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ArbitrLikvidat [17]
3 years ago
8

I need help with question 2

Chemistry
1 answer:
krek1111 [17]3 years ago
8 0

Answer:

here's the answer to your question

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Objects that are able to fall have what type of energy?
ASHA 777 [7]

Answer:

kinetic energy

Explanation:

because of the movement of gravity and space

4 0
3 years ago
uh id.k how to fill this out, i mean the type of reaction part is obvious but id.k what to put in the box and the other blue par
Studentka2010 [4]

Answer:

Explanation:

Strontium chlorate appears as a moist solid or semi-solid slurry of white crystals. May explode under exposure to heat or fire. Used in pyrotechnics

strontium chlorate | Sr(ClO3)2 - PubChem.

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8 0
3 years ago
A 0.35 L balloon found at 19 C is heated in an oven to 250 C. What is the new volume of the balloon?
Svet_ta [14]

Answer:

V = 0.63 L

Explanation:

To solve this problem, we need to use the Charle's law which is a law that involves temperature and volume, assuming we have a constant pressure. The problem do not state that the pressure is being altered, so we can safely assume that the pressure is constant (Maybe 1 atm).

Now, as the pressure is constant, the Charle's law is the following:

V₁ / T₁ = V₂ / T₂   (1)   V is volume in Liter, and T is temperature in Kelvin.

Using this law with the given data, we solve for V₂:

V₂ = V₁T₂ / T₁

Before we use this expression, let's convert the temperatures to Kelvin:

T₁ = 19 + 273 = 292 K

T₂ = 250 + 273 = 523 K

Now, let's calculate the volume of the balloon:

V₂ = 0.35 * 523 / 292

<h2>V₂ = 0.63 L</h2>
5 0
3 years ago
Read 2 more answers
The amplitude of a wave is most directly related to what wave characteristic? Crest length
galina1969 [7]

The amplitude of a wave is most directly related to crest length.

Amplitude means the height of a crest or a trough.



7 0
3 years ago
A sample consisting of 1.00 mol of perfect gas molecules at 27 °C is expanded isothermally from an initial pressure of 3.00 atm
Evgesh-ka [11]

Answer:

a) reversibly

ΔU = 0

q = 2740.16 J

w = -2740.16 J

ΔH = 0

ΔS(total) = 0

ΔS(sys)  =9.13 J/K

ΔS(surr) = -9.13 J/K

b) against a constant external pressure of 1.00 atm

ΔU = 0

w = -1.66 kJ

q = 1.66 kJ

ΔH = 0

ΔS(sys) = 9.13 J/K

ΔS(surr) = -5.543 J/K

ΔS(total) = 3.587 J/K

Explanation:

<u>Step 1</u>: Data given:

Number of moles = 1.00 mol

Temperature = 27.00 °C = 300 Kelvin

Initial pressure = 3.00 atm

Final pressure = 1.00 atm

The gas constant = 8.31 J/mol*K

<u>(a) reversibly</u>

<u>Step 2:</u> Calculate work done

For ideal gases ΔU depends only on temperature. So as it is an isothermal (T constant).

Since the temperature remains constant:

ΔU = 0

ΔU = q + w

q = -w

w = -nRT ln (Pi/Pf)

⇒ with n = the number of moles of perfect gas = 1.00 mol

⇒ with R = the gas constant = 8.314 J/mol*K

⇒ with T = the temperature = 300 Kelvin

⇒ with Pi = the initial pressure = 3.00 atm

⇒ with Pf = the final pressure = 1.00 atm

w =- 1*8.314 *300 * ln(3)

w = -2740.16 J

q = -w

q = 2740.16 J

<u>Step 3:</u> Calculate change in enthalpy

Since there is no change in energy, ΔH = 0

<u>Step 4:</u> Calculate ΔS

for an isothermal process

ΔS (total) = ΔS(sys) + ΔS(surr)  

ΔS(sys) = -ΔS(surr)

ΔS(sys) = n*R*ln(pi/pf)

ΔS(sys) = 1.00 * 8.314 * ln(3)

ΔS(sys) = 9.13 J/K

ΔS(surr) = -9.13 J/K

ΔS (total) = ΔS(sys) + ΔS(surr) = 0

<u>(b) against a constant external pressure of 1.00 atm</u>

<u>Step 1</u>: Calculate the work done

w = -Pext*ΔV

w = -Pext*(Vf - Vi)

⇒ with Vf = the final volume

⇒ with Vi = the initial volume

We have to calculate the final and initial volume. We do this via the ideal gas law P*V=n*R*T

V = (n*R*T)/P

Initial volume = (n*R*T)/Pi

⇒ Vi = (1*0.08206 *300)/3

   ⇒ Vi = 8.206 L

Final volume = (n*R*T)/Pf

     ⇒ Vf = (1*0.08206 *300)/1

      ⇒ Vf = 24.618 L

The work done w = -Pext*(Vf - Vi)

w = -1.00* ( 24.618 - 8.206)

w = -16.412 atm*L

w = -16 .412 *(101325/1atm*L) *(1kJ/1000J)

w = -1662.9 J = -1.66 kJ

<u>Step 2:</u> Calculate the change in internal energy

ΔU = 0

q = -w

q = 1.66 kJ

ΔH = 0 because there is no change in energy

<u>Step 3: </u>Calculate ΔS

ΔS(sys) = n*R*ln(3)

ΔS(sys) = 1.00 * 8.314 * ln(3)

ΔS(sys) = 9.13 J/K

ΔS(surr) = -q/T

ΔS(surr) = -1662.9J/300K

ΔS(surr) = -5.543 J/K

ΔS(total) = ΔS(surr) +ΔS(sys) = -5.543 J/K + 9.13 J/K = 3.587 J/K

4 0
3 years ago
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