Gaseous ethane CH3CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 0.60 g of ethane is mixed with 3.52 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 2 significant digits.

Answers

Answer 1

Answer:

1.8 g

Explanation:

Step 1: Write the balanced equation

CH₃CH₃(g) + 3.5 O₂(g) ⇒ 2 CO₂(g) + 3 H₂O(g)

Step 2: Determine the limiting reactant

The theoretical mass ratio of CH₃CH₃ to O₂ is 30.06:112.0 = 0.2684:1.

The experimental mass ratio of CH₃CH₃ to O₂ is 0.60:3.52 = 0.17:1.

Thus, the limiting reactant is CH₃CH₃

Step 3: Calculate the mass of CO₂ produced

The theoretical mass ratio of CH₃CH₃ to O₂ is 30.06:88.02.

0.60 g CH₃CH₃ × 88.02 g CO₂/30.06 g CH₃CH₃ = 1.8 g


Related Questions

công thức phân tử của glucozo

Answers

C₆H₁₂O₆ is the molecular formula of gulcozo.

Calculate the mass percent of each component in the following solution.
159 g NiCl2 in 500 g water
% Nicla
% water

Answers

Answer:

% NiCl2 = 24.13%

% water = 78.57%

Explanation:

Mass percentage = mass of solute/mass of solution × 100

According to this question, a solution contains 159 g of NiCl2 in 500 g of water. Hence, mass of the solution is calculated as follows:

Mass of solution = 159g + 500g

Mass of solution = 659g

Therefore;

A) % Mass of NiCl2 in solution = mass of NiCl2/mass of solution × 100

% Mass of NiCl2 in solution = 159/659 × 100

% Mass of NiCl2 in solution = 0.2413 × 100

= 24.13%

B) % Mass of water in solution = mass of water/mass of solution × 100

% Mass of water in solution = 500/659 × 100

% Mass of water in solution = 0.7587 × 100

% Mass of water in solution = 75.87%

In the reoxidation of QH2 by purified ubiquinone-cytochrome c reductase (Complex III) from heart muscle, the overall stoichiometry of the reaction requires 2 mol of cytochrome c per mole of QH2 because:

Answers

Answer: Options related to your question is missing below are the missing options

a. cytochrome c is a one-electron acceptor, whereas QH2 is a two-electron donor.

b. cytochrome c is a two-electron acceptor, whereas QH2 is a one-electron donor.

c. cytochrome c is water soluble and operates between the inner and outer mitochondrial membranes

d. heart muscle has a high rate of oxidative metabolism, and therefore requires twice as much cytochrome c as QH2 for electron transfer to proceed normally.

e. two molecules of cytochrome c must first combine physically before they are catalytically active.

answer:

cytochrome c is a one-electron acceptor, whereas QH2 is a two-electron donor. ( A )

Explanation:

The overall stoichiometry of the reaction requires 2 mol of cytochrome per mole of QH2 because a cytochrome is simply a one-electron acceptor while QH2 is not a one-electron donor ( i.e. it is a two-electron donor )

An electron donor in a reaction is considered a reducing agent because it donates its electrons to another compound thereby self oxidizing itself in the process.

To prepare a standard (calibration) curve for a spectroscopy experiment, start by preparing ___________ with ______________ Then, measure the ______________ of each solution at _____________ and create a plot of ____________ for the measured values. Finally, find the best-fit line of the data set.

Answers

Answer: See explanation

Explanation:

The calibration curve is the method used for the determination of the concentration of a substance such that the unknown sample will be compared to some standard samples of the known concentration.

To prepare a standard (calibration) curve for a spectroscopy experiment, start by preparing (multiple solutions) with (different known concentrations). Then, measure the (absorbance) of each solution at (thesame wavelength) and create a plot of (absorbance vs. concentration) for the measured values. Finally, find the best-fit line of the data set.

using the balanced equation below how many grams of lead(||) sulfate would be produced from the complete reaction of 23.6 g lead (|V) oxide

Answers

Answer:

59.8 g of PbSO₄.

