Answer:
Any element in group 18 has eight valence electrons (except for helium, which has a total of just two electrons
The oceanic crust is destroyed at convergent boundaries because it_______________________ *
a) diverges to form a rift valley
b) goes into a subduction zone where it is melted by the hot magma
c) transforms
d) the rock crumbles at an ocean ridges
Answer: I believe the answer is d) the rock crumbles at an ocean ridges
Explanation:
Calculate the maximum work available from 50.0 g of aluminum in the following cell when the cell potential is 1.15 V. Al(s) |Al3+(aq) || H+(aq) | O2(g) |Pt. Note that O2 is reduced to H2O.
Answer:
W = 615.91 kJ
Explanation:
We need to use the following expression:
W = ΔG° * mol (1)
But in order to determine the ΔG° we need the following expression:
ΔG° = -n * F * E° (2)
Where F is a constant and is 96,500 J/V mol
n is the number of transferred electrons in the reaction. As we are passing from Al to Al³⁺ we can say that the number of electrons are 3.
Finally to get the moles, we need the the atomic weight of aluminum which is 26.98 g/mol, so the moles:
moles = m/MM (3)
Let's calculate the moles of aluminum:
moles = 50 / 26.98 = 1.85 moles of aluminum
Now let's calculate the gibbs energy using (2):
ΔG° = -3 * 96,500 * 1.15
ΔG° = -332,925 J or simply -332.925 kJ/mol
Finally, using (1) we can determine the work done:
W = 332.925 * 1.85
W = 615.91 kJHope this helps
The maximum work available from 50.0 g of aluminum in the following cell when the cell potential is 1.15 V will be 615.91 kJ
What is cell potential ?The cell potential, Ecell, is the measure of the potential difference between two half cells in an electrochemical cell.
Expression for work done ;
W = ΔG° x mol (1)
But, to determine the ΔG°
ΔG° = -n x F x E° (2)
Where,
F is a constant and is 96,500 J/V moln is the number of transferred electrons in the reaction.(As there is variation from Al to Al³⁺ the change in number of electrons are 3)
To get the moles,
moles = m/MM (3)
moles = 50 / 26.98 = 1.85 moles of aluminum
Now,
ΔG° = -3 x 96,500 x 1.15
ΔG° = -332,925 J
= -332.925 kJ/mol
Now, using (1) we can determine the work done:
W = 332.925 x 1.85
W = 615.91 kJ
Hence, The maximum work available from 50.0 g of aluminum in the following cell when the cell potential is 1.15 V will be 615.91 kJ
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For many purposes we can treat butane as an ideal gas at temperatures above its boiling point of . Suppose the pressure on a sample of butane gas at is cut in half. Is it possible to change the temperature of the butane at the same time such that the volume of the gas doesn't change? yes no If you answered yes, calculate the new temperature of the gas. Round your answer to the nearest °C.
Answer:
A. Yes
B. The new temperature of the gas is -116 °C
Note: The question is incomplete. The complete question is given below :
For many purposes we can treat butane C H10) as an ideal gas at temperatures above its boiling point of - 1. °C. Suppose the pressure on a 500 mL sample of butane gas at 41.0°C is cut in half. Iyes Is it possible to change the temperature of the butane at the same time such that the volume of the gas doesn't change? yes no If you answered yes, calculate the new temperature of the gas. Round your answer to the nearest °C.
Explanation:
A. According to the pressure law of gases,for a fixed mass of gas the pressure of a gas is directly proportional to its Kelvin temperature once the volume is kept constant. This means that a change in temperature can bring about a change in pressurein a gas at constant volume.
B. From the pressure law of gasese: P1/T1 = P2/T2
Where initial pressure = P1, final pressure = P2
Initial temperature = T1, final temperature = T2
For the butane gas;
P1 = P
P2 = P/2
T1 = 41°C = (273 + 41 ) K = 314 K
T2 = ?
