What is denser a iron sheet or iron powder??

If this energy were used to vaporize water at 100.0 ∘C, how much water (in liters) could be vaporized? The enthalpy of vaporizat

Zanzabum

Answer:

0.429 L of water

Explanation:

First to all, you are not putting the value of the energy given to vaporize water, so, to explain better this problem, I will assume a value of energy that I took in a similar exercise before, which is 970 kJ.

Now, assuming that the water density is 1 g/mL, this is the same as saying that 1 g of water = 1 mL of water

If this is true, then, we can assume that 1 kg of water = 1 L of water.

Knowing this, we have to use the expression to get energy which is:

Q = m * ΔH

Solving for m:

m = Q / ΔH

Now "m" is the mass, but in this case, the mass of water is the same as the volume, so it's not neccesary to do a unit conversion.

Before we begin with the calculation, we need to put the enthalpy of vaporization in the correct units, which would be in grams. To do that, we need the molar mass of water:

MM = 18 g/mol

The enthalpy in mass:

ΔH = 40.7 kJ/mol / 18 g/mol = 2.261 kJ/g

Finally, solving for m:

m = 970 / 2.261 = 429 g

Converting this into volume:

429 g = 429 mL

429 / 1000 = 0.429 L of water

3 0

3 years ago

Una partícula se mueve en el plano XY efectúa un desplazamiento mientras actúa sobre ella una fuerza constante. X= (4i + 3j) m,

dsp73

Answer:

a) La magnitud del desplazamiento es de 5 m

La magnitud de la fuerza es 20 N

b) El trabajo realizado por la fuerza es de 100 J

c) El ángulo entre la fuerza y el plano es 0 °

Explanation:

a) La magnitud del desplazamiento se encuentra por la relación;

$\left | X \right | = \sqrt{X_{x}^{2}+X_{y}^{2}}$

Lo que da;

$\left | X \right | = \sqrt{4^{2}+3^{2}} = 5 \ m$

De manera similar, la magnitud de la fuerza, F, se encuentra como sigue;

$\left | F \right | = \sqrt{F_{x}^{2}+F_{y}^{2}}$

Lo que da;

$\left | F \right | = \sqrt{16^{2}+12^{2}} = 20 \ N$

b) El trabajo, W, realizado por la fuerza = Fuerza, F × Distancia, X

∴ Ancho = 20 N × 5 m = 100 N · m = 100 J

c) La dirección de la fuerza viene dada por la siguiente fórmula;

$tan^{-1} \left (\dfrac{F_y}{F_x} \right ) = tan^{-1} \left (\dfrac{12}{16} \right ) = 38.9^{\circ}$

La dirección del plano viene dada por la siguiente fórmula;

$tan^{-1} \left (\dfrac{X_y}{X_x} \right ) = tan^{-1} \left (\dfrac{3}{4} \right ) = 38.9^{\circ}$

Por tanto, el ángulo entre la fuerza y el plano = 0 °

La fuerza actúa a lo largo del plano.

6 0

3 years ago

An automobile having a mass of 2000 kg deflects its suspension springs 0.02 m under static conditions. Determine the nafural fre

alexandr1967 [171]

Answer:Frequency = 3.525 Hertz

Explanation:In static equilibrium, kd =mg

Where k= effective spring constant of the spring.

mg= The weight of the car.

d= static deflection.

Therefore, w =SQRTg/d

w = SQRT 9.81/0.02

w= 22.15 rad/sec

Converting to Hertz unit for frequency

1 rad/s = 0.1591

22.15rad/s=?

22.15 × 0.1591= 3.525 hertz

7 0

3 years ago

what kind of weather is associated with high pressure?How does density,humidity, and air motion compare to that low pressure sys

frutty [35]

Usually nice weather , i dont know the answer to the second part

6 0

3 years ago

Find acceleration given the following:

Lorico [155]

Answer:

Explanation:df=378

Di=0

Vi=0

T=6s

Formula

Df=Di+Vit+1/2(at)2

378=0+0+6+1/2×a×t2

378=6+1/2×a×36

378=6+18a

18a=378+6=384

a=385/18=21.33(cm/s)/s

5 0

3 years ago