Explanation:

The balanced equation for the reaction is given below:

Pb + PbO₂ + 2H₂SO₄ —> 2PbSO₄ + 2H₂O

Next, we shall determine the mass of PbO₂ that reacted and the mass of PbSO₄ produced from the balanced equation. This can be obtained as follow:

Molar mass of PbO₂ = 207 + (16×2)

= 207 + 32

= 239 g/mol

Mass of PbO₂ from the balanced equation = 1 × 239 = 239 g

Molar mass of PbSO₄ = 207 + 32 + (16×4)

= 207 + 32 + 64

= 303 g/mol

Mass of PbSO₄ from the balanced equation = 2 × 303 = 606 g

SUMMARY:

From the balanced equation above,

239 g of PbO₂ reacted to produce 606 g of PbSO₄.

Finally, we shall determine the mass of PbSO₄ that will be produced by the reaction of 23.6 g of PbO₂. This can be obtained as follow:

From the balanced equation above,

239 g of PbO₂ reacted to produce 606 g of PbSO₄.

Therefore, 23.6 g of PbO₂ will react to produce = (23.6 × 606) / 239 = 59.8 g of PbSO₄.

Thus, 59.8 g of PbSO₄ were obtained from the reaction.

Lead of mass 0.75kg is heated from 21°c to its melting point and continues to be heated unit it has all melted. Calculate how much energy is supplied to the lead. [Melting point of lead 372.5°c specific latent heat of fusion of lead = 23000 Jkg 'k '] ​

Answers

Answer:

65.5J

Explanation:

ML=Q

ML=MC(change in temperature)

0.75 X 23000 =0.75 X 351 X C

C= 65.5J

The energy supplied to the lead to melt from 21°c to its melting point is 51521 Joules.

What is the specific heat capacity?

Specific heat is the amount of heat energy supplied to change the temperature of one unit mass of a substance by 1 °C. The SI unit of the specific heat capacity of a substance is J/Kg.

The mathematical expression for the specific heat capacity can be written as:

Q = mCΔT      Where C is the specific heat of the substance.

The specific heat capacity depends upon the starting temperature and is an intensive characteristic of the material.

Given, the melting point of the lead T₂ = 327.5° C

The initial temperature of the lead, T₁ = 21° C

The latent heat of the lead given, L = 230000 J/Kg K

The specific heat of the lead, C = 130 J/Kg K

The heat required to melt the lead from 12°C to 327.5 °C is :

Q = m× [C × (T₂ - T₁) + L ]

Q = 0.75 × [0.130 (327.5 - 21)  + 23000]

Q = 51521 J

Learn more about specific heat, here:

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Given the equation representing a nuclear reaction in
which X represents a nuclide:
232Th → He + x
Which nuclide is represented by X?
A) 236
B) 228
Ra
SS
C) 236
Ra
92
U
92
D) 228
.
Ss U

Answers

The nuclide represented as X is thorium and this is an alpha decay.

The equation shown represents an alpha decay. In an alpha decay, an alpha particle is given off.

The atomic number of the parent nuclide is greater than that of the daughter nuclide by two units while the mass number of the parent is greater than that of the daughter nuclide by four units.

Hence the equation occurs as follows;

[tex]\frac{232}{92} Th ------> \frac{228}{88} Ra + He[/tex]

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calculate the hydrogen ion concentration of a solution who's pH is 2.4​

Answers

Answer:

I don't know sorry yyyyyyy6yyyyyyyyyyyyyyyyyyyyyyyyyyy

Choose the correct statement
a) The maximum value of principal quantum number (n) is 7
b) The angular quantum number (l) can receive value from 1 to (n-1)
c) The magnetic quantum number (ml) shows the energy of electron
d) The magnetic quantum number (ml) show how many orbitals in each subshell

Answers

Answer:

maybe number b is correct. ...thank you

A sample of a compound is found to consist of 0.44g H and 6.92g O what’s its formula

Answers

Answer:

Explanation:

H = 0.44/1.01 = 0.4356

O = 6.92/16 = 0.4319

This gives a 1:1 ratio. So the closest thing you could say is the formula is 0H

Going to your chemical storage room, you could justify that it is H2O2 or hydrogen peroxide. The question needs one more fact to make the answer certainty.

