alkaline earth metal?
A. Silicon (Si)
B. Magnesium (Mg)
C. Carbon (C)
D. Aluminum (AI)
Answer:
B
Explanation:
The alkaline earth metals are the elements located in Group 2. The only element out of our choices that is in Group 2 is Magnesium.
3. There are many different primary standards that could be used in a standardization titration. What are the criteria for a primary standard
Answer:
High purity.
Stability (low reactivity)
Low hygroscopicity (to minimize weight changes due to humidity)
Explanation:
There are different primary standards that could be used in a standardization titration in order to achieve the best and accurate result possible. These standards include high purity,stability and low hygoscropicity .
A high purity means the reactants lack impurities which could affect the result. Stability also ensures that there is non reactivity with elements/compounds in the atmosphere while low hygroscopicity ensures weight changes are minimized due to humidity.
write the IUPAC name OF THE FOLLOWING COMPOUNDS
Answer:
Explanation:
a) 2 chloro butane
b) 2-3 dimethyl butane
c) 2 bromo 3 nitro pentane
d) 2-3 trimethyl pentane
e) 2-bromo,3-methyl,4-nitro hexane
f) 2-methyl cyclo butane
A reaction has the rate law, rate = k[A]3[B]. Which change will cause the greatest increase in the reaction rate?
Answer:
Increasing the concentration of A will cause the greatest change over the rate.
Explanation:
Hello,
In this case, considering the given rate law, which is third-order with respect to A, changing its concentration, the rate will be significantly modified. For instance, suppose a concentration of A and B of 1M and a symbolic rate constant (k), this causes the rate to be:
[tex]r=k[1M]^3[1M]=1k\frac{M^4}{s}[/tex]
Then, if we change the concentration of A to 2 M holding the concentration of B in 1 M, the new rate constant will be:
[tex]r=k[2M]^3[1M]=8k\frac{M^4}{s}[/tex]
Nevertheless, if we hold the concentration of A in 1 M and the concentration of B is now 2 M (same change), the new rate constant is:
[tex]r=k[1M]^3[2M]=2k\frac{M^4}{s}[/tex]
It means that increasing the concentration of A will cause the greatest change over the rate.
Best regards.
The change in concentration of A will cause the greatest increase in the reaction rate.
Rate of reaction :Given reaction has rate law ,
[tex]rate=k[A]^{3} [B][/tex]
which is third-order with respect to concentration of A
From rate law equation, It is observed that the degree of concentration of A is three.
It means that small change in concentration of A result large change in rate of reaction.Thus, the change in concentration of A will cause the greatest increase in the reaction rate.Learn more about the rate of reaction here :
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A calibration curve constructed from absorbance values of solutions containing a known concentration of permanganate ions has the following best-fit line:
y = (3.62× 10^3 L/mol) x
where y is the absorbance of the solution at 525 nm and x is the concentration of MnO4- (aq) in mol/L. The path length of the cuvettes used in the experiment is 1 cm. Based on this information, what is the molar absorptivity of MnO4- (aq) at 525 nm?
Answer:
3.62×10³ L/mol
Explanation:
Beer-Lambert law relates the absorbance of a sample and its concentration. Its formula is:
A = ε×C×l
Where A is absorbance of the sample, ε is molar absorptivity (A constant f each sample), C its concentration and l is path length
Now, the formula obtained was:
y = (3.62×10³ L/mol) x
Where Y ia absorbance = A, x its concentration = C and 1cm is path length.
You can write:
A = (3.62×10³ L/mol)×C×l
That means, molar absorptivity of your sample under the meaured conditions is:
3.62×10³ L/mol
Calculate the molality of a solution prepared by dissolved 19.9 g of kcl in 750ml of water
First, we find the molar mass of KCl which is AK+ACl=39+35.5=74.5g/mole
now we find out the number of moles by dividing the given mass to the molar mass n=m/M=19.9/74.5=0.26 moles. The molarity of a solution is equal to the number of moles divided by the volume of the solution. =0.26moles/0.75liters=0.346M
Generally, systems move spontaneously in the direction of increasing entropy. TRUE FALSE
Answer:
true
Explanation:
2
22. A sodium chloride solution is 15.0% m/m%. Calculate mass of sodium chloride in 219 g solution.
14.2g
80.38
11.2 g
32.9 g
Answer: The mass of sodium chloride in 219 g solution is 32.9 g
Explanation:
To calculate the mass percent of element in a given compound, we use the formula:
[tex]\text{Mass percent of A}=\frac{\text{Mass of A}}{\text{mass of A +mass of B}}\times 100[/tex]
To find mass of sodium chloride in solution:
[tex]\text{Mass percent of sodium chloride}=\frac{\text{Mass of sodium chloride}}{\text{mass of solution}}\times 100[/tex]
Mass percent of sodium chloride= 15.0 %
Mass of solution = 219g
[tex]15=\frac{\text{Mass of sodium chloride}}{219}\times 100[/tex]
[tex]{\text{Mass of sodium chloride}=32.9g[/tex]
Thus mass of sodium chloride in 219 g solution is 32.9 g
Identify the energy transformations in the following actions.
