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2 years ago

Give the name of one of the 3 common acids used in chemistry

Chemistry

2 answers:

FrozenT [24]2 years ago

7 0

I think one is Hydrochloric acid

rjkz [21]2 years ago

7 0

Answer:

Common strong acids include hydrochloric acid, sulfuric acid, phosphoric acid, and nitric acid. Common weak acids include acetic acid, boric acid, hydrofluoric acid, oxalic acid, citric acid, and carbonic acid.

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2. A cell with few energy needs would most

sattari [20]

A Cell with few energy needs would most likely contain a small number of Mitochondria.

All cells require energy to function, but cells typically have significant energy needs that can only be met by the mitochondria, the cell's powerhouse.
They transform glucose into ATP, a chemical with a huge energy storage capacity.
Muscles have a large number of mitochondria, allowing them to react rapidly and powerfully to the body's ongoing need for energy.
Macromolecules, defunct cell components, and microbes are all digested by lysosomes.
Vacuoles are typically tiny and aid in the sequestration of waste.
The ribosome, an intercellular structure consisting of both RNA and protein, is where a cell produces new proteins.

Therefore out of all these cell organelles, the cell has fewer mitochondria for less energy need.

Learn more about cell organelles here:

brainly.com/question/13408297

#SPJ9

5 0

2 years ago

Calculate for the following electrochemical cell at 25°C, Pt H2(g) (1.0 atm) H (0.010 M || Ag (0.020 M) Ag if E (H) - +0.000 V a

viva [34]

Answer : The correct option is, (b) +0.799 V

Solution :

The values of standard reduction electrode potential of the cell are:

$E^0_{[H^{+}/H_2]}=+0.00V$

$E^0_{[Ag^{+}/Ag]}=+0.799V$

From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : $H_2\rightarrow 2H^{+}+2e^-$

Reaction at cathode (reduction) : $Ag^{+}+1e^-\rightarrow Ag$

The balanced cell reaction will be,

$H_2+2Ag^{+}\rightarrow 2H^{+}+2Ag$

Now we have to calculate the standard electrode potential of the cell.

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}$

$E^o_{cell}=(+0.799V)-(+0.00V)=+0.799V$

Therefore, the standard cell potential will be +0.799 V

4 0

3 years ago

Which statements about fresh-water sources are true.? (Multiple Choice) (please help asap thanks!:)

Kay [80]

Answer:

The correct answers are:
"Only about 3 percent of Earth's water is fresh water."

"About 75% percent of the fresh water on Earth is frozen in ice sheets."

"The largest source of usable fresh water is groundwater."

Explanation:

3 percent of Earth's water is most certainly fresh water. Confirmed with a few fact checks.

The largest source of usable fresh water on Earth is groundwater. It's more difficult to access but it's there and much more usable than water say frozen in ice on the sea.

The most correct option left would be 75% of Earth's freshwater being in ice sheets even though it's about 70%.

3 0

2 years ago

What is the concentration of a solution containing 1.11 g sugar (sucrose, C12H22O11, MW = 342.3 g/mol, d = 1.587 g/cm3) in 432 m

djyliett [7]

Answer:

0.0075 M

0.0060 m

Explanation:

Our strategy here is to use the definition of molarity and molality to solve this question.

The molarity is the number of moles of solute, sucrose in this case, per liter of solution.

The molality is the number of moles of solute per kilogram of solvent.

So the molarity of the solution is

M = moles of solute/ V solution

As we see we need the volume of solution since we are only given the volume of solvent, but this will be easy to compute since we have the density of sucrose.

So determine the moles of sucrose , and the volume of solution:

Moles sucrose = 1.11 g/342.3 g/mol = 3.24 x 10⁻³ M

Volume of solution = Vol Sucrose + Vol glycerine

d = m/V ⇒ Vsucrose = m / d = 1.11 g/ 1.587 g/cm³ = 0.70 cm³

Vol solution = 432 mL + 0.70 mL = 432.7 mL (1cm³ = 1 mL)

Vol solution = 432.7 mL x 1 L / 1000 mL = 0.4327 L

⇒ M = 3.2 x 10⁻³ mol / 0.4327 L = 0.0075 M

For the molarity what we need is to first calculate the kilograms of glycerine from the given density:

d = m/v ⇒ m = d x v = 1.261 g/cm³ x 432 cm³ = 544.75 g

Converting to Kg:

544.75 g x 1 Kg/ 1000 g = 0.544 kg

Now the molality is

m = mol sucrose/ kg solvent = 3.24 x 10⁻³ mol / 0.544 Kg = 0.0060 m

Note: In the calculation for volume of solution we could have approximated it to that of just glycerine, but since the density of sucrose was given we calculated the total volume of solution to be more rigorous.

8 0

3 years ago

The pKa of chloroacetic acid is 2.9. If an organic chemist has a 0.10 M aqueous solution of chloroacetic acid at pH 2.4, what pe

Trava [24]

Answer:

The answer is given below.

Explanation:

We will consider the acid as HA and will set up an ICE table with the equilibrium dissociation of α.

AT pH 2.4 the initial H+ concentration will be 3.98^10-3 M

HA → H+ + A-

Initial concentration: 0.1 → 3.98 ^10-3 + 0

equilibrium concentration: 0.1(1-α) → 3.98 * 10-3 + 0.1α 0.1α

pKa of chloroacetic acid is 2.9

-log(Ka) = 2.9

Ka = 1.26 * 10-3

From the equation, Ka = [H+] * [A-] / [HA]

1.26 * 10-3 = (3.98 * 10-3 + 0.1α )* 0.1α / 0.1(1-α)

Since α<<1, we assume 1-α = 1

Solving the equation, we have: α = 0.094

Since this is the fraction of acid that has dissociated, we can say that % of base form = 100 * α= 9.4%

6 0

3 years ago