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3 years ago

15

How much does a gold bar weigh and how much is it worth?

1 answer:

lord [1]3 years ago

6 0

A gold bar is worth 506,400 each. And they weigh 12.4 KG

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A mass of 0.1 kg of helium fills a 0.2 m3 rigid tank at 350 kPa. The vessel is heated until the pressure is 700 kPa. Calculate t

QveST [7]

Explanation:

(a) First, we will calculate the number of moles as follows.

No. of moles = $\frac{\text{mass}}{\text{molar mass}}$

Molar mass of helium is 4 g/mol and mass is given as 0.1 kg or 100 g (as 1 kg = 1000 g).

Putting the given values into the above formula as follows.

No. of moles = $\frac{\text{mass}}{\text{molar mass}}$

= $\frac{\text{100 g}}{4 g/mol}$

= 25 mol

According to the ideal gas equation,

PV = nRT

or, $(P_{2} - P_{1})V = nR (T_{2} - T_{1})$

$(6.90 atm - 3.45 atm) \times 200 L = 25 \times 0.0821 L atm/mol K \Delta T$

$\Delta T$ = 336.17 K

Hence, temperature change will be 336.17 K.

(b) The total amount of heat required for this process will be calculated as follows.

q = $mC \Delta T$

= $100 g \times 5.193 J/g K \times 336.17 K$

= 174573.081 J/K

or, = 174.57 kJ/K (as 1 kJ = 1000 J)

Therefore, the amount of total heat required is 174.57 kJ/K.

3 0

3 years ago

When a liquid has vapor pressure equal to atmospheric pressure, it?

Olenka [21]

When vapor pressure equals atmospheric pressure, it's called boiling point of that liquid.

5 0

3 years ago

Read 2 more answers

A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho

d1i1m1o1n [39]

Answer:

Part a)

$KE_r = 8 J$

Part b)

v = 3.64 m/s

Part c)

$KE_f = 12.7 J$

Part d)

$v = 2.9 m/s$

Explanation:

As we know that moment of inertia of hollow sphere is given as

$I = \frac{2}{3}mR^2$

here we know that

$I = 0.0484 kg m^2$

R = 0.200 m

now we have

$0.0484 = \frac{2}{3}m(0.200)^2$

$m = 1.815 kg$

now we know that total Kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

$20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2$

$20 = 1.5125 v^2$

$v = 3.64 m/s$

Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

$KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2$

$KE_r = 8 J$

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

$mgh = (KE_i) - KE_f$

$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

$v = 2.9 m/s$

8 0

3 years ago

1. When an object's distance from another object is changing it must be

slega [8]

An object is in motion if its distance from another object is changing. An object is in motion if it changes position relative to a reference point. An reference point is a place or object used for comparison to determine if something is moving.

3 0

3 years ago

If Phoenix, Arizona, experiences a cool, wet day in June, does that mean the regions climate is changing?

jenyasd209 [6]

It doesn't mean that the climate is changing, it is probably just the morning dew.

5 0

3 years ago