<h3><u>Answer;</u></h3>
It contains only water.
<h3><u>Explanation;</u></h3>
- The pH of a solution is a measure of the molar concentration of hydrogen ions in the solution and as such is a measure of the acidity or basicity of the solution.
- A pH of 7 is neutral. A solution with a pH of 7 is known as neutral solution, such a solution has equal concentration of H+ and OH- concentration. The OH- concentration balances the H+ concentrations in the solution making it neutral.
- Pure water is an example of a solution that is neutral, has a pH of 7.
To solve this problem it is necessary to apply the concepts of Work. Work is understood as the force applied to travel a determined distance, in this case the height. The force in turn can be expressed by Newton's second law as the ratio between mass and gravity, as well

Where,
m = mass
h = height
g = Gravitational constant
When it ascends to the second floor it has traveled the energy necessary to climb a height, under this logic, until the 4 floor has traveled 3 times the height h of each of the floors therefore

Replacing in our equation we have to

The correct answer is 4.
Answer:

Explanation:
The resistance of a conductor is directly proportional to its length and is inversely proportional to its cross-sectional area, this dependence is given by:

is the material's resistance, L is the legth and A is the cross-sectional area.
For the first and second coils, we have:

For the third and fourth coils, we have:

Answer
given,
mass of block (m)= 6.4 Kg
spring is stretched to distance, x = 0.28 m
initial velocity = 5.1 m/s
a) computing weight of spring
k x = m g


k = 224 N/m
b) 




c) 


d) 


e)


A = 0.682 m
Force =
=
F = 94.20 N
Answer:
359 g Mn
General Formulas and Concepts:
- Dimensional Analysis
- Reading the Periodic Table of Elements
Explanation:
<u>Step 1: Define</u>
6.53 mol Mn
<u>Step 2: Find conversion</u>
1 mol Mn = 54.94 g Mn
<u>Step 3: Dimensional Analysis</u>
<u />
= 358.758 g Mn
<u>Step 4: Simplify</u>
<em>We are given 3 sig figs.</em>
358.758 g Mn ≈ 359 g Mn