All categories

3 years ago

In an open system, the vapor pressure is equal to the outside air pressure. true false

Physics

2 answers:

Anvisha [2.4K]3 years ago

8 0

False if the pressure is in an open system it could not be equal to the outside

elena-s [515]3 years ago

4 0

Answer:

A) True

Explanation:

Hello! I can assure you that this answer is true. In an open system, the vapor pressure is equal to the outside air pressure at a liquid's boiling point. Hope this comes to your help! (:

You might be interested in

A rifle is aimed horizontally at a target 50.0 m away. The bullet hits the target 2.90 cm below the aim point. . . Whats the bul

Marina CMI [18]

The problem ask to calculate the bullet's flight time and the bullet's speed as it left the barrel. So base on the problem, the answer would be that the flight time is 0.076 seconds and the speed of the bullet is 657.9 m/s. I hope you are satisfied with my answer and feel free to ask for more if you have questions and further clarifications.

5 0

2 years ago

A car is up on a hydraulic lift at a garage. The wheels are free to rotate, and the drive wheels are rotating with a constant an

masya89 [10]

Answer:

Explanation:

Given

Wheels are rotating with constant angular velocity let say $\omega$

Presence of constant angular velocity show that there is no angular acceleration thus there is no tangential acceleration.

But any particle on the rim will experience a constant acceleration towards center called centripetal acceleration.

(a) yes, there will be tangential velocity which is given by

$v=r\cdot \omega$

where r=radial distance from center

(b)tangential acceleration

there would be no tangential acceleration as velocity is constant

(c)centripetal acceleration

Yes, there will be centripetal acceleration given by

$a_c=\omega ^2\times r$

7 0

3 years ago

A student investigated how the mass of water in an electric kettle affected the time taken for the water to reach boiling point.

Sedbober [7]

Answer:

1.because of the heat produced by the socat

2. they should have control how they placed the heater

3. because the water is to much

4.because is different from the question

5. because that is how the question is

4 0

2 years ago

A 19.12 g mixture of Ca(NO3)2 and KCl is dissolved in 149 g of water. The freezing point of the solution was measured as −5.77 ∘

hichkok12 [17]

Answer:

The mass percentage of calcium nitrate is 31.23%.

Explanation:

Let the the mass of calcium nitrate be x and mass of potassium chloride be y.

Total mass of mixture = 19.12 g

x + y = 19.12 g..(1)

Mass of solvent = 149 g = 0.149 kg

Freezing point of the solution, $T_f$ = -5.77 °C

Molal freezing constant of water = 1.86 °C/m =1.86 °C/(mol/kg)

The van't Hoff factor contribution by calcium nitrate is 3 and by potassium chloride is 2.So:

i = 3

i' = 2

Freezing point of water = T = 0°C

$\Delta T_f=T-T_f=0^oC-(-5.77^oC)=5.77^oC$

$\Delta T_f=i\times K_f\times m$

$Molality=m(mol/kg)=\frac{\text{Moles of solute}}{\text{mass of solvent in kg}}$

$5.77^oC=1.86 ^oC/(mol/kg)\times (\frac{ i\times x}{164 g/mol\times 0.149 kg}+\frac{i'\times y}{74.5 g/mol0.149 kg})$

On solving we get:

$\frac{3x}{164 g/mol}+\frac{2x}{74.5 g/mol}=0.4622 mol$ ....(2)

Solving equation (1)(2) for x and y:

x =5.973 g

y = 13.147 g

Mass percent of $Ca(NO_3)_2$ in the mixture:

$\frac{x}{19.21 g}\times 100=\frac{5.973 g}{19.12 g}=31.23\%$

The mass percentage of calcium nitrate is 31.23%.

5 0

3 years ago

The blank is the core of an atom

alexdok [17]

Answer:

The <u>nucleus</u> is the core of an atom.

Step-by-step explanation:

That's just how it is.

5 0

3 years ago