All categories

2 years ago

Why is the Hubble’s law considered a piece of evidence that supports the Big Bang?

Chemistry

2 answers:

Harrizon [31]2 years ago

6 0

Answer:

Hubble's law is considered the first observational basis for the expansion of the universe, and today it serves as one of the pieces of evidence most often cited in support of the Big Bang model. The motion of astronomical objects due solely to this expansion is known as the Hubble flow

Explanation:

Hubble's law says that the universe is expanding outward. Actually Hubble's law was discovered before the Big Bang theory was formulated. The Big Bang Theory is an attempt to explain the observations that led to Hubble's Law. Before the 1900s the theory was that the universe was eternal and self existent.

The real significance of Hubble's Law is that the Universe is expanding in all directions not just from Earth. This means all matter in the Universe started at one point in space at the time of 'the Big Bang'.

Cosmic microwave background radiation as evidence for the Big Bang and expansion of the Universe. American astronomer Edwin Hubble (the space telescope was named after him) measured the speed of galaxies and their distance from Earth and obtained the following graph.

Almost all astronomers agree on the theory of the Big Bang, that the entire Universe is spreading apart, with distant galaxies speeding away from us in all directions. Run the clock backwards to 13.8 billion years ago, and everything in the Cosmos started out as a single point in space. In an instant,...

Leni [432]2 years ago

4 0

Explanation:

Hubble's Law basically states that the greater the distance of a galaxy from ours, the faster it recedes. It was proof that the Universe is expanding. It was also the first observational support for a new theory on the origin of the Universe proposed by Georges Lemaitre: the Big Bang.

You might be interested in

gogolik [260]

Answer: I2 is the Oxidant; while the 2S2O3(-2) is the reductant.

Explanation:

An Oxidant is any substance that oxidizes, or receives electrons from, another; in so doing, it becomes reduced in oxidation number.

A Reductant thus exactly the opposite.

Note that the equation provided shows that Iodine (I2) received an electron to become NEGATIVELY CHARGED:

I2 --> 2I-.

The oxidation number reduced from 0 to -1.

In contrast, the oxidation number of 2S2O3(-2) increases from -4 to -2.

Thus, I2 is the Oxidant; while the 2S2O3(-2) is the reductant.

7 0

3 years ago

a concentration solution of H2so4 is 59.4% by mass (m/m) and has a density of 1.83 g/mL. How many mL of the solution would be re

Blababa [14]

Answer: 41.5 mL

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

Given : 59.4 g of $H_2SO_4$ in 100 g of solution

moles of $H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=\frac{59.4g}{98g/mol}=0.61$

Volume of solution = $\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.83g/ml}=54.6ml$

Now put all the given values in the formula of molality, we get

$Molality=\frac{0.61\times 1000}{54.6ml}=11.2M$

To calculate the volume of acid, we use the equation given by neutralisation reaction:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of stock acid which is $H_2SO_4$

$M_2\text{ and }V_2$ are the molarity and volume of dilute acid which is $H_2SO_4$

We are given:

$M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL$

Putting values in above equation, we get:

$11.2\times V_1=0.30\times 1550\\\\V_1=41.5mL$

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

4 0

3 years ago

An area has experienced weathering and erosion for many years. Over time, some rock formations have been left behind forming pea

Anni [7]

Answer:

Explanation:

The rock that is left behind is there because it has resisted the forces of erosion.

6 0

3 years ago

80 POINTS!!! ALL PLATO USERS: asap!!!

Crank

This year course engages students in becoming skilled readers of prose written in a variety of periods, disciplines, and

rhetorical contexts and in becoming skilled writers who compose for a variety of purposes. More immediately, the course

prepares the students to perform satisfactorily on the A.P. Examination in Language and Composition given in the spring.

Both their writing and their reading should make students aware of the interactions among a writer’s purposes, audience

expectations, and subjects as well as the way generic conventions and the resources of language contribute to effectiveness

in writing. Students will learn and practice the expository, analytical, and argumentative writing that forms the basis of

academic and professional writing; they will learn to read complex texts with understanding and to write prose of

sufficient richness and complexity to communicate effectively with mature readers. Readings will be selected primarily,

but not exclusively, from American writers. Students who enroll in the class will take the AP examination.

4 0

3 years ago

What is Keq for the reaction N₂ + 3H2 = 2NH3 if the equilibrium

hram777 [196]

The Keq for the reaction N₂ + 3H2 = 2NH3 if the equilibrium concentrations are Keq = 1.5. The correct option is D.

Keq is the ratio of the concentration of reactant to the concentration of the product.

The balanced equation is

N₂ + 3H₂ = 2NH₃

The equilibrium constant is $\rm \dfrac{[NH_3]^2}{[N_2]\; [H_2]^3}$

The given concentrations of the compounds have been:

Ammonia = 3 M

Nitrogen = 1 M

Hydrogen = 2 M

$\rm \dfrac{9}{1\times 8} = 1.5$

Thus, the correct option is D. Keq = 1.5.

Learn more about Keq

brainly.com/question/24059926

#SPJ1

3 0

2 years ago