GM 2 all ,What is an atom define it .Good Day​

Answers

Answer:

An atom is the smallest particle of an element that can take part in chemical reaction.

Explanation:

hope it will help u Amri

ort
Which is a primary alcohol?
0 3-pentanol
2-propanol
1-ethanol
4-octanol
urvey
Lig A Moving to another question will save this response.

Answers

Answer:

1 ethanol is right answer

Explanation:

CH3- CH2-OH

If the balance were not tared prior to weighing out the KHP.... how would you expect this to affect the molarity of NaOH calculated? What type of error is this?

Answers

Answer:

Following are the response to the given question:

Explanation:

In the given scenario, When the balance has never been tainted before the KHP is weighted, which can affect the molar concentration of NaOH because its molarity is directly proportional to the weight including its substance. In this question it is the mistake is systemic because it may be corrected by modifying balancing parameters.

How does the radiation from radioisotopes cause damage to human tissue?
Answers

A.
by ionization knocking electrons away from atoms

B.
by breaking the bonds between atoms in molecules

C.
by causing nuclear chain reactions inside cells

D.
by causing transmutations of atoms within cells

Answers

Answer:

Explanation:

B.

by breaking the bonds between atoms in molecules

Which of the following is classified as paramagnetic?
1.NO(+)
2.Be2
3.B2(+)
4.+N2(+)​

Answers

A not sure though but I think

What would be the name of this compound?

Answers

Answer:

2,3 Dimethyl hexane

Explanation:

First, start the count from which side is given the CH3 smallest number

first; the longest carbon chain in this compound is 6

and you don't have any double and triple bonds or functional groups so it is Hexane

you start to count from the right side to give the branch molecules the smallest number ..

CH3 = methyl

and you have 2 methyl in this compound ..

and 2 mean you must write ( Di )

you write the name in this way

2,3 Dimethyl hexane

hope this helps you.

stay safe ...

Which of the following ligands is not capable of exhibiting linkage isomerism?
a. NCO-
b. -OH
c. -CN
d. -SCN

Answers

Answer:

a

...

........

...........

Which of the following behaviors would best describe someone who is listening and paying attention? a) Leaning toward the speaker O b) Interrupting the speaker to share their opinion c) Avoiding eye contact d) Asking questions to make sure they understand what's being said

Answers

Answer:

a Leaning towards the speaker

If a buffer is composed of 23.34 mL of 0.147 M acetic acid and 33.66 mL of 0.185 M sodium acetate, how many mL of 0.100 M NaOH can be added before the buffer capacity is reached

Answers

Answer:

25.5mL of 0.100M NaOH are needed to reach buffer capacity.

Explanation:

The buffer capacity is reached when the ratio between moles of conjugate base (Sodium acetate) and moles of weak acid (Acetic acid) is 10:

Moles sodium acetate / Moles Acetic acid = 10

The reaction of acetic acid, HA, with NaOH, to produce sodium acetate, NaA is:

HA + NaOH → H2O + NaA

That means the moles of NaOH added = Moles of HA that are being subtracted and moles of NaA that are been produced.

