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2 years ago

Which of the four cars had the least net force applied to it when it was launched down the ramp

Chemistry

1 answer:

castortr0y [4]2 years ago

5 0

The least net force applied : Car 3(12 N)

<h3>Further explanation </h3>

Newton's 2nd law explains that the acceleration produced by the resultant force on an object is proportional and in line with the resultant force and inversely proportional to the mass of the object

∑F = m. a

Car 1 ⇒m=0.5 kg, a=36 m/s²

$\tt \sum F=0.5\times 36=18~N$

Car 2⇒m=0.8 kg, a=50 m/s²

$\tt \sum F=0.8\times 50=40~N$

Car 3⇒m=0.6, a=20 m/s²

$\tt \sum F=0.6\times 20=12~N$

Car 4⇒m=1, a=19~m/s²

$\tt \sum=1\times 19=19~N$

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For the following reaction, 22.6 grams of nitrogen monoxide are allowed to react with 4.64 grams of hydrogen gas . nitrogen mono

antiseptic1488 [7]

Answer:

- 10.5 g of N₂

- Limiting reagent: NO

- 3.13 g of H₂ remains

Explanation:

First of all we state the reaction: 2NO(g) + 2H₂(g) → 2H₂O(l) + N₂(g)

We need to find out the limiting reactant and the excess reagent

Ratio in the reactants is 2:2. Let's convert the mass to moles:

22.6 g / 30 g/mol = 0.753 moles of NO

4.64 g / 2 g/mol = 2.32 moles of H₂

Certainly the limiting reagent is the NO and the excess reactant is the hydrogen:

- For 0.753 moles of NO, we need 0.753 moles of H₂ (we have 2.32 moles)

- For 2.32 moles of H₂, we need 2.32 moles of NO (and we don't have enough NO, because we only have 0.753 moles)

As the H₂ is the excess reagent, some moles still remains after the reaction is complete → 2.32 mol - 0.753 mol = 1.567 moles

We convert the moles to mass: 1.567 mol . 2g /1mol = 3.13 g of H₂ remains

As the NO is the limiting reagent, we can work with the equation:

We propose this rule of three: 2 moles of NO can produce 1 mol of N₂

Then, 0.753 moles of NO must produce (0.753 . 1) /2 = 0.376 moles of N₂

We convert the moles to mass 0.376 mol . 28 g / 1 mol = 10.5 g

3 0

2 years ago

Describe Rutherford's contribution to the atomic model.

Nady [450]

Answer:

Rutherford's experiment, also known as

$\alpha - scattering \: experiment$

supports the existence of neutrons and the nucleus.

Explanation:

In the above diagram, Rutherford was trying to explain his contributions using thin foils of gold and other metals as targets for alpha particles from a radioactive source.

He observed that the majority of particles penetrated the foil either undeflected or with only a slight deflection. But, every now and then an alpha particle was scattered(or deflected) at a large angle..

According to Rutherford, most of the atoms must be empty space. This explains why the majority of alpha particles passed through through the gold foil with little or no deflection. The atoms positive charges, Rutherford proposed are all concentrated in the Nucleus, which is a dense central core within the atom.

Whenever an alpha particle came close to a nucleus in the scattering experiment, it experienced a large repulsive force and therefore a large deflection. Moreover, an alpha particle coming towards a nucleus would be completely repelled and its direction would be reversed. The positively charged particles in the Nucleus are called Protons.

I hope you find this useful... Have a lovely day. 

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sammy [17]

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alukav5142 [94]

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