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1 year ago

The normal boiling point of cyclohexane is 81.0 oC. What is the vapor pressure of cyclohexane at 81.0 oC?

Physics

1 answer:

Aloiza [94]1 year ago

6 0

Answer:

The vapor pressure of cyclohexane at 81.0°C is 101325 Pa.

Explanation:

Given that,

Boiling point = 81.0°C

Atmospheric pressure :

Atmospheric pressure is the force per unit area exerted by the weight of the atmosphere.

The value of atmospheric pressure is

$P=101325\ Pa$

Vapor pressure :

Vapor pressure is equal to the atmospheric pressure.

Hence, The vapor pressure of cyclohexane at 81.0°C is 101325 Pa.

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A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C.

Sedaia [141]

Answer:

335°C

Explanation:

Heat gained or lost is:

q = m C ΔT

where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

Heat gained by the water = heat lost by the copper

mw Cw ΔTw = mc Cc ΔTc

The water and copper reach the same final temperature, so:

mw Cw (T - Tw) = mc Cc (Tc - T)

Given:

mw = 390 g

Cw = 4.186 J/g/°C

Tw = 22.6°C

mc = 248 g

Cc = 0.386 J/g/°C

T = 39.9°C

Find: Tc

(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)

Tc = 335

7 0

1 year ago

A FBD of a rocket launching into space should include:

Vladimir [108]

Answer:

Explanation:

the force of the rocket engine pushing it up, the force of gravity pulling it down, maybe some force of air resistance as the rocket goes fast, hmmm Free Body Diagrams (FBD) should have any and all forces on the model, unless they are negligible . or so slight they really make little difference in the total outcome.

3 0

1 year ago

An airplane flying parallel to the ground undergoes two consecutive dis- placements. The first is 75 km 30.0° west of north, and

torisob [31]

Answer: displacement of airplane is 172 km in direction 34.2 degrees East of North

Explanation:

In constructing the two displacements it is noticed that the angle between the 75 km vector and the 155 km vector is a right angle (90 degrees).

Hence if the plane starts out at A, it travels to B, 75 km away, then turns 90 degrees to the right (clockwise) and travels to C, 155 km away from B. Angle ABC is 90 degrees, hence we can use Pythagoras theorem to solve for AC

AC2 = AB2 + BC2 ; AC^2 = 752 + 1552 ; from this we get AC = 172 km (3 significant figures)

Angle BAC = Tan-1(155/75) ; giving angle BAC = 64.2 degrees

Hence AC is in a direction (64.2 - 30) = 34.2 degrees East of North

Therefore the displacement of the airplane is 172 km in a direction 34.2 degrees East of North

5 0

1 year ago

Calculate the buoyant force in air on a kilogram of titanium (whose density is about 4.5 grams per cubic centimeter). compare wi

aleksklad [387]

1) The buoyant force acting on an object immersed in a fluid is:
$B=d_f V_d g$
where $d_f$ is the density of the fluid, $V_d$ is the volume of displaced fluid, and $g=9.81~m/s^2$ is the gravitational acceleration.

2) We must calculate the volume of displaced fluid. Since the titanium object is completely immersed in the fluid (air), this volume corresponds to the volume of 1 Kg of titanium, whose density is $d=4.5~g/cm^3 = 4.5\cdot10^3~Kg/m^3$ . Using the relationship between density, volume and mass, we find
$V_d= \frac{m}{d}= \frac{1~Kg}{4.5\cdot10^3Kg/m^3}=2.22\cdot10^{-4}~m^3$

3) Now we can recall the formula written at step 1) and calculate the buoyant force. The air density is $d_f = 1~Kg/m^3$ , so we have
$B=d_f V_d g=1~Kg/m^3 \cdot 2.22\cdot10^{-4}~m^3 \cdot 9.81~m/s^2=2.22\cdot10^{-3}~N$

4) The weight of 1 Kg of titanium is instead:
$W=mg=1~Kg \cdot 9.81~m/s^2=9.81~N$
So, the buoyant force is negligible compared to the weight.

7 0

1 year ago

An owl has a mass of 4.00 kg. It dives to catch a mouse, losing 800.00 J of its GPE. What was the starting height of the owl, in

vesna_86 [32]

Answer:

height =20m

Explanation:

gpe=mgh

800=4×10×x

40x=800

x=20

3 0

1 year ago