From the equation, T2 = T1 × P2 / P1
T2 = 314 × P/2 /P
T2 = 157 K
T2 = (157 - 273) °C = -116 °C
Therefore, the new temperature of the gas is -116 °C
Metal plating is done by passing current through a metal solution. For example, an item can become gold plated by attaching the item to a power source and submerging it into a Au3 solution. The item itself serves as the cathode, at which the Au3 ions are reduced to Au(s). A piece of solid gold is used as the a node and is also connected to the power source, thus completing the circuit.
Required:
What mass of gold is produced when 15.0A of current are passed through a gold solution for 15.0min ?
Answer:
9.18g
Explanation:
Step 1: Write the reduction half-reaction
Au³⁺(aq) + 3 e⁻ ⇒ Au(s)
Step 2: Calculate the mass of gold is produced when 15.0A of current are passed through a gold solution for 15.0min
We will use the following relationships:
1 min = 60 s1 A = 1 C/s1 mole of electrons has a charge of 96486 C (Faraday's constant).1 mole of Au is produced when 3 moles of electrons circulate.The molar mass of Au is 196.97 g/mol.The mass of gold produced is:
[tex]15.0 min \times \frac{60s}{1 min} \times \frac{15.0C}{1s} \times \frac{1 mol e^{-} }{96486C} \times \frac{1molAu}{3 mol e^{-} } \times \frac{196.97gAu}{1molAu} = 9.18gAu[/tex]
A gas has a volume of 6.00 liters at a temperature of 300 K and a pressure of 1.00 atm what is the volume of the gas in liters at a temperature of 600 K and a pressure of 3.00 atm
Answer:
V₂ = 4.00 L
Explanation:
Given that:
Volume (v1) = 6.00 L
Temperature (T1) = 300 K
Pressure (P1) = 1.00 atm
VOlume (V2) = unknown???
Temperature (T2) = 600 K
Pressure (P2) = 3.00 atm
Using combined gas law equation:
[tex]\dfrac{P_1V_1}{T_1}= \dfrac{P_2V_2}{T_2}[/tex]
[tex]\dfrac{1 \times 6}{300} = \dfrac{3 \times V_2}{600}[/tex]
[tex]\dfrac{1}{50} = \dfrac{1 \times V_2}{200}[/tex]
200 = 50V₂
V₂ = 200/50
V₂ = 4.00 L
Calculate the kinetic energy of a mole of oxygen gas molecules that have a speed
of 10.0 m/s.
Answer:
1600J
Explanation:
1 mole of oxygen gas (O2) has a mass of 32g i.e. the molar mass = 32g/mol
Kinetic energy (K.E) = ½ × m × v²
Where;
m = mass (g)
v = speed or velocity (m/s)
From the information given in this question;
m = 32g
V = 10m/s²
K.E = ½ × 32 × 10²
K.E = 16 × 100
K.E = 1600J
Qué uso le darías en la vida diaria ayuda
Answer:
ako po kailangan ko po nga ayuda
Explanation:
thanks for the points ❤️
Which substance can be identified as a homogeneous mixture?
chocolate milk
pepperoni pizza
chocolate chip cookie
peanut butter sandwich
Answer:
A) Chocolate Milk
Explanation:
A homogeneous mixture is one that is made of one substance. Technically, none of these are, but chocolate milk would be the closest. Pizza has cheese, pepperoni, and crust. Chocolate chip cookies have chips and the cookie. Peanut butter sandwiches have peanut butter and bread. Therefore, chocolate milk is right, as it has no other parts.
by what process does water vapor become a cloud?
a. evaporation
b. transpiration
c. condensation
d. precipitation
8. Sulfur has a first ionization energy of 1000 kJ/mol. Photons of what frequency are required to ionize one mole of Sulfur?
Answer:
the frequency of photons [tex]v = 1.509\times10^{39}Hz[/tex]
Explanation:
Given: first ionization energy of 1000 kJ/mol.