a. Turning on a space heater
b. Dropping an apple core into the garbage
c. Climbing up a rope ladder
d. Starting a car
e. Turning on a flashlight
Think about each action as a before and after. For instance, before the space heater is turned on, what kind of energy is there? After the space heater is turned on, what happens? This change from before to after will help you identify the two types of energy involved.
Identify five more types of energy transformations that you see at home, at school, or outdoors. Make sure to name the action, such as turning on a light, as well as the two types of energy involved. Remember, for energy to be transformed, the type of energy before and after a task must be different.
Answer:
to. Turn on a heater (electrical energy that is transformed into heat energy)
yes. Throwing an apple core in the trash (chemical energy that is transformed into nutritional energy for decomposers)
C. Climbing a rope ladder (chemical energy from our food into mechanical energy that allows us to climb the ladder)
re. Starting a car (electrochemical energy in mechanical energy)
me. Turning on a flashlight (chemical energy from the battery into light energy)
Five daily actions that exemplify the transformation of energy:
Do physical activity (change of metabolic energy in mechanics)
Turn on a heater (electrical energy that is transformed into heat energy)
Lighting a stove (Chemical energy product of combustion that results in the transformation of that energy into heat)
Turn on a cell phone (chemical energy characteristic of the battery that is transformed into sound and light energy)
Riding a bicycle (nutritional energy or energy of the metabolism that is transformed into mechanical energy)
Explanation:
The energy is never lost, it is always transformed.
It is one of the great principles of physics and is the reason why the universe is infinite.
The volume of ammonia gas at 1.14 atm of pressure is gradually decreased from 90.3 mL to 43.4 mL. What is the final pressure of ammonia if there is no change in temperature
Answer:
THE FINAL PRESSURE OF AMMONIA IF THERE IS NO CHANGE IN TEMPERATURE AND A DECREASE IN VOLUME FROM 90.3 mL TO 43.4 mL IS 2.91 atm.
Explanation:
At constant temperature, the pressure of a given mass of gas is inversely proportional to the volume. This question follows Boyle's law of gas laws.
Mathematically written as:
P1V1 = P2V2
Re-arranging the formula by making P2 the subject of the formula;
P2 = P1V1 / T2
P1 = 1.4 atm
V1 = 90.3 mL
V2 = 43.4 mL
P2 = unknown
So therefore, we have:
P2 = 1.4 * 90.3 / 43.4
P2 = 2.91 atm
The final pressure of ammonia is therefore 2.91 atm.
Answer:
2.37 atm
Explanation:
Step 1: Given data
Initial pressure of ammonia (P₁): 1.14 atmInitial volume of ammonia (V₁): 90.3 mLFinal pressure of ammonia (P₂): ?Final volume of ammonia (V₂): 43.4 mLConstant temperatureStep 2: Calculate the final pressure of ammonia
Since the temperature is kept constant, we can calculate the final pressure of ammonia using Boyle's law.
[tex]P_1 \times V_1 = P_2 \times V_2\\P_2 = \frac{P_1 \times V_1}{V_2} = \frac{1.14atm \times 90.3mL}{43.4mL} = 2.37 atm[/tex]
BEARINGS
Question 12 (SSCE 1994 May/June)
(a) A village P is 10km from a lorry station, Q, on
a bearing 065º. Another village R, is 8km from
Q on a bearing 155º. Calculate
(i) the distance of R from P to the nearest
kilometer
(ii) the bearing of R from P to the nearest degree
(b) M is a village on PR such that QM is
perpendicular to PR. Find the distance of M
from P to the nearest kilometer.
Answer:
a. (i) the distance of R from P is 13 Km to the nearest kilometer
(ii) the bearing of R from P = 360 - (38 + 25) = 297° to the nearest degree
b. the distance of m from P is 11 Km to the nearest kilometre
Explanation:
a) A triangle PQR is formed. Q = 90°, p = 8 km; r = 10 km; distance of R from P is q is to be found.
