All categories

3 years ago

What is the current scientific thinking about the formation of the solar system?

Chemistry

2 answers:

vitfil [10]3 years ago

8 0

Answer:

D. The solar system formed from the dust and gases in a nebula. A center bulge in the nebula became the sun. The area of flat rings in the nebula became the individual orbital paths in which the planets revolve around the sun in the plane of the ring the path came from. Dust and gas particles found in each area combined to form the planets and their moons.

Explanation:

Stars are born in molecular clouds known as nebula.

According to current most believed theory, solar system was born in solar nebula. It is composed of dust and gases. This cloud of dust and gases collapsed under its own weight and condensed to form the core. It started spiralling and gathering more mass till temperature rose to start the nuclear fusion reaction and the sun was born. The rest of the matter accreted to form planets and their moons. The planets orbit the sun in elliptical paths.

Thus, options D is correct.

Eva8 [605]3 years ago

6 0

The solar system formed dust and many gases in a nebula. The center of the nebula eventually became the sun. The flat rings of outside the nebula became the orbit paths, and that is where the planets stay and rotate around the sun. After that dust particles and gases that were found were all combined up to create planets and moons. so its D

MARK AS BRAINLIEST

You might be interested in

Identify the balanced equation for the following reaction: SO2(g) + O2(g) → SO3(g)

sergiy2304 [10]

Answer: The balanced equation for the given reaction is

$2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g)$ .

Explanation:

A chemical equation which contains same number of atoms on both reactant and product side.

For example, $SO_{2}(g) + O_{2}(g) \rightarrow SO_{3}(g)$

Here, number of atoms on reactant side are as follows.

S = 1

O = 4

Number of atoms on product side are as follows.

S = 1

O = 3

To balance this equation, multiply $SO_{2}$ by 2 on reactant side and multiply $SO_{3}$ by 2. Hence, the equation will be re-written as follows.

$2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g)$

Here, number of atoms on reactant side are as follows.

S = 2

O = 6

Number of atoms on product side are as follows.

S = 2

O = 6

Now, there are same number of atoms on both reactant and product side. So, this equation is balanced.

Thus, we can conclude that the balanced equation for the given reaction is $2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g)$ .

3 0

3 years ago

Help me this 2 . I’m not good at this sorry

emmainna [20.7K]

Left is respiratory

Right is circulatory

6 0

3 years ago

4. Energy can be conserved by -

mart [117]

Energy can be conserved by efficient energy use.

Answer: Option A

Explanation:

Energy can be transferred from one form to another, but it cannot be destroyed or created. So it can be conserved if efficiently used. Thus efficient usage of energy lead to conservation of energy. Due to conservation of energy, the forces can be renewable and non-renewable.

So, we should know how the input energy can be completely converted to another form of energy leading to efficient usage of energy without any loss. As if there is no loss, input energy will be equal to output energy leading to 100% efficiency.

4 0

3 years ago

A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.

irga5000 [103]

Answer:

1. pH = 1.23.

2. $H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

$H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+$

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the pKa is:

$pKa=-log(Ka)=-log(5.90x10^{-2})=1.23$

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

$pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23$

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

$n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol$

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

$H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Which is also shown in net ionic notation.

Best regards!

4 0

3 years ago

Calculate the number of moles found in 3.045x1024 atoms of helium. PLS HELP

Kisachek [45]

Explanation:

so for this u have to use this equation where

Moles = number of particle/6.02×10^23

= 3.045 × 10^24/6.02×10^23

= 5.0581

write it to 3 S.F so 5.06 moles

4 0

2 years ago