No. of moles of sulfur = 1 mole
[tex]\Delta E_1 = 1000KJ/mol[/tex]
We know that plank's constant
[tex]h = 6.626\times10^{-34} Js[/tex]
Let the frequency of photons be ν
Also we know that ΔE = hν
this implies ν = ΔE/h
[tex]= \frac{10^6J}{6.626\times10^{-34} Js}[/tex]
[tex]v = 1.509\times10^{39}Hz[/tex]
Hence, the frequency of photons [tex]v = 1.509\times10^{39}Hz[/tex]
A sample of air was collected on a day when the total atmosphere pressure was
592 mmHg. The sample contained only oxygen and nitrogen gas. If the oxygen in the sample had a pressure of 261 mmHg, how much pressure did the nitrogen
have? NOTE: You must show your calculation on the attached scratch paper,
including which of the Gas Law formulas you used. *
A. 853 mm Hg
B. 0.206 mm Hg
C. 4.76 mm Hg
D. 331 mm Hg
(Please show your work)
Answer:
1. The gas law used: Dalton's law of partial pressure.
2. Pressure of nitrogen = 331 mmHg
Explanation:
From the question given above, the following data were obtained:
Total pressure (Pₜ) = 592 mmHg
Pressure of Oxygen (Pₒ) = 261 mmHg
Pressure of nitrogen (Pₙ) =?
The pressure of nitrogen in the sample can be obtained by using the Dalton's law of partial pressure. This is illustrated below:
Pₜ = Pₒ + Pₙ
592 = 261 + Pₙ
Collect like terms
592 – 261 = Pₙ
331 = Pₙ
Pₙ = 331 mmHg
Therefore, the pressure of nitrogen in the sample is 331 mmHg
Suppose that you must go into court and testify as to the level of X in the river water. Your lawyer is concerned that your analytical results are lower than they should be because you cannot extract with benzene (distribution Ratio K=4) all of X from your samples. Assume the concentration of X in the 1.00 L river water sample is 1.06 x 10-4 M.
a. How many extracts would be required to remove all but 1000 molecules of X from the sample with repeated 10.00 mL portions of extract?
b. How many extractions are required for a 99.99% chance of extraction of the last molecule remaining (0.01% chance of a single molecule left behind)
Answer:
The appropriate solution is:
(a) n ≈ 900
(b) n ≈ 1165
Explanation:
According to the question,
(a)
The final number of molecules throughout water will be:
= [tex](\frac{1000}{1000}\times 4\times 10 )^n[/tex]
where, n = number of extractions
Now,
The initial number of molecules will be:
= [tex]1.06\times 10^{-4}\times 6.023\times 10^{23}[/tex]
= [tex]6.387\times 10^{19}[/tex]
Final number of molecule,
⇒ [tex]1.566\times 10^{-16}=(\frac{1000}{1040} )^n[/tex]
[tex]n \approx 900[/tex]
(b)
Final molecules of X = left (0.01%)
hence,
⇒ [tex]initial = 6.384\times 10^{19}[/tex]
[tex]\frac{1}{6.384\times 10^{19}} =(\frac{1000}{1040} )^2[/tex]
[tex]n \approx 1165[/tex]
PLZ ANSWER QUICK I AM IN TIMER HELP WILL GIVE BRAINLIEST
Which is greater, the moon's period of rotation or its period of revolution?
They are equal.
the moon's revolution period around Earth
Neither are known.
the moon's rotational period
Answer:
B
Explanation:
HELP ME PLEASE I AM STRUGGLING
maybe umm think so pick think so
Why are fossil fuels considered nonrenewable resources?
Answer:
They are considered nonrenewable because they are prehistoric animals that has been extinct therefore can't be renew.