(i) Using the cosine rule: q² = p² + r² - 2prCosQ
q² = 8² + 10² - 2 * 8 * 10 * Cos90
q² = 64 + 100 + 0
q² = 164
q = 13 Km to the nearest kilometre
Therefore, the distance of R from P is 13 Km to the nearest kilometer
(ii) the bearing of R from P
The angle at P is found using the formula Cos P = (q² + r² - p²)/2qr
Cos P = 13² + 10² - 8²/2 * 13 *10
Cos P = 0.7884
P = Cos⁻¹ 0.7884
P = 38°
Therefore, the bearing of R from P = 360 - (38 + 25) = 297° to the nearest degree
Note : 25° is alternate (Northwest) to 65°at P
b) A right-angled triangle QMP is formed
using the trigonometrical ratios; cos Θ = adjacent/hypotenuse
where the hypotenuse side = 10 km, adjacent side = distance of M from P, x
cos P = x/10
x = cos 38 * 10
x = 11 Km to the nearest kilometre
Therefore, the distance of m from P is 11 Km to the nearest kilometre
g Air contains nitrogen, oxygen, argon, and trace gases. If the partial pressure of nitrogen is 592 mm Hg, oxygen is 160 mm Hg, argon is 7 mm Hg, and trace gas is 1 mm Hg, what is the atmospheric pressure
Answer:
Explanation:
Atmospheric pressure = partial pressure of nitrogen + partial pressure of oxygen + partial pressure of argon + partial pressure of trace element
putting the given values
Atmospheric pressure = 592 + 160 + 7 + 1
= 760 mm of Hg .
A glass cylinder contains 2 gases at a pressure of 106 kPa. If one gas is at 7 kPa, what is the pressure of attributed to the other gas? a) 9 kPa b) 99 kPa c) 113 kPa d) 7 kPa e) 2 kPa (URGENT)
Answer:
b) 99 kPa
Explanation:
According to Daltons law of partial pressure, the total pressure of a mixture of two or more non reactive gases is the sum of their individual pressures. Let the total pressure of a mixture of n number of gases be [tex]P_{total}[/tex] and their individual pressure be [tex]P_1,P_2,P_3,\ .\ .\ .\ ,\ P_n[/tex], According to Daltons partial pressure law:
[tex]P_{total}=P_1+P_2+P_3+.\ .\ .+P_n[/tex]
Since A glass cylinder contains 2 gases at a pressure of 106 kPa, therefore n = 2. Also one gas ([tex]P_1[/tex]) is at 7 kPa. Using Daltons partial pressure law:
[tex]P_{total}=P_1+P_2+P_3+.\ .\ .+P_n\\P_{total}=P_1+P_2\\106\ kPa=7\ kPa+P_2\\P_2=106\ kPa-7\ kPa\\P_2=99\ kPa[/tex]
According to the collision theory, all collisions do not lead to reaction. Which choice gives both reasons why not all collisions between reactant molecules lead to reaction? 1. The total energy of two colliding molecules is less than some minimum amount of energy. 2. Molecules cannot react with each other unless a catalyst is present. 3. Molecules that are improperly oriented during collision will not react. 4. Solids cannot react with gases.
Answer: 3. Molecules that are improperly oriented during collision will not react.
Explanation:
According to the collision theory , the number of collisions that take place per unit volume of the reaction mixture is called collision frequency. The effective collisions are ones which result into the formation of products.
It depends on two factors:-
1. Energy factor: For collision to be effective, the colliding molecules must have energy more than a particular value called as threshold energy.
2. Orientation factor: Also the colliding molecules must have proper orientation at the time of collision to result into formation of products.
Thus not all collisions between reactant molecules lead to reaction because molecules that are improperly oriented during collision will not react.
How long should you hold the iron on the hair to heat the strand and set the base ?
A) 5 seconds
B) 15 seconds
C) 30 seconds
D) 1 minute
Charcoal from the dwelling level of the Lascaux Cave in France gives an average count of 0.97 disintegrations of ^14 C per minute per gram of sample. Living wood gives 6.68 disintegrations per minute per gram. Estimate the date of occupation and hence the probable date of the wall painting in the Lascaux Cave. Hint: Disintegrations per minute per gram" has the same units as the time-derivative of concentration for a radioactive decay model. (You may use the fact that the half-life of ^14C is 5568 years.)