The initial moles of each species is:

Acetic acid:

23.34mL = 0.02334L * (0.147mol / L) = 0.00343 moles Acetic Acid

Sodium Acetate:

33.66mL = 0.03366L * (0.185mol / L) = 0.00623 moles Sodium Acetate

We can write the moles of each species when NaOH is added as:

Moles sodium acetate / Moles Acetic acid = 10

0.00623 moles + X / 0.00343 moles - X = 10

Where X are moles of NaOH added

Solving for X:

0.00623 moles + X = 0.0343 moles - 10X

11X = 0.0281

X = 0.00255 moles of NaOH are needed

In Liters:

0.0255mol NaOH * (1L / 0.100mol) = 0.0255L of 0.100M NaOH are needed =

25.5mL of 0.100M NaOH are needed to reach buffer capacity

An ordinary gasoline can measuring 30.0 cm by 20.0 cm by 15.0 cm is evacuated with a vacuum pump.
1a. Assuming that virtually all of the air can be removed from inside the can, and that atmospheric pressure is 14.7 psi, what is the total force (in pounds) on the surface of the can?
1b. Do you think that the can could withstand the force?

Answers

Answer:

Explanation:

From the given information:

The surface area of the can = (30 × 20 × 2) +(20× 15 × 2) +(30 × 15 × 2)

= 1200 + 600 + 900

= 2700 cm²

Since 1 inch² = 0.155 inch²

The surface area in inches² = 2700 × 0.155 inch²

= 418.5 inches²

The total force can be determined by using the expression:

Force = Pressure ×Area

Force = 14.7 psi ×  418.5 inches²

Force = 6151.95 lbs

Yes, the gasoline can will be able to withstand the force.

Please help answering 11)

Answers

Answer:

the answer is C

Explanation:

In what areas of the periodic table do you find the most highly reactive elements?

Answers

Answer:

The elements toward the bottom left corner of the periodic table are the metals that are the most active in the sense of being the most reactive.

The most highly reactive elements are typically found at the far left (Group 1) and far right (Group 17) of the periodic table.

Highly reactive elements in the periodic table

Group 1 elements, also known as alkali metals, are located on the far left of the periodic table. They have one electron in their outermost energy level and are highly reactive due to their tendency to lose that electron to achieve a stable electron configuration. This makes them very reactive with water and other substances.

Group 17 elements, known as halogens, are located on the far right of the periodic table. They have seven electrons in their outermost energy level and are highly reactive due to their strong tendency to gain one electron to achieve a stable electron configuration. This makes them reactive with metals and other elements.

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does anyone know how to solve this and what the answer would be?

Answers

Dynamic equilibrium is showed at the point at which solid liquid and gas intersect.

At the point at which solid liquid and gas intersect represents a system that shows dynamic equilibrium. There is equal amount of reactants and products at the point of dynamic equilibrium because the transition of substances occur between the reactants and products at equal rates, means that there is no net change. Reactants and products are formed at the rate that no change occur in their concentration.

https://brainly.com/question/24310467

What process occurs during the corrosion of iron?
Answers

A.
Iron is oxidized.

B.
Iron is reduced.

C.
Iron (III) is oxidized.

D.
Iron (III) is reduced.

Answers

Answer:

A

Explanation:

The iron corrodes so it oxidized

What is the molarity of an HCl solution if 25.0 mL of this solution required 17.80 mL of 0.108 M NaOH to reach the end point in a titration?

Answers

Answer:

[tex]\boxed {\boxed {\sf 0.0769 \ M}}[/tex]

Explanation:

We are asked to find the molarity of an acid given the details of a titration experiment. The formula for titration is as follows:

[tex]M_AV_A= M_B V_B[/tex]

In this formula, M is the molarity of the acid or base and V is the volume of the acid or base. The molarity of the hydrochloric acid (HCl) is unknown and the volume is 25.0 milliliters.

[tex]M_A * 25.0 \ mL = M_BV_B[/tex]

The molarity of the sodium hydroxide (NaOH) is 0.108 molar and the volume is 17.80 milliliters.

[tex]M_A * 25.0 \ mL = 0.108 \ M * 17.80 \ mL[/tex]

We are solving for the molarity of the acid and we must isolate the variable [tex]M_A[/tex]. It is being multiplied by 25.0 milliliters. The inverse operation of multiplication is division, so we divide both sides of the equation by 25.0 mL.