Use the scenario to answer the following question. A group of students working in a chemistry lab are planning a procedure to neutralize 10.0mL of 5.0 M hydrochloric acid (strong acid) with 5.0 M potassium hydroxide (strong base). In their procedure they plan on adding an equal volume of the base to the acid. What would be the expected outcome of carrying out this step as planned?
a- The resulting pH will be less than 7 because potassium hydroxide is less concentrated than the hydrochloric acid.
b- The resulting pH will not be able to be determined because the concentrations of the acid and base are not the same.
c- The resulting pH will be greater than 7 because potassium hydroxide is less concentrated than the hydrochloric acid.
d- The resulting pH will be equal to 7 because a strong base will neutralize a strong acid.
Answer:
d- The resulting pH will be equal to 7 because a strong base will neutralize a strong acid.
Explanation:
The reaction between potassium hydroxide and hydrochloric acid of equal volume and equal concentration yields a solution of pH 7 at equivalence point. We must note that KOH is a strong base while HCl is a strong acid. This fact influences the pH of the system at equivalence point.
Owing to the fact that the acid is exactly neutralized by the base; at the equivalence point of such titration, it is expected that hydrogen ions(H+) and hydroxide ions (OH-) must have reacted to form water, this leads to a final pH of 7.
For the following reaction conditions select the correct statement regarding the reaction below.
OH Cl OTs OTs N TsCl W X Y Z N
a. W is the correct product.
b. X is the correct product.
c. Y is the correct product.
d. Z is the correct product since pyridine acts as a nucleophile.
e. Both X and Y are both formed in this reaction as a racemic mixture
Answer:
B
Explanation:
The appropriate diagram of the question is shown in the first image attached below.
From the diagram, we see the reaction of Cyclopentanol taking place under Tscl pyridine. We are to show the reaction mechanism and determine from the options, which appropriate product fits in.
So, from the reaction, the hydroxyl substituent reacts with Tscl where cl is being lost. This process is followed by an attack of N substituent on the pyridine with the Hydrogen atom and cleaves off for the structure to form a stable structure. The stereochemistry of the compound remains unchanged and it maintains its stick formula.
Thus, X is the appropriate and the correct product.
In this experiment, you will need to prepare 250.0 mL of 0.100 M KCl(aq) solution. Determine the mass in grams of potassium chloride required to prepare this solution. Type your numerical answer (no units) in the box below. Even though the number of significant figures is limited to three by the inputs, please report your answer to four significant figures.
Answer: The mass in grams of potassium chloride required to prepare this solution is 1.862 grams
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]
where,
n = moles of solute
[tex]V_s[/tex] = volume of solution in ml
moles of [tex]KCl[/tex] = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{xg}{74.5g/mol}[/tex]
Now put all the given values in the formula of molality, we get
[tex]0.100=\frac{x\times 1000}{74.5g/mol\times 250.0ml}[/tex]
[tex]x=1.862g[/tex]
Therefore, the mass in grams of potassium chloride required to prepare this solution is 1.862 grams
What're the differences between oxidation and reduction
Answer:
Oxidation:Oxidation is the loss of electron or Hydrogen atoms or gain of Oxygen atomsRemoval or loss of electronsA ⟶ A⁺ + e⁻
Removal of HydrogenAH + B ⟶ A + BH
Addition of OxygenA + BO ⟶ AO + B
All the above reactions release energy Reduction:Reduction is the gain of electron or Hydrogen atoms or loss of Oxygen atomsAddition or gain of electronsA + e⁻ ⟶ A⁻
Addition of HydrogenA + BH ⟶ AH + B
Removal of OxygenAO + B ⟶ A + BO
All the above reactions stores energy-TheUnknownScientist
Calculate the concentration of NO2NO2 in an equilibrium mixture given that the concentration of N2O4=0.200MN2O4=0.200M and Keq=67.3Keq=67.3 for the reaction: 2NO2(g)⇌N2O4(g)2NO2(g)⇌N2O4(g). Calculate the concentration of in an equilibrium mixture given that the concentration of and for the reaction: . 18.3 M 0.00297 M 337 M 0.0545 M none of the above
Answer: The concentration of [tex]NO_{2}[/tex] in the given equilibrium mixture is 0.0545 M.
