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2 years ago

If a car travels 1255 miles how many kilometers is that i mile = 1.609km

Chemistry

1 answer:

ANTONII [103]2 years ago

3 0

Answer:

2,019 km

Explanation:

Step 1: Given data

Distance traveled by the car (D): 1,255 mi

Step 2: Convert the distance traveled by the car to kilometers

To convert one unit into another, we use a conversion factor. In this case, the appropriate conversion factor between miles and kilometers is 1 mile = 1.609 km. The distance traveled by the car, in kilometers, is:

D = 1,255 mi × (1.609 km/1 mi) = 2,019 km

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A 20.0 -\,L volume of an ideal gas in a cylinder with a piston is at a pressure of 3.2 atm. Enough weight is suddenly removed fr

zzz [600]

Answer:

1. ΔE = 0 J

2. ΔH = 0 J

3. q = 3.2 × 10³ J

4. w = -3.2 × 10³ J

Explanation:

The change in the internal energy (ΔE) and the change in the enthalpy (ΔH) are functions of the temperature. If the temperature is constant, ΔE = 0 and ΔH = 0.

The gas initially occupies a volume V₁ = 20.0 L at P₁ = 3.2 atm. When the pressure changes to P₂ = 1.6 atm, we can find the volume V₂ using Boyle's law.

P₁ × V₁ = P₂ × V₂

3.2 atm × 20.0 L = 1.6 atm × V₂

V₂ = 40 L

The work (w) can be calculated using the following expression.

w = - P . ΔV

where,

P is the external pressure for which the process happened

ΔV is the change in the volume

w = -1.6 atm × (40L - 20.0L) = -32 atm.L × (101.325 J/1atm.L) = -3.2 × 10³ J

The change in the internal energy is:

ΔE = q + w

0 = q + w

q = - w = 3.2 × 10³ J

6 0

2 years ago

Explain why you hear a “whoosh” sound when you open a can containing a carbonated drink. Which gas law applies?

Lana71 [14]

Carbonated drinks have the air under pressure so that carbon bubbles are forced into the drink, keeping it carbonated. So when you open a can, the air under pressure in the can comes out of the can at a high speed, making a "whooshing" sound. The gas law that applies to this concept is the Boyle's Law (PV=k or P1V1=P2V2).

6 0

2 years ago

Read 2 more answers

A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at

Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0

2 years ago

C2H5OH(aq) + MnO− 4 (aq) → Mn2+(aq) + CH3COOH(aq) of acetic acid from ethanol by the action of permanganate ion in acidic soluti

Andreas93 [3]

Answer :

Ethanol $(C_2H_5OH)$ act as reducing agent.

The smallest possible integer coefficient of $MnO_4^-$ in the combined balanced equation is, 4

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.

Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced. Thus, it will undergo reduction reaction.

The given chemical reaction is,

$C_2H_5OH(aq)+MnO_4^-(aq)\rightarrow CH_3COOH(aq)+Mn^{2+}(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $C_2H_6O\rightarrow C_2H_4O_2$

Reduction : $MnO_4^-\rightarrow Mn^{2+}$

Now balance oxygen atom on both side.

Oxidation : $C_2H_6O+H_2O\rightarrow C_2H_4O_2$

Reduction : $MnO_4^-\rightarrow Mn^{2+}+4H_2O$

Now balance hydrogen atom on both side.

Oxidation : $C_2H_6O+H_2O\rightarrow C_2H_4O_2+4H^+$

Reduction : $MnO_4^-+8H^+\rightarrow Mn^{2+}+4H_2O$

Now balance the charge.

Oxidation : $C_2H_6O+H_2O\rightarrow C_2H_4O_2+4H^++4e^-$

Reduction : $MnO_4^-+8H^++5e^-\rightarrow Mn^{2+}+4H_2O$

In order to balance the electrons, we multiply the oxidation reaction by 5 and reduction reaction by 4 and then added both equation, we get the balanced redox reaction.

Oxidation : $5C_2H_6O+5H_2O\rightarrow 5C_2H_4O_2+20H^++20e^-$

Reduction : $4MnO_4^-+32H^++20e^-\rightarrow 4Mn^{2+}+16H_2O$

The balanced chemical equation in acidic medium will be,

$5C_2H_6O+4MnO_4^-+12H^+\rightarrow 5C_2H_4O_2+4Mn^{2+}+11H_2O$

In the redox reaction ethanol act as reducing agent and permanganate ion act as an oxidizing agent.

4 0

2 years ago

A mixture of CH4 and H2O is passed over a nickel catalyst at 1000 K. The emerging gas is collected in a 5.00L flask and is found

Harman [31]

Answer:

Kc =<u> 3.74*10⁻³ </u>

Kp = 25.21

Explanation:

Step 1: Data given

Temperature = 1000 K

Volume = 5.00 L

Mass of CO = 8.62 grams

Mass of H2 = 2.60 grams

Mass of CH4 = 43.0 grams

Mass of H2O = 48.4 grams

Kc = [CO]*[H₂]³ / ([CH₄]∙*H₂O])

Kp = p(CO)*p(H₂)³ / (p(CH₄)*p(H₂O) )

Step 2: The balanced equation

CH₄ + H₂O ⇄ CO + 3 H₂

Step 3: Calculate number of moles

The number of moles of each compund in the equilibrium mixture are:

Moles = mass / molar mass

n(CH₄) = 43.0g / 16g/mol = 2.688mol

n(H₂O) = 48.4g / 18g/mol = 2.689mol

n(CO) = 8.62g/28g/mol = 0.308mol

n(H₂) = 2.60g / 2g/mol = 1.3mol

Step 4: Calculate concentrations at equilibrium

So the equilibrium concentrations are:

Concentration = moles / volume

[CH₄] = 2.688mol/5L = 0.5376 M

[H₂O] = 2.689mol/5L = 0.5378M

[CO] = 0.308mol/5L = 0.0616M

[H₂) = 1.3mol/5L = 0.26M

Step 5: Calculate Kc

Kc = 0.0616 ∙ (0.26)³ / (0.5376∙0.5378) = <u>3.74*10⁻³ </u>

Step 5: Calculate partial pressure

Partial pressures in equilibrium can be found from ideal gas law:

p(X) = n(X)∙R∙T/V = [X]∙R∙T

=> p(CH₄) = [CH₄]∙R∙T = 0.5376mol/L * 0.082 06Latm/molK ∙ 1000K = 44.11 atm

p(H₂O) = [H₂O]∙R∙T = 0.5738mol/L * 0.082 06Latm/molK * 1000K = 44.13 atm

p(CO) = [CO]∙R∙T = 0.0616mol/L * 0.082 06Latm/molK * 1000K = 5.05atm

p(H₂) = [CO]∙R∙T = 0.26mol/L * 0.082 06Latm/molK * 1000K = 21.34atm

Step 5: Calculate Kp

Kp = p(CO)*p(H₂)³ / (p(CH₄)*p(H₂O) )

Kp = 5.05*21.34³ / (44.11*44.13 ) = 25.21

8 0

2 years ago