What is the density of an object with mass of 60 g and a volume of 2cm3

A 10.0 L tank contains 0.329 kg of helium at 28.0 ∘C. The molar mass of helium is 4.00 g/mol . Part A How many moles of helium a

nadya68 [22]

Answer:

82.25 moles of He

Explanation:

From the question given above, the following data were obtained:

Volume (V) = 10 L

Mass of He = 0.329 Kg

Temperature (T) = 28.0 °C

Molar mass of He = 4 g/mol

Mole of He =?

Next, we shall convert 0.329 Kg of He to g. This can be obtained as follow:

1 Kg = 1000 g

Therefore,

0.329 Kg = 0.329 Kg × 1000 g / 1 Kg

0.329 Kg = 329 g

Thus, 0.329 Kg is equivalent to 329 g.

Finally, we shall determine the number of mole of He in the tank. This can be obtained as illustrated below:

Mass of He = 329 g

Molar mass of He = 4 g/mol

Mole of He =?

Mole = mass / molar mass

Mole of He = 329 / 4

Mole of He = 82.25 moles

Therefore, there are 82.25 moles of He in the tank.

8 0

2 years ago

một hòn đá ném theo phương nằm ngang với vân tốc v0=15 m/s. tính gia tốc tiếp tuyến và gia tốc pháp tuyến của hòn đá sau lúc ném

Lady_Fox [76]

Answer:

what language is this???

5 0

2 years ago

Acceleration and Force

olga55 [171]

Answer:

I'm pretty sure its 3m/s^2 for the acceleration but I don't know the force part sorry .

Explanation:

15m/s - 0m/s divided by 5 s = 3m/s

I'm no expert or anything so I could be wrong but this is the best I can give you. Sorry

6 0

1 year ago

A 45.0-kg girl is standing on a 168-kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat,

muminat

Answer:

The speed of the plank relative to the ice is:

$v_{p}=-0.33\: m/s$

Explanation:

Here we can use momentum conservation. Do not forget it is relative to the ice.

$m_{g}v_{g}+m_{p}v_{p}=0$ (1)

Where:

m(g) is the mass of the girl
m(p) is the mass of the plank
v(g) is the speed of the girl
v(p) is the speed of the plank

Now, as we have relative velocities, we have:

$v_{g/b}=v_{g}-v_{p}=1.55 \: m/s$ (2)

v(g/b) is the speed of the girl relative to the plank

Solving the system of equations (1) and (2)

$45v_{g}+168v_{p}=0$

$v_{g}-v_{p}=1.55$

$v_{p}=-0.33\: m/s$

I hope it helps you!

8 0

2 years ago

A mass is oscillating with amplitude A at the end of a spring.

Dmitry_Shevchenko [17]

A) $x=\pm \frac{A}{2\sqrt{2}}$

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

$E=\frac{1}{2}kA^2$ (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

$E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2$ (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

$U=\frac{1}{3}K$

Using (2) we can rewrite this as

$U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}$

And using (1), we find

$U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2$

Substituting $U=\frac{1}{2}kx^2$ into the last equation, we find the value of x:

$\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}$

B) $x=\pm \frac{3}{\sqrt{10}}A$

In this case, the kinetic energy is 1/10 of the total energy:

$K=\frac{1}{10}E$

Since we have

$K=E-U$

we can write

$E-U=\frac{1}{10}E\\U=\frac{9}{10}E$

And so we find:

$\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A$

3 0

2 years ago