1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
MAVERICK [17]
3 years ago
7

An unknown solution has a pH of 7. How would you classify this solution? acidic basic neutral There is not enough information t

o answer the question.
Physics
2 answers:
Fudgin [204]3 years ago
7 0
There's exactly enough information to answer the question! xD
pH stands for power of Hydrogen. It represents the proportion of Hydrogen ions in your body. the values of the pH ranges from 1 to 14.
When the pH = 7, the solution is considered as neutral. When the pH > 7, the solution is considered as Basic. And when the pH < 7, the solution is considered as Acidic.
By hypothesis, we know nothing about the solution except of its pH which is 7. Which means that this is a neutral solution.

Hope this Helps! :)
Schach [20]3 years ago
6 0
It would be a neutral,

The acidity of water is 7 as well

You might be interested in
Some of the highest tides in the world occur in the Bay of Fundy on the Atlantic Coast of Canada. At Hopewell Cape the water dep
Nady [450]

Answer:

(a) 1.939 m/h

(b) 0.926 m/h

(c) -0.315 m/h

(d) -1.21 m/h

Explanation:

Here, we have the water depth given by the function of time;

D(t) = 7 + 5·cos[0.503(t-6.75)]

Therefore, to find the velocity of the depth displacement with time, we differentiate the given expression with respect to time as follows;

D'(t) = \frac{d(7 + 5\cdot cos[0.503(t-6.75)])}{dt}

= 5×(-sin(0.503(t-6.75))×0.503

= -2.515×(-sin(0.503(t-6.75))

= -2.515×(-sin(0.503×t-3.395))

Therefore we have;

(a) at 5:00 AM = 5 -  0:00 = 5

D'(5) =  -2.515×(-sin(0.503×5-3.395)) = 1.939 m/h

(b) at 6:00 AM = 6 -  0:00 = 6

D'(5) =  -2.515×(-sin(0.503×6-3.395)) = 0.926 m/h

(c) at 7:00 AM = 7 -  0:00 = 7

D'(5) =  -2.515×(-sin(0.503×7-3.395)) = -0.315 m/h

(d) at Noon 12:00 PM = 12 -  0:00 = 12

D'(5) =  -2.515×(-sin(0.503×12-3.395)) = -1.21 m/h.

4 0
2 years ago
Thanks + BRAINLIST <br><br> Please need correct answer asappp
nirvana33 [79]

Answer:

  1. Standing waves can be thought of as a sin wave and a cos wave overlapping each other. They go in different direction hence C is correct
  2. Wave interference can be thought of as the opposite of destructive ---> constructive anda hencd meet and interact on the same medium such that answer B is correct
6 0
2 years ago
A spring is hung from the ceiling. A 0.442-kg block is then attached to the free end of the spring. When released from rest, the
siniylev [52]

Answer:

K=58.8N/m

Explanation:

From the question we are told that:

Mass M=0.442

Drop distance d=0.150

Generally the equation for Spring Constant is mathematically given by

 K=\frac{2mg}{x}

 K=\frac{2*0.442*9.8}{1.150}

 K=58.8N/m

6 0
3 years ago
A large, cylindrical water tank with diameter 3.60 m is on a platform 2.00 m above the ground. The vertical tank is open to the
zysi [14]

To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.

The volume of a tank is given by

V = \frac{\pi d^2}{4}h

Where

d = Diameter

h = Height

Considering that there are two stages, let's define the initial and final volume as,

V_0 = \frac{\pi d^2}{4}H

V_f = \frac{\pi d^2}{4}h

We know as well by definition that

1gal = 3.785*10^{-3}m^3

Then we have for the statement that

V_f = V_0 -1gal

V_f = V_0 - 3.785*10^{-3}

Replacing the previous data

\frac{\pi d^2}{4}h = \frac{\pi d^2}{4}H- 3.785*10^{-3}

\frac{\pi (3.6)^2}{4}h = \frac{\pi (3.6)^2}{4}(2)- 3.785*10^{-3}

Solving to get h,

h = 1.99963m

Therefore the change is

\Delta h = H-h

\Delta h = 2- 1.99963

\Delta h = 3.7*10^{-4}m=0.37mm

Therefore te change in the height of the water in the tank is 0.37mm

4 0
3 years ago
A sailboat moves north for a distance of 10.00 km when blown by a wind from the exact south with a force of 5.00 x 10^4 n. how m
Nataliya [291]
Work is defined as the force times the distance which is mathematically expressed W = Fxd. The given force is 5x10^4 and the distance is 10000 m (the distance is converted as meter because Nm = J) the work done by the wind is W = 5x10^4 N (10000) = 500 x 10^6 Joules. I hope it answered your question
8 0
2 years ago
Other questions:
  • Volume of sound is controlled by the _____ of a wave
    10·1 answer
  • A jet is travelling at a speed of 1200 km/h and drops cargo from a height of 2.5 km above the ground Calculate the time it takes
    13·1 answer
  • find the period of a simple pendulum of 1m length placed on earth and on moon g on moon =1.67m/s² g on earth=10m/s²
    12·1 answer
  • A series of evenly timed pulses create a wave that can be described as a _______<br><br> wave.
    15·1 answer
  • What is the momentum of a 546,540 kg train that is travelling at 7.8 m/s​
    6·1 answer
  • How to calculate motion​
    8·2 answers
  • Which of the following are true about renewable resources?
    13·1 answer
  • According to you who was more genius Albert Einstein or Newton?​
    8·2 answers
  • Which of the following scenarios represents a safe situation for the child?
    9·2 answers
  • Draw the well labelled diagram of thermo flask.​
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!