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3 years ago

Why does salt dissollve in water act as an antifreeze

Chemistry

2 answers:

Svetlanka [38]3 years ago

7 0

It dissolves the hydrogen in water

AnnyKZ [126]3 years ago

3 0

Salt disrupt the bonds of hydrogen in water. This makes the freezing point of water decrease. In other words, as the saltiness increases, the freezing point decreases. This is why salt dissolved in water acts as antifreeze. Best of luck to you, my buddy.

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Give the quantum number set for one electron in the 3p sub level of a sulfur (S)

vesna_86 [32]

Answer:- Atomic number for sulfur is 16 and it's electron configuration is $1s^2 2s^2 2p^6 3s^2 3p^4$ . Here, there are total for electrons in 3p and the set of quantum numbers for these 4 electrons would be as..

For the first electron of 3p-

n = 3, l = 1, ml = -1 and ms = +(1/2)

for the second electron of 3p-

n = 3, l = 1, ml = 0 and ms = +(1/2)

for the third electron of 3p-

n = 3, l = 1, ml = +1 and ms = +(1/2)

and for the fourth electron of 3p-

n = 3, l = 1, ml = -1 and ms = -(1/2)

3 0

3 years ago

Nitrogen dioxide is a red-brown gas that is responsible for the color of photochemical smog. What is the volume of 1 mol of nitr

Ray Of Light [21]

B ideal gas has a volume of 22.4

5 0

3 years ago

Indicate which solution in each pair has the lower pH. Your response should be a four letter "word". The first letter should be

JulijaS [17]

Answer:

bcfh

Explanation:

HClO₄ reacts with water thus:

HClO₄ + H₂O → H₃O⁺ + ClO₄⁻

That means HClO₄ produce H₃O⁺ that decreases pH. That means the higher concentration of HClO₄ decreases pH. Thus, lower pH will be:

b) 0.2 M HClO4

The reaction of NaClO₄ is:

NaClO₄ + H₂O → OH⁻ + HClO₄ + Na⁺

The higher concentration of NaClO₄ the higher production of OH⁻ that increase pH, that means the lower concentration of NaClO₄ the lower pH, thus, the answer is:

<em>c) 0.1 M NaClO or</em>

HF reacts with water thus;

HF ⇄ H⁺ + F⁻

The equilibrium constant is:

k = [H⁺] [F⁻] / [HF] = 3,5x10⁻⁴

For HNO₂ equilibrium is:

HNO₂ ⇄ H⁺ + NO₂⁻

k = [H⁺] [NO₂⁻] / [HNO₂] = 4,5x10⁻⁴

As k value is higher for HNO₂, the concentration of H⁺ will be higher in this system doing the HNO₂ with the lower pH.

f) 0.1 M HNO2

NaOH is a strong base that produce OH⁻ that increase pH, pure water is neutral, thus, the lowe pH is:

h) pure water

I hope it helps!

7 0

3 years ago

Name the following:<br> Simple chemical molecules made of single unit

Stolb23 [73]

Answer:

atom is the answer I think

7 0

3 years ago

A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 64.0 mL of 0.0600 M EDTA . Titration of the excess unreacted EDTA req

tigry1 [53]

Answer:

the concentration of $Cd^{2+}$ in the original solution= 0.0088 M

the concentration of $Mn^{2+}$ in the original solution = 0.058 M

Explanation:

Given that:

The volume of the sample containing Cd2+ and Mn2+ = 50.0 mL; &

was treated with 64.0 mL of 0.0600 M EDTA

Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+

i.e the strength of the Ca2+ = 0.0310 M

Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+

To determine the concentrations of Cd2+ and Mn2+ in the original solution; we have the following :

Volume of newly freed EDTA = $\frac{Volume\ of \ Ca^{2+}* Sample \ of \ strength }{Strength \ of EDTA}$

= $\frac{14.2*0.0310}{0.0600}$

= 7.3367 mL

concentration of $Cd^{2+}$ = $\frac{volume \ of \ newly \ freed \ EDTA * strength \ of \ EDTA }{volume \ of \ sample}$

= $\frac{7.3367*0.0600}{50}$

= 0.0088 M

Thus the concentration of $Cd^{2+}$ in the original solution = 0.0088 M

Volume of excess unreacted EDTA = $\frac{volume \ of \ Ca^{2+} \ * strength \ of Ca^{2+} }{Strength \ of \ EDTA}$

= $\frac{16.1*0.0310}{0.0600}$

= 8.318 mL

Volume of EDTA required for sample containing $Cd^{2+}$ and $Mn^{2+}$ = (64.0 - 8.318) mL

= 55.682 mL

Volume of EDTA required for $Mn^{2+}$ = Volume of EDTA required for

sample containing $Cd^{2+}$ and

$Mn^{2+}$ -- Volume of newly freed EDTA

Volume of EDTA required for $Mn^{2+}$ = 55.682 - 7.3367

= 48.3453 mL

Concentration of $Mn^{2+}$ = $\frac{Volume \ of EDTA \ required \ for Mn^{2+} * strength \ of \ EDTA}{volume \ of \ sample}$

Concentration of $Mn^{2+}$ = $\frac{48.3453*0.0600}{50}$

Concentration of $Mn^{2+}$ in the original solution= 0.058 M

Thus the concentration of $Mn^{2+}$ = 0.058 M

6 0

3 years ago