I’m confused on question 13

3. Astronomers use various tools to gather evidence in support of the Big Bang

katrin [286]

The Big Bang theory is one of the ones used most frequently nowadays to describe how the cosmos came into being.

<h3>What does big bang theory state?</h3>

According to this theory, the universe formed 13.8 billion years ago as a result of the primordial atom, a single particle, exploding and generating an unprecedented cosmic disaster. The universe is always expanding, according to the same theory.

The key supporting data for the big bang theory are:

The galaxies to getting further and distant from us. This distance is taken at high speeds. In other words, the distance between galaxies is inversely correlated with their spacing velocity. Our perception of galaxy dysthanization occurs because the universe is continually expanding, according to this data, which is known as the "Hubble Law," and this hypothesis also explains that The cosmos has already been unified.

According to the big bang theory, the cosmos was extremely hot at the time of its creation; if this is accurate, we should find some proof of this warmth. Radio astronomers discovered this proof when they noticed that microwaves with a temperature of 2,725 degrees Kelvin included cosmic background radiation. This radiation is perhaps proof of the universe's high temperature.

Last but not least, the Big Bang most likely contributed to the abundance of hydrogen and helium elements.

Learn more about big bang theory here:

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7 0

2 years ago

A 0.15 m solution of chloroacetic acid has a ph of 1.86. What is the value of ka for this acid?

dem82 [27]

Answer: $1.67\times 10^{-3}$

Explanation:

$ClCH_2COOH\rightarrow ClCH_2COO^-+H^+$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Given: c = 0.15 M

pH = 1.86

$K_a$ = ?

Putting in the values we get:

Also $pH=-log[H^+]$

$1.86=-log[H^+]$

$[H^+]=0.01$

$[H^+]=c\times \alpha$

$0.01=0.15\times \alpha$

$\alpha=0.06$

As $[H^+]=[ClCH_2COO^-]=0.01$

$K_a=\frac{(0.01)^2}{(0.15-0.15\times 0.06)}$

$K_a=1.67\times 10^{-3]$

Thus the vale of $K_a$ for the acid is $1.67\times 10^{-3}$

4 0

3 years ago

I am a nonmetal.I am in the Oxygen family and in row 3.I have 6 valence electrons.I am yellow and have a stinky smell.Who am i

Dimas [21]

Answer:

sulfur

Explanation:

In oxygen family sulfur has yellow color and also having stinky smell. Thus given statements are about sulfur.

It is present in oxygen family.

It has six valance electrons.

Its atomic number is 16.

Its atomic weight is 32 amu.

The electronic configuration of sulfur is given below,

S₁₆ = 1s² 2s² 2p⁶ 3s² 3p⁴

We can see the valance shell is third shell and it have six electrons thus sulfur have six valance electrons. (3s² 3p⁴ )

Sulfur is used in vulcanisation process.

It is used in bleach and also as a preservative for many food.

it is used to making gun powder.

8 0

3 years ago

17 points help asap!! :) thank youuuu :) Jeff is in a stationary school bus. Which is the best reference point for him to use to

3241004551 [841]

Answer:

The answer would be A

Hope this helps!!!

5 0

3 years ago

1.33 dm3 of water at 70°C are saturated by 2.25

astraxan [27]

Given that 4.50 dm³ of Pb(NO₃)₂ is cooled from 70 °C to 18 °C, the

amount amount of solute that will be deposited is 1,927.413 grams.

<h3>How can the amount of solute deposited be found?</h3>

The volume of water 1.33 dm³ of water 70 °C.

The number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water at 70 °C = 2.25 moles

At 18 °C, the number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water = 0.53 moles

Therefore;

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C is therefore;

1.33 dm³ contains 2.25 moles.

$Number \ of \ moles \ in \ 4.50 \ dm^3 = \dfrac{2.25}{1.33} \times 4.50 \approx \mathbf{7.613 \, moles}$

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C ≈ 7.613 moles

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C is therefore;

1.33 dm³ contains 0.53 moles

$Number \ of \ moles \ in \ 4.50 \ dm^3 = \dfrac{0.53}{1.33} \times 4.50 \approx \mathbf{1.79 \, moles}$

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C ≈ 1.79 moles

The number of moles that precipitate out = The amount of solute deposited

Which gives;

Amount of solute deposited = 7.613 moles - 1.79 moles = 5.823 moles

The molar mass of Pb(NO₃)₂ = 207 g + 2 × (14 g + 3 × 16 g) = 331 g

The molar mass of Pb(NO₃)₂ = 331 g/mol

The amount of solute deposited = Number of moles × Molar mass

Which gives;

The amount of solute deposited = 5.823 moles × 331 g/mol =<u> 1,927.413 g </u>

Learn more about saturated solutions here:

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5 0

3 years ago