All categories

3 years ago

Which contains the most moles: 10 g of hydrogen gas, 100 g of carbon, or 50 g of lead?

Physics

1 answer:

Stolb23 [73]3 years ago

5 0

Answer:

Carbon

Explanation:

We know that

Number of moles = mass/molecular mass

Hydrogen: given as H2 with 10g an atomic number of 1 so MW( 1x 2= 2)

Number of moles = Weight/MW

= 10/2

= 5

Carbon: atomic number 12 C and mass 100

Number of moles = Weight/MW

= 100/12

= 8.33

Lead: atomic num 207 and mW 207 pb

Number of moles = Weight/MW

= 50/207

= 0.241

Therefore Carbon has the most moles.

You might be interested in

How far away is the furthest artificial satellite orbiting earth?

rusak2 [61]

Real-time distance and velocity data is provided by NASA and JPL. At a distance of 153.88 AU (23.020 billion km; 14.304 billion mi) from Earth as of September 5, 2021, it is the most distant artificial object from Earth. The probe made successful flybys of Jupiter, Saturn, and Saturn's largest moon, Titan.

8 0

3 years ago

A helicopter pulls upward by means of a rope on a 250 kg crate to lift it UNIFORMLY. What is the net force on the crate?

Cloud [144]

Answer:

The net force = 0

Explanation:

The given information includes;

The mass of the crate = 250 kg

The way the helicopter lifts the crate = Uniformly (constant rate (speed), no acceleration)

In order to pull the crate upwards, the helicopter has to provide a force equivalent to the weight of the crate keeping the helicopter on the ground.

The weight of the crate = The mass of the crate × The acceleration due gravity acting on the crate

The weight of the crate, $F_w$ ↓ = 250 kg × 9.81 m/s² = 2,452.5 N

The force the helicopter should provide to just lift the crate, $F_{(helicopter)}$ ↑ = The weight of the crate = 2,452.5 N

The net force, $F_{(net)}$ = $F_{(helicopter)}$ ↑ - $F_w$ ↓ = 2,452.5 N - 2,452.5 N = 0

The net force = 0.

3 0

3 years ago

Consider two conducting spheres with one having a larger radius than the other. Both spheres carry the same amount of excess cha

RSB [31]

Answer: Option (a) is the correct answer.

Explanation:

It is known that potential energy is the energy occupied by an object or substance due to its position is known as potential energy.

Therefore, more is the space occupied by an object more will be its position at a particular location. Hence, more will be its potential energy. On the other hand, smaller is the space occupied by an object, smaller will be the position holded by it.

Hence, smaller will be its potential energy.

Thus, we can conclude that for the given situation the statement, potential energy of the larger sphere is greater than that of the smaller sphere, is true.

6 0

4 years ago

Car A (mass 1100 kg) is stopped at a traffic light when it is rearended by car B(mass 1400 kg). Both cars then slide with locke

viva [34]

Answer:

Part a)

$v_a = 3.94 m/s$

Part b)

$v_b = 3.35 m/s$

Part C)

$v_b = 6.44 m/s$

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

Explanation:

Part a)

As we know that car A moves by distance 6.1 m after collision under the frictional force

so the deceleration due to friction is given as

$a = -\frac{F_f}{m}$

$a = -\frac{\mu mg}{m}$

$a = - \mu g$

now we will have

$v_f^2 - v_i^2 = 2ad$

$0 - v_i^2 = 2(-\mu g)(6.1)$

$v_a = \sqrt{(2(0.13)(9.81)(6.1)}$

$v_a = 3.94 m/s$

Part b)

Similarly for car B the distance of stop is given as 4.4 m

so we will have

$v_b = \sqrt{2(0.13)(9.81)(4.4)}$

$v_b = 3.35 m/s$

Part C)

By momentum conservation we will have

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1400 v_b = 1100(3.94) + 1400(3.35)$

$v_b = 6.44 m/s$

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

3 0

4 years ago

El coeficiente de variación de la resistencia con la temperatura del carbón es -0.0005/°c.Si la resistencia de una resistencia d

Doss [256]

Answer:

R (120) = 940Ω

Explanation:

The variation in resistance with temperature is linear in metals

ΔR (T) = R₀ α ΔT

where α is the coefficient of variation of resistance with temperature, in this case α = -0,0005 / ºC

let's calculate

ΔR = 1000 (-0,0005) (120-0)

ΔR = -60 Ω

ΔR = R (120) + R (0) = -60

R (120) = -60 + R (0)

R (120) = -60 + 1000

R (120) = 940Ω

3 0

4 years ago