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2 years ago

How many protons are there in 2.5 l of liquid water?

1 answer:

4 0

The density of liquid water is 1 kg/L.

Mass of water = (1 kg/L)(2.5 L)(1000 g/1 kg) = 2500 g water

The molar mass of water is 18 g/mol.

Moles of water = 2500 g / 18 g/mol = 138.889 mol water

According to Avogadro's number, in one mole of any substance, there are 6.022×10²³ atoms.

Atoms of water = 138.889 mol * 6.022×10²³ atoms/mol = 8.364×10²⁵ atoms of water

In every atom of water, there are 10 protons: 8 from one oxygen atom, and 1 each from the two hydrogen atoms.

Protons of water = 8.364×10²⁵ atoms * 10 protons/atom
<em>Protons of water = 8.364×10²⁶ </em>

You might be interested in

How are mixtures and solutions different?

borishaifa [10]

Answer:

The answer is 4.

Explanation:

Take trail mix as a mixture. It can easily be separated by hand. Take sugar in hot coca as a solution. You cannot separate sugar from the hot coca with your hands.

3 0

2 years ago

A gas has a volume of 5.0 L at a pressure of 50 KPa. What happens to the volume when the pressure is increased to 125?

Alexeev081 [22]

The volume becomes two. You have to use the equation P1 x V1 = P2 x V2
P is pressure and V is volume.
P1 = 50 P2 = 125
V1 = 5 V2 = v (we don't know what it is)
Then set up the equation:
50 times 5 = 125 times v
250 = 125v
the divide both sides by 125 and isolate v
2 = v
Therefore the volume is decreased to 2.
Also, Boyle's Law explains this too: Volume and pressure are inversely related, This means that when one goes up the other goes down (ie when pressure increases volume decreases and vice versa). Becuase the pressure went up from 50 KPa tp 125 KPa the volume had to decrease.

7 0

2 years ago

What is hydrogen ion concentration of an acid solution whose PH is 3.4?

meriva

It’s easy, if the PH of any acidic solution = -Log[H+], where [H+] is hydrogen ion concentration, multiply each term by (-1) then raise each term as a power to (10), so it will become like this:-
[H+] = 10^(-PH)

4 0

2 years ago

3) In the Hydrolysis of Disaccharides and Polysaccharides portion of the lab, starch should give one positive iodine test and on

AfilCa [17]

Answer: potassium iodide is the basic test for starch,and the positive test is blue-black coloration, any other test substance which is not starch will give a negative results.

Explanation:

Starch is an example of polysaccharide and since the standard test for it is potassium iodide solution, it gives a positive test.

Diasaccharides e.g maltose are reducing sugars.their standard test is BENEDICT test .

Therefore, in the hydrolysis; starch should give a positve test, while Diasaccharides should give negative rest.

3 0

2 years ago

1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L

anastassius [24]

Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol $(P^o_1)$ = 20.9 torr

Vapor presume of 2‑Propanol $(P^o_2)$ = 45.2 torr

Mole fraction of 1‑Propanol $(x_1)$ = 0.540

Mole fraction of 2‑Propanol $(x_2)$ = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

$p_1=x_1\times p^o_1$

where,

$p_1$ = partial vapor pressure of 1‑Propanol

$p^o_1$ = vapor pressure of pure substance 1‑Propanol

$x_1$ = mole fraction of 1‑Propanol

$p_1=(0.540)\times (20.9torr)=11.3torr$

and,

$p_2=x_2\times p^o_2$

where,

$p_2$ = partial vapor pressure of 2‑Propanol

$p^o_2$ = vapor pressure of pure substance 2‑Propanol

$x_2$ = mole fraction of 2‑Propanol

$p_2=(0.46)\times (45.2torr)=20.8torr$

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

$\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352$

and,

$\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648$

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

3 0

2 years ago