[tex]\frac {M_A * 25.0 \ mL }{25.0 \ mL}= \frac{0.108 \ M * 17.80 \ mL }{25.0 \ mL}[/tex]

[tex]M_A= \frac{0.108 \ M * 17.80 \ mL }{25.0 \ mL}[/tex]

The units of milliliters cancel.

[tex]M_A= \frac{0.108 \ M * 17.80 }{25.0 }[/tex]

[tex]M_A= \frac{1.9224}{25.0 } \ M[/tex]

[tex]M_A= 0.076896 \ M[/tex]

The original measurements have 3 and 4 significant figures. We must round our answer to the least number of sig figs, which is 3. For the number we calculated, that is the ten-thousandth place. The 9 to the right of this place tells us to round the 8 up to a 9.

[tex]M_A \approx 0.0769 \ M[/tex]

The molarity of the hydrochloric acid is 0.0769 Molar.

You perform a distillation to separate a mixture of propylbenzene and cyclohexane, and you obtain 2.9949 grams of cyclohexane (density -0.779 g/mL, MW - 84.16 g/mol) and 1.6575 grams of propylbenzene (density = 0.862 g/mL, MW = 120.2 g/mol). What is the volume percent composition of cyclohexane in the mixture?

Answers

Answer:

66.67%

Explanation:

From the given information:

mass of cyclohexane = 2.9949 grams

density of cyclohexane = 0.779 g/mL

Recall that:

Density = mass/volume

Volume = mass/density

So, the volume of cyclohexane = 2.9949 g/ 0.779 g/mL

= 3.8445 mL

Also,

mass of propylbenzene = 1.6575 grams

density of propylbenzene = 0.862 g/mL

Volume of propylbenzene =  1.6575 g/ 0.862 g/mL

= 1.9229 mL

The volume % composition of cyclohexane from the mixture is:

[tex]= (\dfrac{v_{cyclohexane}}{v_{cyclohexane}+v_{propylbenzene}})\times 100[/tex]

[tex]= (\dfrac{3.8445}{3.8445+1.9229})\times 100[/tex]

[tex]= (\dfrac{3.8445}{5.7674})\times 100[/tex]

= 66.67%

The sample concentration was measured at 50mg/ml. The loading concentration needs to be 10mg/ml. The final volume needs to be 25ul. What is the volume of sample needed and the amount of buffer needed to reach 25ul

Answers

Answer:

a)  [tex]V_1=5ul[/tex]

b)  [tex]v=20ul[/tex]

Explanation:

From the question we are told that:

initial Concentration [tex]C_1=50mg/ml[/tex]

Final Concentration [tex]C_2=10mg/ml[/tex]

Final volume needs [tex]V_2 =25ul[/tex]

Generally the equation for Volume is mathematically given by

[tex]C_1V_1=C_2V_2[/tex]

[tex]V_1=\frac{C_1V_1}{C_2}[/tex]

[tex]V_1=\frac{10*25}{50}[/tex]

[tex]V_1=5ul[/tex]

Therefore

The volume of buffer needed is

[tex]v=V_2-V_1\\\\v=25-5[/tex]

[tex]v=20ul[/tex]

Adding more than one equivalent of HCl to pent-1-yne will lead to which product:______.
a. 1,2-dichloro-1-butene.
b. 1,1-dichloropentane.
c. 2,2-dichloropentane.
d. 2,2-dichlorobutane.

Answers

Answer:

c. 2,2-dichloropentane.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to firstly draw the structure of the reactant, pent-1-yne:

[tex]CH\equiv C-CH_2-CH_2-CH_2[/tex]

Now, we infer the halogen is added to the carbon atom with the most carbon atoms next to it, in this case, carbon #2, in order to write the following product:

[tex]CH\equiv C-CH_2-CH_2-CH_2+2HCl\rightarrow CH_3- CCl_2-CH_2-CH_2-CH_2[/tex]

Whose name is 2,2-dichloropentane.

Regards!

