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2 years ago

Which diagram shows a pattern similar to the emission spectrum of hydrogen?

Chemistry

1 answer:

rosijanka [135]2 years ago

3 0

Hello!

the correct answer would be b!

explanation: with increasing wavelength, it goes from higher energy to lower energy. since energy levels converge at higher energy levels, the answer b shows that the image of line becoming more separated than the following other answers

if this helps, click “thanks” :)

- emily

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How many grams of sulfur must be burned to give 100.0 g of So2

andriy [413]

Answer:

50 g of S are needed

Explanation:

To star this, we begin from the reaction:

S(s) + O₂ (g) → SO₂ (g)

If we burn 1 mol of sulfur with 1 mol of oxygen, we can produce 1 mol of sulfur dioxide. In conclussion, ratio is 1:1.

According to stoichiometry, we can determine the moles of sulfur dioxide produced.

100 g. 1mol / 64.06g = 1.56 moles

This 1.56 moles were orginated by the same amount of S, according to stoichiometry.

Let's convert the moles to mass

1.56 mol . 32.06g / mol = 50 g

4 0

2 years ago

A beach has a supply of sand grains composed of calcite, ferromagnesian silicate minerals, and non-ferromagnesian silicate miner

bezimeni [28]

Ferromagnesian silicate minerals (i looked it up)

4 0

2 years ago

The number of p atoms in 1.0 g of ba3(po4)2 is:

marta [7]

First, we determine the number of moles of barium phosphate. This is done using:

Moles = mass / Mr
Moles = 1 / 602
Moles = 0.002

Now, we see from the formula of the compound that each mole of barium phosphate has 2 moles of phosphorus (P). Therefore, the moles of phosphorus are:
0.002 * 2 = 0.004

The number of particles in one mole of substance is 6.02 x 10²³. The number of phosphorus atoms will be:
0.004 * 6.02 x 10²³

2.41 x 10²¹ atoms of phosphorus are present

3 0

2 years ago

A gas is contained in a thick-walled balloon. When the pressure changes from 319 mm Hg to 215 mm Hg, th volume changes from 0.55

Dmitry_Shevchenko [17]

In this question we need to find the new volume of the gas. Since we have been given the pressure and temperature change, we can used to combined gas law equation.
$\frac{P1V1}{T1} = \frac{P2V2}{T2}$
the parameters for 1st instance are given on the left side and parameters for the second instance are given on the right side of the equation
(319 mmHg x 0.558 L)/ 115 K = (215 mmHg x V)/387 K
V = 2.79 L

7 0

2 years ago

Read 2 more answers

A converging lens with a focal length of 70.0cm forms an image of a 3.20-cm-tall real object that is to the left of the lens. Th

Setler [38]

Answer: (i)The object is at a distance of 120cm from the lens while the image is at a distance of 160cm from the lens. (ii)The image is Real

Note: This is the complete question; A converging lens with a focal length of 70.0cm forms an image of a 3.20-cm-tall real object that is to the left of the lens. The image is 4.50cm tall and inverted. Where are the object and image located in relation to the lens? Is the image real or virtual?

Explanation:

The required formulas are;

1. lens formula given by, 1/s' + 1/s = 1/f

2. magnification = h'/h = s'/s

where h' is image height, h is object height, s' is image distance, s is object distance, f is focal length of lens.

h' = 4.50cm, h = 3.20cm, f = 70cm ( f of a converging lens is positive)

solving for s' and s in eqn (2)

4.50/3.20=s'/s

1.4=s'/s

s'=1.4s

to find the values of s' and s, we use equation (1) and substitute s' = 1.4s

1/1.4s + 1/s = 1/70

2.4/1.4s =1/70

s = 2.4*70/1.4 = 120cm

s' = 1.4*120 = 160cm

Therefore, the object is on the left, 120cm from the lens while the image is 160cm on the right of the lens

Since the object is at a distance between f and 2f from the lens, the image formed is real, inverted and magnified

3 0

2 years ago