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3 years ago

Explain the following statement: A concentrated acid is not necessarily a strong acid.

2 answers:

8 0

Well, we know that a concentrated acid has a high concentration (obviously). And a strong acid is where the solute is completely dissolved into the solvent. That means that not all concentrated acids are strong acids.

andrezito [222]3 years ago

3 0

Concentration involves the relative amounts of solvent and solute in a solution, when strength is related to the extent to which a substance dissociates :))
i hope this is helpful
have a nice day

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Please, pretty please help me!

Andrews [41]

Answer:

umm

Explanation:

1223565 578633 =334675

8 0

2 years ago

Read 2 more answers

if the vapor's volume were to be incorrectly recorded as 125ml, how will this error affect the calculated molar mass of the unkn

diamong [38]

So base on your question that as if the vapors volume were to incorrectly recorded as 125ml, the effect of the error to calculate the molar mass is the same as the error in measuring the volume of the vapor. I hope you are satisfied with my answer and feel free to ask for more

7 0

3 years ago

Describe two experiments to determine the speed of propagation of a transverse wave on a rope. You have the following tools to u

AnnZ [28]

Answer:

#See solution for details.

Explanation:

Tools: $stopwatch, \ meter \ stick, \ mass \ measuring \ scale , \ force \ measuring \ device$ .

$Experiment \ 1$ :Calculate the speed of the wave using the time, $t$ it takes to travel along the rope. Rope's length, $L$ is measured using the meter stick.

-Attach one end of rope to a wall or post, shake from the unfixed end to generate a pulse. Measure the the time it takes for the pulse to reach the wall once it starts traveling using the stopwatch.

-Speed of the pulse can then be obtained as:

$v=\frac{L}{t}$

$Experiment \ 2$ : Apply force of known value to the rope then use the following relation equation to find the speed of a pulse that travels on the rope.

$v=\sqrt{\frac{F}{\mu}}\ ,\mu=\frac{m}{L}$

-Use the measuring stick and measuring scale to determine $L,m$ values of the rope then obtain $\mu$ .

-Use the force measuring constant to determine $F$ . These values can the be substituted in $Experiment \ 1$ to obtain $v.$

4 0

3 years ago

Example No. 10

Alexxx [7]

The force constant of the spring is determined as 14,222.2 N/m.

<h3>Force constant of the spring</h3>

Apply the principle of conservation of energy,

K.E = U

where;

K.E kinetic energy of the elevator
U is elastic potential energy of the spring

¹/₂mv² = ¹/₂kx²

mv² = kx²

k = mv²/x²

Where;

m is mass of the elevator
v is speed
x is compression of the spring

k = (2000 x 8²)/(3²)

k = 14,222.2 N/m

Thus, the force constant of the spring is determined as 14,222.2 N/m.

Learn more about force constant here: brainly.com/question/1968517

#SPJ1

5 0

1 year ago

Assume the Earth is a ball of perimeter 40, 000 kilometers. There is a building 20 meters tall at point a. A robot with a camera

torisob [31]

Answer:

Approximately 21 km.

Explanation:

Refer to the not-to-scale diagram attached. The circle is the cross-section of the sphere that goes through the center C. Draw a line that connects the top of the building (point B) and the camera on the robot (point D.) Consider: at how many points might the line intersects the outer rim of this circle? There are three possible cases:

No intersection: There's nothing that blocks the camera's view of the top of the building.
Two intersections: The planet blocks the camera's view of the top of the building.
One intersection: The point at which the top of the building appears or disappears.

There's only one such line that goes through the top of the building and intersects the outer rim of the circle only once. That line is a tangent to this circle. In other words, it is perpendicular to the radius of the circle at the point A where it touches the circle.

The camera needs to be on this tangent line when the building starts to disappear. To find the length of the arc that the robot has travelled, start by finding the angle $\angle \mathrm{B\hat{C}D}$ which corresponds to this minor arc.

This angle comes can be split into two parts:

$\angle \mathrm{B\hat{C}D} = \angle \mathrm{B\hat{C}A} + \angle \mathrm{A\hat{C}D}$ .

Also,

$\angle \mathrm{B\hat{A}C} = \angle \mathrm{D\hat{A}C} = 90^{\circ}$ .

The radius of this circle is:

$\displaystyle r = \frac{c}{2\pi} = \rm \frac{4\times 10^{7}\; m}{2\pi}$ .

The lengths of segment DC, AC, BC can all be found:

$\rm DC = \rm \left(1.75 \displaystyle + \frac{4\times 10^{7}\; m}{2\pi}\right)\; m$ ;
$\rm AC = \rm \displaystyle \frac{4\times 10^{7}}{2\pi}\; m$ ;
$\rm BC = \rm \left(20\; m\displaystyle +\frac{4\times 10^{7}}{2\pi} \right)\; m$ .

In the two right triangles $\triangle\mathrm{DAC}$ and $\triangle \rm BAC$ , the value of $\angle \mathrm{B\hat{C}A}$ and $\angle \mathrm{A\hat{C}D}$ can be found using the inverse cosine function:

$\displaystyle \angle \mathrm{B\hat{C}A} = \cos^{-1}{\rm \frac{AC}{BC}}$

$\displaystyle \angle \mathrm{D\hat{C}A} = \cos^{-1}{\rm \frac{AC}{DC}}$

$\displaystyle \angle \mathrm{B\hat{C}D} = \cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}$ .

The length of the minor arc will be:

$\displaystyle r \theta = \frac{4\times 10^{7}\; \rm m}{2\pi} \cdot (\cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}) \approx 20667 \; m \approx 21 \; km$ .

5 0

3 years ago