Suppose you are studying the Ksp of CaCl2, which has a molar mass of 110.98 g/mol, at multiple temperatures. You dissolve 4.99 g of CaCl2 in 10.0 mL of water at 100 oC and cool the solution. At 90 oC, a solid begins to appear. What is the Ksp of CaCl2 at 90 oC

Answers

Answer:

Hence the Solubility product,  

Ksp = [Ca2+] [Cl-]2  

or, Ksp = (4.5) (9)2  

or, Ksp = 364.5

Explanation:  

Mass of CaCl2 = 4.99 g  

Molar mass of CaCl2 = 110.98 g/mol  

Moles of CaCl2  

= given mass/ molar mass  

= 4.99/ 110.98  

= 0.045  

Volume = 10.0 mL = 0.01 L  

CaCl2 dissociates into its ion as:  

CaCl2 (s)  \rightleftharpoons Ca2+ (aq) + 2 Cl- (aq)  

At 90°C, the solution is saturated with Ca2+ and Cl- ions.

Moles of Ca2+ = Moles of CaCl2 dissolved = 0.045  

Moles of Cl- = 2 x ( Moles of CaCl2 dissolved) = 2 x 0.045 = 0.09

[Ca2+] = Moles/ Volume = 0.045/ 0.01 = 4.5 M  

[Cl-] = 0.09/ 0.01 = 9 M  

Solubility product,  

Ksp = [Ca2+] [Cl-]2  

or, Ksp = (4.5) (9)2  

or, Ksp = 364.5

Nitric acid can be formed in two steps from the atmospheric gases nitrogen and oxygen, plus hydrogen prepared by reforming natural gas. In the first step, nitrogen and hydrogen react to form ammonia: N2 (g) + 3H2 (g) â 2NH3 (g) =ÎHâ92.kJ In the second step, ammonia and oxygen react to form nitric acid and water:

NH3 (g) + 2O2 (g) â HNO3 (g) + H2O (g) =ÎHâ330.kJ

Required:
Calculate the net change in enthalpy for the formation of one mole of nitric acid from nitrogen, hydrogen and oxygen from these reactions.

Answers

Answer:

-376 kJ

Explanation:

The first step equation:

[tex]\mathsf{N_{2(g)} + 3H_2{(g)} \to 2NH_3{(g)} \ \ \ \Delta H = -92\ kJ}[/tex]    ---- (1)

The second step equation:

[tex]\mathsf{NH_{3(g)} + 2O_2{(g)} \to HNO_3{(g)} +H_2O_{(g)} \ \ \ \Delta H = -330\ kJ}[/tex]      ---- (2)

To determine the enthalpy of formation for 1 mole of HNO₃ (nitric acid), we have the following.

From the above equations; let multiply equation (1) by 1 and equation (2) by 2.

[tex]\mathsf{N_{2(g)} + 3H_2{(g)} \to 2NH_3{(g)} \ \ \ \Delta H = -92\ kJ}[/tex]     ---- (3)

[tex]\mathsf{2NH_{3(g)} + 4O_2{(g)} \to 2HNO_3{(g)} +2H_2O_{(g)} \ \ \ \Delta H = 2(-330)\ kJ}[/tex]      ----- (4)

adding the above two equations, we have:

[tex]\mathsf{N_{2(g)} + 3H_2{(g)}+ 2NH_{3(g)} + 4O_{2(g)} \to 2HNO_{3(g)} + 2NH_3{(g)} +2H_2O_{(g)} \ \ \ \Delta H = (-660 \ kJ -92\ kJ)}[/tex][tex]\mathsf{N_{2(g)} + 3H_2{(g)} + 4O_{2(g)} \to 2HNO_{3(g)} +2H_2O_{(g)} \ \ \ \Delta H = (-752 \ kJ)}[/tex]

Now, from the recent equation, we have:

2 moles of nitric acid = -752 kJ

1 mole of nitric acid will be: = (1 mole × (-752 kJ)) ÷ 2 moles

1 mole of nitric acid will be: = -376 kJ

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