All categories

3 years ago

What is a device that changes electrical energy into mechanical energy?

1 answer:

8 0

<span>A generator converts mechanical energy into electrical energy, while a motor does the opposite - it converts electrical energy into mechanical energy.</span>

You might be interested in

Write the chemical symbol for three different atoms or atomic cations with 22 electrons

damaskus [11]

Answer:

Explanation:

To write the chemical symbol for an atom or cation with 22 electrons, we must understand some few things about an atom.

An atom is made up of positively charged paticles called protons which are located in the nucleus. The neutrons do not have any charges and are they reside together in the nucleus with the protons.

Electrons are negatively charged particles which orbits the nucleus.

Titanium in its neutral state is the 22nd element on the periodic table and it has 22 electrons.

Vanadium is the the 23rd element on the periodic table. If it loses an electron, it becomes positively charged and it is a cation.

Chromium is the 24th element on the periodic table. A loss of two electrons makes the net electrons remaining on it 22.

The symbols are:

₂₂Ti

₂₃V¹⁺

₂₄Cr²⁺

6 0

4 years ago

GIVING BRAINLISTED

Mice21 [21]

Answer: one molecule of O2.

Explanation: sweet i just took a guess but I believe that if 3 o2 molecules - 2 h2 molecules I think that its just basic maths and it is C because 3-2 = 1 and its o2 remaining, sorry if I’m wrong.

6 0

3 years ago

5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf

mr_godi [17]

Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and $pK_{a}$ is as follows.

pH = $pK_{a} + log \frac{base}{acid}$

where, pH = 7.4 and $pK_{a}$ = 7.21

As here, we can use the $pK_{a}$ nearest to the desired pH.

So, 7.4 = 7.21 + $log \frac{base}{acid}$

0.19 = $log \frac{base}{acid}$

$\frac{base}{acid}$ = 1.55

1 mM phosphate buffer means $[HPO_{4}]$ + $[H_{2}PO_{4}]$ = 1 mM

Therefore, the two equations will be as follows.

$\frac{HPO_{4}}{H_{2}PO_{4}}$ = 1.55 ............. (1)

$[HPO_{4}]$ + $[H_{2}PO_{4}]$ = 1 mM ........... (2)

Now, putting the value of $[HPO_{4}]$ from equation (1) into equation (2) as follows.

1.55 $[H_{2}PO_{4}] + [tex][H_{2}PO_{4}]$ = 1 mM

2.55 $[H_{2}PO_{4}]$ = 1 mM

$[H_{2}PO_{4}]$ = 0.392 mM

Putting the value of $[H_{2}PO_{4}]$ in equation (1) we get the following.

0.392 mM + $[HPO_{4}]$ = 1 mM

$[HPO_{4}]$ = (1 - 0.392) mM

$[HPO_{4}]$ = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

7 0

3 years ago

What effect does rapid exhalation of CO2 during exercise have on the concentration of H2CO3 in the blood?

BARSIC [14]

If there is a constant loss of $CO_{2}~(carbon~dioxide)$ the concentration of $H_{2} CO_{3}~(carbonic~acid)$ will be affected and decrease since $CO_{2}$ is a main component of $H_{2}CO_{3}$ .