Answer:
Explanation:
count given by old sample = .97 disintegrations per minute per gram
count given by fresh sample = 6.68 disintegrations per minute per gram
Half life of radioactive carbon = 5568 years
rate of disintegration
dN / dt = λ N
In other words rate of disintegration is proportional to no of radioactive atoms present . As number reduces rate also reduces .
Let initial no of radioactive be N₀ and after time t , number reduces to N
N₀ / N = 6.68 / .97
Now
[tex]N=N_0e^{-\lambda t}[/tex]
[tex]\frac{N}{N_0} =e^{-\lambda t}[/tex]
[tex]\frac{6.68}{.97} = e^{\lambda t}[/tex]
λ is disintegration constant
λ = .693 / half life
= .693 / 5568
= .00012446 year⁻¹
Putting the values in the equation above
[tex]\frac{6.68}{.97} = e^{.00012446\times t}[/tex]
[tex]6.8866 = e^{.00012446\times t}[/tex]
1.929577 = .00012446 t
t = 15503.6 years .
Select the correct answer from each drop-down menu.
Pictures that we receive from space are of the
✓ because it takes time for
to reach Earth.
Answer:
Pictures that we receive from space are of the past because it takes time for light to reach Earth.
Explanation:
For example, Mars is so far away that, depending on its position in orbit, a picture from Mars takes between 4 min and 24 min to reach Earth.
Answer: Pictures that we receive from space are of the
past
because it takes time for
light
to reach Earth.
Explanation:
The final overall chemical equation is Upper Ca upper O (s) plus upper C upper O subscript 2 (g) right arrow upper C a upper C upper O subscript 3 (s).. When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed. is halved. has its sign changed. is unchanged.
Answer:
the enthalpy of the second intermediate equation is halved and has its sign changed.
Explanation:
Let us take a look at the first and second intermediate reactions as well as the overall reaction equation for the process under review;
First reaction;
Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ
Second reaction;
2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ
Hence the overall equation is now;
CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH = ?
According to the Hess law of constant heat summation, the enthalpy of the overall reaction is supposed to be obtained as a sum of the enthalpy of both reactions but this will not give the enthalpy of the overall reaction in this case. The enthalpy of the overall reaction is rather obtained by halving the enthalpy of the second intermediate reaction and reversing its sign before taking the sum as shown below;
Enthalpy of Intermediate reaction 1 + ½(- Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction
Answer:
A.
Explanation:
Did it on Edge.
An amphoteric salt is one that contains an anion that can act as either an acid or a base in water. Baking soda, NaHCO3, is an example. By combining the ionization and hydrolysis reactions of the anion, you get the principle reaction that occurs when this salt is dissolved in water.
2HCO3-(aq) CO32-(aq) + H2CO3(aq)
The pH for such, a solution is given by
pH = pKa1 + pKa2/2
where Ka1 and Ka2 are the ionization constants of the acid (in this case, H2CO3). Note that the pH of the solution is independent of the salt concentration. Calculate the pH of a NaHCO3 solution.
Answer:
pH = 8.34
Explanation:
The equilbriums of the amphoteric HCO₃⁻ (Ion of NaHCO₃) are:
H₂CO₃ ⇄ HCO₃⁻ + H⁺ Ka1 -Here, HCO₃⁻ is acting as a base-
HCO₃⁻⇄ CO₃²⁻ + H⁺ Ka2 -Here, is acting as an acid-
Where Ka1 = 4.3x10⁻⁷ and Ka2 = 4.8x10⁻¹¹. As pKa = -log Ka:
pKa1 = 6.37; pKa2 = 10.32
As the pH of amphoteric salts is:
pH = (pKa1 + pKa2) / 2
pH = 8.34How many molecules of CaCl2 are equivalent to 75.9g CaCl2 (Ca=40.08g/mol, CL=35.45g/mol)
Answer:
[tex]\large \boxed{4.12 \times 10^{23}\text{ formula unis of CaCl}_{2}}$}[/tex]
Explanation:
You must calculate the moles of CaCl₂, then convert to formula units of CaCl₂.
