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4 years ago

In this nuclear reaction equation, oxygen decays to form nitrogen. Which equation correctly describes this decay?

Chemistry

2 answers:

aalyn [17]4 years ago

8 0

Answer:

The answer is A

Explanation:

i just took the asignment

Maru [420]4 years ago

7 0

Answer:

A on edge and beta plus decay

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Before entering the cyclotron, the particles are accelerated by a potential difference V. Find the speed v with which the partic

vesna_86 [32]

Answer:

Speed, $v=\sqrt{\dfrac{2qV}{m}}$

Explanation:

The device which is used to accelerate charged particles to higher energies is called a cyclotron. It is based on the principle that the particle when placed in a magnetic field will possess a magnetic force. Just because of this Lorentz force it moves in a circular path.

Let m, q and V are the mass, charge and potential difference at which the particle is accelerated.

The work done by the particles is equal to the kinetic energy stored in it such that,

$qV=\dfrac{1}{2}mv^2$

v is the speed with which the particles enter the cyclotron

So,

$v=\sqrt{\dfrac{2qV}{m}}$

So, the speed with which the particles enter the cyclotron is $v=\sqrt{\dfrac{2qV}{m}}$ . Hence, this is the required solution.

7 0

3 years ago

Consider the following reaction:

zzz [600]

Answer:

Question 1: Both the rate of reaction of H₂O₂ and the rate of formation of H₂O are:

$6.6\cdot 10^{-3}mol/(liter.s)$

Question 2: 0.60 moles of O₂ are formed in the first 50 s of reaction.

Explanation:

The coefficients of the balanced chemical equation tells the relation of formation of the reactants into the products, which is also the relation between their rates of reaction.

The balanced chemical equation (given) is:

$2H_2O_2(aq)\rightarrow 2H_2O(l)+O_2(g)$

Then, in the same time that 2 moles of H₂O₂ react, 2 moles of H₂O and 1 mol of O₂ are formed.

1. Question 1.

That means that the rate of formation of O₂ is half the rate of reaction of H₂O₂ and half the rate of formation of H₂O.

Hence, if the instantaneous rate of formation of O₂ is 3.3×10⁻³moles/(liters×seconds), then the rate of reaction of H₂O₂ and the rate of formation of H₂O is twice:

$2\times 3.3\cdot 10^{-3}mol/(liter.s)=6.6\cdot 10^{-3}mol/(liter.s)$

2. Question 2.

First you must find how much the concentration has changed.

You can read the initial concentration of H₂O₂ on the graph. It is the y-intercept. It is 1.0M.

You can read the approixmate concentration of H₂O₂ at the time 50 s. It is about 0.2M.

Then the change in concentration in the first 50 s of reaction is about 1.0M - 0.2M = 0.8M.

Now you can convert the concentration 0.8M into number of moles, using the volume (1.5 liters), assuming the liquid volume does not change, and the molarity definition (molarity is the number of moles of solute in one liter of solution)

$\Delta [H_2O_2]=\Delta moles/Volume(liters)$