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Sonja [21]
3 years ago
10

These celestial bodies are thought to originate from two different sources. Some are called long-period and originate from the O

ort Cloud. These bodies orbit the Sun but may take as long as 200 years to complete one orbit. Others are called short-period and originate from the Kuiper Belt in our solar system.
What is this celestial body called?

A) Asteroid

B) Comet

C) Meteor

D) Planet
Chemistry
2 answers:
Whitepunk [10]3 years ago
7 0
Yes it would be metror
Alexus [3.1K]3 years ago
6 0
It’s c I believe because it says 209 years to complete an orbit so hint yourself ?.....
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The figure above shows the Earth at two different positions in its orbit around the Sun. Which position corresponds to summer in
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A. Position B.

Explanation:

In the summer the northern hemisphere is closer to the Sun so it's Position B.

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Which of the following is a correct inference that one can make about the burning of gasoline?
VLD [36.1K]

Answer:

It’s a physical change

Explanation:

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Wassup everyone?????
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Acid & Base Calculations Calculate the hydronium ion concentration for each. Tell whether it is an acid or a base. 1. pH = 5
Anuta_ua [19.1K]

<u>Explanation:</u>

pH is the negative logarithm of hydronium ion concentration present in a solution.

  • If the solution has high hydrogen ion concentration, then the pH will be low and the solution will be acidic. The pH range of acidic solution is 0 to 6.9
  • If the solution has low hydrogen ion concentration, then the pH will be high and the solution will be basic. The pH range of basic solution is 7.1 to 14
  • The solution having pH equal to 7 is termed as neutral solution.

To calculate the pH of the solution, we use equation:

pH=-\log[H_3O^+]     ......(1)

To calculate the pOH of the solution, we use the equation:

pH + pOH = 14                ........(2)

  • <u>For 1:</u>

We are given:

pH = 5.54

Putting values in equation 1, we get:

5.54=-\log[H_3O^+]

[H_3O^+]=2.88\times 10^{-6}M

Now, putting values in equation 2, we get:

14 = 5.54 + pOH

pOH = 8.46

The solution is acidic in nature.

  • <u>For 2:</u>

We are given:

pOH = 9.7

Putting values in equation 2, we get:

14 = 9.7 + pH

pH = 4.3

Now, putting values in equation 1, we get:

4.3=-\log[H_3O^+]

[H_3O^+]=5.012\times 10^{-5}M

The solution is acidic in nature.

  • <u>For 3:</u>

We are given:

pH = 7.0

Putting values in equation 1, we get:

7.0=-\log[H_3O^+]

[H_3O^+]=1.00\times 10^{-7}M

Now, putting values in equation 2, we get:

14 = 7.0 + pOH

pOH = 7.0

The solution is neither acidic nor basic in nature.

  • <u>For 4:</u>

We are given:

pH = 12.9

Putting values in equation 1, we get:

12.9=-\log[H_3O^+]

[H_3O^+]=1.26\times 10^{-13}M

Now, putting values in equation 2, we get:

14 = 12.9 + pOH

pOH = 1.1

The solution is basic in nature.

  • <u>For 5:</u>

We are given:

pOH = 1.2

Putting values in equation 2, we get:

14 = 1.2 + pH

pH = 12.8

Now, putting values in equation 1, we get:

12.8=-\log[H_3O^+]

[H_3O^+]=1.58\times 10^{-13}M

The solution is basic in nature.

  • <u>For 6:</u>

We are given:

[H_3O^+]=1\times 10^{-5}M

Putting values in equation 1, we get:

pH=-\log(1\times 10^{-5})

pH=5

Now, putting values in equation 2, we get:

14 = 5 + pOH

pOH = 9

The solution is acidic in nature.

5 0
3 years ago
I have 50.00 mL of 0.100 M ethyl amine (C2H5NH2). I gradually add a solution of 0.025 M nitric acid (HNO3) to the ethyl amine so
sergeinik [125]

Answer:

4.00 is the pH of the mixture

Explanation:

The ethyl amine reacts with HNO3 as follows:

C2H5NH2 + HNO3 → C2H5NH3⁺ + NO3⁻

To solve this question we need to find the moles of ethyl amine and the moles of HNO3:

<em>Moles C2H5NH2:</em>

0.0500L * (0.100mol/L) = 0.00500 moles ethyl amine

<em>Moles HNO3:</em>

0.201L * (0.025mol/L) = 0.005025 moles HNO3

That means HNO3 is in excess. The moles in excess are:

0.005025 moles HNO3 - 0.00500 moles ethyl amine =

2.5x10⁻⁵ moles HNO₃

In 50 + 201mL = 251mL = 0.251L:

2.5x10⁻⁵ moles HNO₃ / 0.251L = 9.96x10⁻⁵M = [H+]

As pH = -log [H+]

pH = -log 9.96x10⁻⁵M

pH = 4.00 is the pH of the mixture

6 0
3 years ago
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