1. Molar mass of CaCl₂
CaCl₂ = 40.08 + 2×35.45 = 40.08 + 70.90 = 110.98 g/mol
2. Moles of CaCl₂ [tex]\text{Moles of CaCl}_{2} = \text{75.9 g CaCl}_{2} \times \dfrac{\text{1 mol CaCl}_{2}}{\text{110.98 g CaCl}_{2}} = \text{0.6839 mol CaCl}_{2}[/tex]
3. Formula units of CaCl₂
[tex]\text{No. of formula units} = \text{0.6839 mol CaCl}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules CaCl}_{2}}{\text{1 mol P$_{2}$O}_{5}}\\\\= \mathbf{4.12 \times 10^{23}}\textbf{ formula units CaCl}_{2}\\\text{There are $\large \boxed{\mathbf{4.12 \times 10^{23}}\textbf{ formula units of CaCl}_{2}}$}[/tex]
Convert the following measurement
Answer:
6.9 Kg/mol•dL
Explanation:
To convert 6.9×10⁴ g/mol•L to kg/mol•dL,
First, we shall convert to kg/mol•L.
This can be achieved by doing the following:
Recall: 1 g = 1×10¯³ Kg
1 g/mol•L = 1×10¯³ Kg/mol•L.
Therefore,
6.9×10⁴ g/mol•L = 6.9×10⁴× 1×10¯³
6.9×10⁴ g/mol•L = 69 Kg/mol•L
Finally, we shall convert 69 Kg/mol•L to Kg/mol•dL.
This is illustrated below:
Recall: 1 L = 10 dL
1 Kg/mol•L = 1×10¯¹ Kg/mol•dL
Therefore,
69 Kg/mol•L = 69 × 1×10¯¹
69 Kg/mol•L = 6.9 Kg/mol•dL
Therefore, 6.9×10⁴ g/mol•L is equivalent to 6.9 Kg/mol•dL.
An atom with 19 protons and 18 neutrons is a(n)
A. Isotope of potassium(K)
B. Standard atom of argon(Ar)
C. Standard atom of (K)
D. Isotope of argon (Ar)
Answer:
A
Explanation:
The number of protons indicates the element so we know it's potassium. To get the number of neutrons you subtract the number of protons (19) from the mass number which for potassium is 39.
39-19=20 neutrons
Because you have 18 neutrons then yours would be an isotope.
Answer: A. Isotope of potassium(K)
Explanation: Founders Educere answer.
A chemist studies the reaction below. 2NO(g) + Cl2(g) 2NOCl(g) He performs three experiments using different concentrations and measures the initial reaction rates (The data from the three experiments is in the table). 1. Write the rate law 2. Solve for k.
Answer:
1. [tex]Rate =k [NO]^{2}[Cl_{2}][/tex]
2. [tex]k= 0.42 \frac{L^{2}}{mol^{2}*s}[/tex]
Explanation:
[tex]Rate =k [NO]^{m}[Cl_{2}]^{n}[/tex]
[tex]Rate1 = k[0.4]^{m}[0.3]^{n}=0.02\\Rate 2=k [0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate1}{Rate2}=\frac{0.02}{0.08} =\frac{k[0.4]^{m}[0.3]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{1}{4} =(\frac{1}{2} )^{m},\\m=2[/tex]
[tex]Rate3 =k [0.8]^{m}[0.6]^{n}=0.16\\Rate 2= k[0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate3}{Rate2}=\frac{0.16}{0.08} =\frac{k[0.8]^{m}[0.6]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{2}{1} =(\frac{2}{1} )^{n},\\n=1[/tex]
[tex]Rate =k [NO]^{2}[Cl_{2}]^{1}[/tex]
[tex]Rate =k [NO]^{2}[Cl_{2}]^{1}\\Rate 1=k [0.4]^{2}[0.3]^{1} =0.02\\k*0.16*0.3=0.02\\k=\frac{0.02}{0.16*0.3}=\frac{1}{8*(\frac{3}{10} )}=\frac{5}{12} = 0.42 \frac{L^{2}}{mol^{2}*s}[/tex]
Amanda is doing a report for her Earth Science class about the four seasons. Which of the following would be an effective scientific model to incorporate in her project? a. A calendar indicating the first days of autumn, winter, spring, and summer b. Four earth models, each with a different tilt to represent the earth’s position relative to the sun, lined up in order of season c. A poster board with pictures of weather characteristic of the four seasons d. A bar graph indicating average high and low temperatures for Amanda’s area in the autumn, winter, spring, and summer Please select the best answer from the choices provided
Answer:
(B)
Explanation:
edg 2020
The four seasons in earth is originating from the earth's revolution around the sun. Therefore, the most suitable model for Amanda is Four earth models, each with a different tilt to represent the earth’s position relative to the sun, lined up in order of season.
What are seasons?Seasons in earth is originating from the difference in the distance from the sun over each time period. Hence, revolution of earth around sun make these seasons.
The time period at which earth comes closer to sun more hot will be earth's surface and we experience summer season. When we far from sun winter season occurs.
Therefore, different season are coming based on the distance of earth from sun at each revolution point. This is also affected by the tilt of earth's in its own axis.
Hence, different earth models with different distance from sun is most suitable model here for Amanda. Thus option B is correct.
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Determine the volume occupied by 10 mol of helium at 27 ° C and 82 atm
please.
Answer:
3.00 L
Explanation:
Convert the pressure to Pascals.
P = 82 atm × (101325 Pa/atm)
P = 8,308,650 Pa
Convert temperature to Kelvins.
T = 27°C + 273
T = 300 K
Use ideal gas law:
PV = nRT
(8,308,650 Pa) V = (10 mol) (8.314 J/mol/K) (300 K)
V = 0.00300 m³
If desired, convert to liters.
V = (0.00300 m³) (1000 L/m³)
V = 3.00 L
Answer:
[tex]\large \boxed{\text{3.0 L}}[/tex]
Explanation:
[tex]\begin{array}{rcl}pV &=& nRT\\\text{82 atm} \times V & = & \text{10 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{300.15 K}\\82V & = & \text{246 L}\\V & = & \textbf{3.0 L} \\\end{array}\\\text{The volume of the balloon is $\large \boxed{\textbf{3.0 L}}$}[/tex]
Which of the following sets of quantum numbers is NOT allowed? a. n = 5, l= 4, ml= –2, ms = +1/2 b. n = 2, l = 1, ml= 0, ms = +1/2 c. n = 4, l= 4, ml= 0, ms = –1/2 d. n = 3, l= 2, ml= –1, ms = +1/2 e. n = 2, l= 0, ml= 0, ms = -1/2
Answer:
C
Explanation:
the n value must always be greater than the l value
Out of the following set of quantum numbers ,set C is not allowed as the azimuthal and principal quantum numbers are same.
What are quantum numbers?Quantum numbers are the numbers which describe the values of conserved quantities with respect to the dynamics of a quantum system.They correspond to the Eigen values of operators which commute with the Hamiltonian quantities.
The Hamiltonian quantities can be known with precision simultaneously with the system's energy.Quantum numbers can take values of discrete sets of integers or even half-integers even though they can approach infinity in some cases.
They can specifically describe energy levels of electrons, and can also explain angular momentum,spin,etc.These are used to describe the path of an electron in an atom ,when the quantum numbers of all atoms are combined they must comply with the Schrodinger equation.
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Use your periodic table and calculator as needed for the following question.
How much stock solution is needed to make 250 mL of a 6.0M solution. The molarity of the stock solution is 18M.
Selections may be rounded so choose the best answer.
56 mL
83 mL
2.3 mL
4.7 ml
For a single substance at atmospheric pressure, classify the following as describing a spontaneous process, a nonspontaneous process, or an equilibrium system.
A) Solid melting below its melting point
B) Gas condensing below its condensation point
C) Liquid vaporizing above its boiling point
D) Liquid freezing below its freezing point
E) Liquid freezing above its freezing point
F) Solid melting above its melting point
G) Liquid and gas together at boiling point with no net condensation or vaporization
H) Gas condensing above its condensation point
I) Solid and liquid together at the melting point with no net freezing or melting
Answer:
Spontaneous process- This is the process that occurs on its own without the application of any external energy or other factor. They include
B) Gas condensing below its condensation point
C) Liquid vaporizing above its boiling point
D) Liquid freezing below its freezing point
F) Solid melting above its melting point
Non spontaneous - This is the process that doesn’t occurs on its own and requires the application of any external energy or factor. They include
A) Solid melting below its melting point
E) Liquid freezing above its freezing point
H) Gas condensing above its condensation point
Equilibrium system
G) Liquid and gas together at boiling point with no net condensation or vaporization
I) Solid and liquid together at the melting point with no net freezing or melting
A) Solid melting below its melting point - nonspontaneous process
B) Gas condensing below its condensation point - spontaneous process
C) Liquid vaporizing above its boiling point - spontaneous process
D) Liquid freezing below its freezing point - spontaneous process
E) Liquid freezing above its freezing point - nonspontaneous process
F) Solid melting above its melting point - spontaneous process
G) Liquid and gas together at boiling point with no net condensation or vaporization - Equilibrium system
H) Gas condensing above its condensation point - nonspontaneous process
I) Solid and liquid together at the melting point with no net freezing or melting - Equilibrium system
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