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3 years ago

The equation below shows a general equation for a reaction, and the amounts of the substance are written underneath.

Physics

1 answer:

mash [69]3 years ago

8 0

Explanation:

The amount of CD is 35 g, and the amount of AC is 40 g.

Explanation:

AB + CD \rightarrow AC + BDAB+CD→AC+BD

The law of conservation of mass states that in any chemical reaction, the total mass of the reactants must be equal to the total mass of the products.

In this reaction, we know the mass of one reactant (AB, 15 g) and the mass of one product (BD, 10 g). In order to have the same total mass on the left side and on the right side of the equation, the mass of AC must be 5 g more than the mass of CD. We see that the only choice that satisfies this condition is:

The amount of CD is 35 g, and the amount of AC is 40 g.

In fact, if we assume these masses are correct, we have:

- on the reactant side: m(AB)+m(CD)= 15g + 35g = 50g

- on the product side: m(AC)+m(BD)= 40g + 10g = 50g

so, we have the same mass on both sides of the equation, and the law of conservation of mass is satisfied.

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MODERN PHYSICS

frosja888 [35]

Answer:

D. n=6 to n=2

Explanation:

Given;

energy of emitted photon, E = 3.02 electron volts

The energy levels of a Hydrogen atom is given as; E = -E₀ /n²

where;

E₀ is the energy level of an electron in ground state = -13.6 eV

n is the energy level

From the equation above make n, the subject of the formula;

n² = -E₀ / E

n² = 13.6 eV / 3.02 eV

n² = 4.5

n = √4.5

n = 2

When electron moves from higher energy level to a lower energy level it emits photons;

$E = E_0(\frac{1}{n_1^2}-\frac{1}{n_2^2} )\\\\\frac{1}{n_1^2}-\frac{1}{n_2^2} = \frac{E}{E_o} \\\\\frac{1}{4} -\frac{1}{n_2^2} = \frac{3.02}{13.6} \\\\\frac{1}{4} -\frac{1}{n_2^2} =0.222\\\\\frac{1}{n_2^2} = 0.25 - 0.22\\\\\frac{1}{n_2^2} = 0.03\\\\n_2^2 = 33.33\\\\n_2 = \sqrt{33.33} = 6$

For a photon to be emitted, electron must move from higher energy level to a lower energy level. The higher energy level is 6 while the lower energy level is 2

Therefore, The electron energy-level transition is from n = 6 to n = 2

3 0

3 years ago

Ignoring air resistance and the little friction from the plastic tube, the magnet was a freely-falling object in each trial. If

horsena [70]

Answer:

emf will also be 10 times less as compared to when it has fallen $40 \mathrm{m}$

Explanation:

We know, from faraday's law-

$e m f=-N \frac{\Delta \Phi}{\Delta T}$

and $\Phi=B . A$

So, as the height increases the velocity with which it will cross the ring will also increase. $(v=\sqrt{2 g h})$

Given

$\mathrm{V} 1(\text { Speed at } 40 \mathrm{m})=2 \mathrm{x} \mathrm{V} 2(\text { speed at } 10 \mathrm{m})$

$\sqrt{2 g h_{2}}=2 \times \sqrt{2 g h_{1}}=28.28 \mathrm{m} / \mathrm{s}$

Now, from $40 \mathrm{cm}$

$V_{3}=\sqrt{2 g h_{3}}=\sqrt{2 \times 10 \times 0.4}=2.82 \mathrm{m} / \mathrm{s}$

From equation a and b we see that velocity when dropped from $40 \mathrm{m}$ is 10 times greater when height is 40 $\mathrm{cm}$ so, emf will also be 10 times less as compared to when it has fallen $40 \mathrm{m}$