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2 years ago

Explain the current theory of what a quasar is and where quasars can be found.

Chemistry

1 answer:

Kay [80]2 years ago

8 0

Answer:

Current understanding

It is now known that quasars are distant but extremely luminous objects, so any light which reaches the Earth is redshifted due to the metric expansion of space. ... Light and other radiation cannot escape from within the event horizon of a black hole.

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After an investigation, what do scientists often do?

den301095 [7]

Record observations, pose a question, create a test for an individual variable, test, come to a conclusion

7 0

3 years ago

Draw the structure(s) of all of the possible monochloro derivatives of 2,4-dimethylpentane, c7h15cl.

pishuonlain [190]

The monochloroderivatives will be obtained by substituting chemically non equivalent hydrogen with chlorine atom, one by one

So the possible monochloro derivatives of 2,4-dimethylpentane (figure 1) are shown in figure (2)

7 0

3 years ago

The compound known as diethyl ether, commonly referred to as ether, contains carbon, hydrogen, and oxygen. A 1.376 g sample of e

hoa [83]

Answer:

The answer to your question is: C₄H₁₀O

Explanation:

Data

CxHyOz

mass sample : 1.376 g

mass CO₂ = 3.268 g

mass H₂O = 1.672 g

Process

Reaction

CxHyOz + O₂ ⇒ CO₂ + H₂O

1.- Calculate the moles and mass of carbon

Molecular mass CO₂ = 44g

44 g of CO₂ -------------- 12 g of C

3.268 g of CO₂ -------- x

x = (3.268 x 12) / 44

x = 0.891 g of Carbon

12 g of carbon ----------- 1 mol

0.891 g of C ---------- x

x = (0.891 x 1) / 12

x = 0.0743 moles of carbon

2.- Calculate the moles and mass of hydrogen

18 g of water --------------- 2 g of H

1.672 g of H₂O ------------ x

x = (1.672 x 2) / 18

x = 0.186 g of hydrogen

1 g of hydrogen ------------ 1 mol of H

0.186 g of H ------------ x

x = (0.186 x 1) / 1

x = 0.186 moles of H

3.- Calculate the mass of Oxygen and its moles

Mass of Oxygen = 1.376 - 0.891 - 0.186

= 0.299 g of O₂

Moles of Oxygen

16 g of Oxygen ---------------- 1 mol

0.299 g of O ----------------- x

x = (0.299 x 1) / 16

x = 0.019 moles of Oxygen

4.- Divide by the lowest number of moles

Carbon 0.0743/ 0.019 = 3.9 ≈ 4.0

Hydrogen 0.186/ 0.019 = 9.7 = 10

Oxygen 0.019/ 0.019 = 1

5.- Write the empirical formula

C₄H₁₀O

4 0

3 years ago

Contrast a mixture and a pure substance. In your answer be sure to explain what each is, how you can tell them apart, and supply

Wewaii [24]

A pure substance<span> has a definite and constant composition. A pure substance can be either an element or a compound, but the composition of a pure substance doesn’t vary.</span> Ex: salt or sugar<span>

</span>Mixtures are physical combinations of pure substances that have no definite or constant composition — the composition of a mixture varies according to who prepares the mixture. Ex: sand mixed with salt

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3 years ago

Round the mmHg readings to the nearest whole number. Round the kPa readings to the nearest tenth. Round the atm readings to the

denpristay [2]

The pressure of 2.29 atm can be converted to $1740.4 mm Hg.$