Explanation:
The ratio of concentration of products and reactants raised to the power of their coefficients is called equilibrium constant. The symbol used to denote equilibrium constant is [tex]K_{eq}[/tex].
As the given reaction equation is as follows.
[tex]2NO_{2}(g) \rightleftharpoons N_{2}O_{4}(g)[/tex]
The expression for equilibrium constant of this reaction is as follows.
[tex]K_{eq} = \frac{[N_{2}O_{4}]}{[NO_{2}]^{2}}[/tex]
Now, substitute the given values into above formula as follows.
[tex]K_{eq} = \frac{[N_{2}O_{4}]}{[NO_{2}]^{2}}\\67.3 = \frac{(0.2)}{[NO_{2}]^{2}}\\So, [NO] = \sqrt{\frac{0.2}{67.3}}\\= 0.0545 M[/tex]
Thus, we can conclude that the concentration of [tex]NO_{2}[/tex] in the given equilibrium mixture is 0.0545 M.
What is the name of Na
- Standard state refers to
a) 1 atm and 25°C
b) 1 atm and 100°C
c) 1 atm and gaseous reactants
d) None of these
Answer: The standard state refers to 1 atm and [tex]25^{o}C[/tex].
Explanation:
It is known that a chemical/substance can either be present in a solid, liquid or gaseous state.
So, when the phase of a substance like solid, liquid or gas is present at 1 atmosphere pressure and at a temperature of [tex]25^{o}C[/tex] then it known as standard state of substance.
Thus, we can conclude that standard state refers to 1 atm and [tex]25^{o}C[/tex].
Why would you want to slow down a chemical reaction?
Answer:
We often want to decrease the rates of certain reactions rather than speeding them up. For example, to prolong the shelf lives of certain foods, the chemical reactions by which they spoil must be slowed down.
which shows a disaccharide
Answer: It is :B:
Explanation: also sub to technoblade
_____ the irregular break of a mineral.
a. Habit
b. arrangement
c. cleavage
d. Fracture.
Answer:
D Fracture
Explanation:
jus took the test
How many atoms are there in 24 moles of Fe?
Answer: There are [tex]145 \times 10^{23}[/tex] atoms in 24 moles of Fe.
Explanation:
According to the mole concept, 1 mole of a substance contains [tex]6.022 \times 10^{23}[/tex] atoms.
Therefore, atoms present in 24 moles of Fe are calculated as follows.
[tex]24 mol \times 6.022 \times 10^{23} atoms/mol\\= 144.52 \times 10^{23} atoms\\or, 145 \times 10^{23} atoms[/tex]
Thus, we can conclude that there are [tex]145 \times 10^{23}[/tex] atoms in 24 moles of Fe.
A solution is prepared by mixing 2.50g of CaCl2 with 50.0g H2O, what is the mass percent of CaCl2?
A. 3.76%
B. 4.76%
C. 5.76%
D. 6.76%
Write the balanced half-reaction that occurs at the anode in a lead-acid (storage) battery during discharge. Phases are optional. anode half-reaction: Write the balanced half-reaction that occurs at the cathode in a lead-acid (storage) battery during discharge. Phases are optional. cathode half-reaction: Write the balanced overall cell reaction that occurs in the lead-acid (storage) battery during discharge. Phases are optional. overall cell reaction:
Answer: Anode: [tex]Pb+SO_4^{2-}\rightarrow PbSO_4+2e^-[/tex]
Cathode: [tex]PbO_2+4H^++SO_4^{2-}+2e^-\rightarrow PbSO_4+2H_2O[/tex]
Overall cell reaction : [tex]Pb+2SO_4^{2-}+PbO_2+4H^+\rightarrow 2PbSO_4+2H_2O[/tex]
Explanation:
Lead storage battery is a secondary cell used in automobiles and invertors. The anode is made up of lead and undergoes oxidation during discharging and cathode is made up of lead oxide and acts as cathode during discharging. The electrolyte used is dilute .
