All categories

2 years ago

How many grams of co2 are in 3 moles of the compound

Physics

2 answers:

Ber [7]2 years ago

7 0

Answer:132.0285

Explanation: Hope this helps!

kati45 [8]2 years ago

4 0

Answer:

132 g

Explanation:

we know,

moler mass of co₂ = 44.01 g/mol

it means,

1 moles of co₂ =44.01 g

so,

3 moles of co₂ =(44.01 x 3) g

=132.03 g

≈ 132 g

You might be interested in

What is the student's kinetic energy at the bottom of the hill if he is moving

soldi70 [24.7K]

Answer:

KE = 10530 J or 10.53 KJ

Explanation:

The formula for kinetic energy is KE = 1/2 mv^2

Let's apply the formula:

KE = 1/2 mv^2

KE = 1/2 (65kg) (18m/s)^2

KE = 10530 J or 10.53 KJ

5 0

2 years ago

An object has a mass of 5.20g. The mass of the object in kg is

vampirchik [111]

Answer:

0.0520kg

Explanation:

divide by 1000

8 0

2 years ago

A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet

dimaraw [331]

Answer:

$v = 2.8898 \frac{m}{s}$

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

$E_{ep} = \frac{1}{2} k (\Delta x)^2$

where $\Delta x$ is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

$\vec{F} = - k \Delta \vec{x}$

as we need 9.12 N to compress 4.60 cm, this means:

$k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}$

$k = 198.26 \ \frac{ N}{m}$

So, the elastic energy of the compressed spring is:

$E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2$

$E_{ep} = 0.209759 \ Joules$

And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

$\Delta E_{gp} = m g \Delta h$

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

$\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m$

$\Delta E_{gp} = 0.00224 \ Joules$

<h3>Kinetic Energy</h3>

We know that the kinetic energy for a mass m moving at speed v is:

$E_k = \frac{1}{2} m v^2$

so, for the pellet will be

$E_k = \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

<h3>All together</h3>

By conservation of energy, we know:

$E_{ep} = \Delta E_{gp} + E_k$

$0.209759 \ Joules = 0.00224 \ Joules + \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.209759 \ Joules - 0.00224 \ Joules$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.207519 \ Joules$

$v = \sqrt{ \frac{ 0.207519 \ Joules}{ \frac{1}{2} \ 4.97 \ 10^{-3} kg } }$

$v = 2.8898 \frac{m}{s}$

7 0

2 years ago

Two men are standing on a frictionless ice surface holding opposite ends of a rope. One man (mass = 80 kg) pulls on the rope wit

ankoles [38]

Answer:

The acceleration of man 1 and 2 is $3.125\ m/s^2$ and $4.167\ m/s^2$ .

Explanation:

Mass of man 1, m₁ = 80 kg

Mass of man 2, m₂ = 60 kg

One man pulls on the rope with a force of 250 N.

Let a₁ is acceleration of man 1,

F = m₁a₁

$a_1=\dfrac{F}{m_1}\\\\a_1=\dfrac{250}{80}\\\\a_1=3.125\ m/s^2$

Let a₂ is acceleration of man 1,

F = m₂a₂

$a_2=\dfrac{F}{m_2}\\\\a_2=\dfrac{250}{60}\\\\a_2=4.167\ m/s^2$

So, the acceleration of man 1 and 2 is $3.125\ m/s^2$ and $4.167\ m/s^2$ .

7 0

2 years ago

How do you do this question?

umka2103 [35]

Explanation:

The moment of inertia of each disk is:

Idisk = 1/2 MR²

Using parallel axis theorem, the moment of inertia of each rod is:

Irod = 1/2 mr² + m (R − r)²

The total moment of inertia is:

I = 2Idisk + 5Irod

I = 2 (1/2 MR²) + 5 [1/2 mr² + m (R − r)²]

I = MR² + 5/2 mr² + 5m (R − r)²

Plugging in values:

I = (125 g) (5 cm)² + 5/2 (250 g) (1 cm)² + 5 (250 g) (5 cm − 1 cm)²

I = 23,750 g cm²

7 0

2 years ago