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Ksju [112]
3 years ago
15

Help pls what is the (H+) of a solution with a pH of 4?

Chemistry
2 answers:
AfilCa [17]3 years ago
6 0
The answer will be a I’m sure

Answer :A
Natalka [10]3 years ago
4 0

Answer:

A

Explanation:

[H+] = 10^-pH

[H+] = 1 x 10^-4 M

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Which of the following characteristics describes aerobic organisms.
horsena [70]

Answer: Option (3) is the correct answer.

Explanation:

Aerobic organisms are the organisms which survive and grow in the presence of oxygen.

When oxidation of glucose occurs in the presence of oxygen then it is known as aerobic respiration.

In aerobic respiration, food releases energy to produce ATP which is necessary for cell activity. There is complete breakdown of glucose in aerobic respiration that is why more energy is released. Therefore, aerobic organisms become active.

Thus, we can conclude that characteristics very active, efficient use of energy describes aerobic organisms.

6 0
3 years ago
A ball is equipped with a speedometer and launched straight upward. The speedometer reading four seconds after launch is shown a
Andrew [12]

Answer:

Question 1: <u>1 s after the motion starts</u>

Question 2: <u>0 (just when the motion starts)</u>

Explanation:

You will need to work with approximates values because the precision of the speedometers is low and you are requested to find approximate times.

<u>1. From the speedometer shown at the right.</u>

You can obtain how long the ball has been falling from the highest altitute it reached using the speed of 10 m/s shown by the speedometer at the right.

  • Free fall equation: Vf = Vo - gt

  • Vo = 0 ⇒ Vf = gt ⇒ t = Vf / g

For this problem, I recommend to work with a rough estimate of g: g = 10 m/s² ( I will tell you why soon)/

  • t = [10 m/s] / [10 m/s²] = 1 s

That is the time falling. Since four seconds after launch have elapsed, the upward time was 3 seconds. This will let you to calculate the launching speed.

<u>2. Time when the speedometer displays a reading of 20 m/s</u>

First, calculate the launching speed:

  • Vf = Vo - gt

Since the ball was 3 seconds going upward and the speed at the maximum altitude is 0 you get:

  • 0 = Vo - gt

   

  • Vo = gt = 10 m/s² × 3 s = 30 m/s

Now, use the initial velocity to calculate when the ball is going upward with the speedometer reading is 20 m/s

  • 20 m/s = 30 m/s - 10 m/s² × t

  • t = [ 30 m/s - 20 m/s] / [10 m/s²] = 1 s

Thus, the first answer is t = 1 s.

<u />

<u>3. Time when the speedometer displays a reading of 30 m/s</u>

This is the same speec estimated for the launching: 30 m/s.

So, this reading corresponds to the moment when the ball was launched.

Thus time is 0, i.e. it is the same instant of the launch.

If you had worked with g = 9.80 m/s², the time had been negative. This is due to the precision of the instruments.

That is why I recommended to work with g = 10 m/s².

6 0
3 years ago
What type of epithelium would be most suited for high levels of diffusion and filtration?
NemiM [27]
Sorry the answer might be a little late but it would be <span>Simple squamous</span>
6 0
3 years ago
What is the poH of a<br> 4.8 x 10-10 M H+ solution?
stiks02 [169]

Answer: pOH = 4.68

Explanation:

pOH = 14 - pH

pH = - Log [H+]

= - Log [4.8 x 10^-10]

= -(-9.32)

pH =+9.32

Therefore, pOH= 14 - 9.32

= 4.68

6 0
3 years ago
What is the density (g/L) of CHCl3 vapor at 1.00 atm and 298 K?
Scrat [10]

Answer:

4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.

4 0
3 years ago
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