Lead storage battery acts as electrochemical cell while discharging.
Discharging reaction for Anode:
Discharging reaction for Cathode: [tex]PbO_2+4H^++SO_4^{2-}+2e^-\rightarrow PbSO_4+2H_2O[/tex]
Overall cell reaction : [tex]Pb+2SO_4^{2-}+PbO_2+4H^+\rightarrow 2PbSO_4+2H_2O[/tex]
A lead storage battery is an energy storage device. At the anode lead and sulfate ion reacts to produce Lead(II) sulfate and release two electrons.
What are anode and cathode?In secondary cells or the recharge-discharge cells during recharge, the positive electrode is the anode, while during discharge cathode is the positive electrode.
The anode (lead) of the cell undergoes an oxidation reaction during the discharge, whereas the lead oxide or the cathode undergoes reduction.
The discharging reaction at the anode of the cell is given as,
[tex]\rm Pb + SO_{4}^{2-} \rightarrow PbSO_{4} + 2e^{-}[/tex]
The discharging reaction at the cathode of the cell is given as,
[tex]\rm PbO_{2} + 4H^{+} + SO_{4}^{2-} + 2 e^{-} \rightarrow PbSO_{4} + 2H_{2}O[/tex]
The overall cell reaction is given as,
[tex]\rm Pb + 2SO_{4}^{2-} + PbO_{2} + 4H^{+} \rightarrow 2PbSO_{4} + 2H_{2}O[/tex]
Therefore, the gain and loss of electrons are represented at the cathode and the anode of the cell.
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8. Calculate (H^+), (OH^-), pOH and the pH for a 0.00024 M solution of calcium hydroxide. Must show work!
Answer:
1. [H⁺] = 2.0×10¯¹¹ M
2. [OH¯] = 4.8×10¯⁴ M
3. pOH = 3.3
4. pH = 10.7
Explanation:
From the question given above, the following data were obtained obtained:
Molarity of Ca(OH)₂ = 0.00024 M
We'll begin by calculating the concentration of the hydroxide ion [OH¯]. This can be obtained as follow:
Ca(OH)₂ (aq) —> Ca²⁺ + 2OH¯
From the balanced equation above,
1 mole of Ca(OH)₂ produced 2 moles of OH¯.
Therefore, 0.00024 M Ca(OH)₂ will produce = 2 × 0.00024 = 4.8×10¯⁴ M OH¯
Thus, the concentration of the hydroxide ion [OH¯] is 4.8×10¯⁴ M
Next, we shall determine the pOH of the solution. This can be obtained as follow:
Concentration of the hydroxide ion [OH¯] = 4.8×10¯⁴ M
pOH =?
pOH = –Log [OH¯]
pOH = –Log 4.8×10¯⁴
pOH = 3.3
Next, we shall determine the pH of the solution. This can be obtained as follow:
pOH = 3.3
pH =?
pH + pOH = 14
pH + 3.3 = 14
Collect like terms
pH = 14 – 3.3
pH = 10.7
Finally, we shall determine the concentration of hydrogen ion [H⁺]. This can be obtained as follow:
pH = 10.7
Concentration of hydrogen ion [H⁺] =?
pH = –Log [H⁺]
10.7 = –Log [H⁺]
Divide both side by –1
–10.7 = Log [H⁺]
Take the antilog of –10.7
[H⁺] = Antilog (–10.7)
[H⁺] = 2.0×10¯¹¹ M
SUMMARY:
1. [H⁺] = 2.0×10¯¹¹ M
2. [OH¯] = 4.8×10¯⁴ M
3. pOH = 3.3
4. pH = 10.7
How do scientists determine the names for so many different living things
Answer:
Taxonomy
Explanation: