What best explains freezing point depression

The labels have fallen off three bottles containing powdered samples of metals; one contains zinc, one lead, and the other plati

Alinara [238K]

Here we have to identify the metal powder by the given disposal.

The identification of Zinc can be done by 1 m nitric acid (HNO₃) and Ni(NO₃)₂ which will produce hydrogen gas by reaction and displacement reaction as shown below.

Zn + 2 HNO₃ = Zn(NO₃)₂ + H₂ (g)

Zn + Ni(NO₃)₂ = Zn(NO₃)₂ + Ni

The identification of lead can be done by the reaction with 1 m nitric acid (HNO₃) which produces lead nitrate.

The reaction is-

Pb + 4HNO₃ = Pb(NO₃)₂ + 2NO₂ + 2H₂O

The identification of platinum can be done by the reaction with all the given disposal as it will not react with any of the compound.

1. Identification of Zinc (Zn):

(a) Zn metal will not react with sodium nitrate (NaNO₃) as sodium remains at the lower position of the activity series than zinc.

Zn + NaNO₃ = No reaction

(b) Zn metal will react with 1 m HNO₃ to form hydrogen gas. The reaction is:

Zn + 2 HNO₃ = Zn(NO₃)₂ + H₂ (g)

(c) Zn will react with nickel nitrate [Ni(NO₃)₂] because it may only cause displacement reaction the reduction potential of Zn²⁺/Zn (-0.76) is less than that of Ni²⁺/Ni (-0.23). Thus the reaction will be:

Zn + Ni(NO₃)₂ = Zn(NO₃)₂ + Ni

2. Identification of lead (Pb):

(a) Pb metal will not react with sodium nitrate (NaNO₃) as sodium remains at the lower position of the activity series than Pb.

Pb + NaNO₃ = No reaction

(b) Pb reacts with HNO₃ to form lead nitrate. The reaction is:

Pb + 4HNO₃ = Pb(NO₃)₂ + 2NO₂ + 2H₂O

(c) The standard reduction potential of Pb²⁺/Pb is more than nickel Ni²⁺/Ni thus there will be no reaction between Pb and NI(NO₃)₂.

Pb + Ni(NO₃)₂ = No reaction.

3. Identification of platinum (Pt)

(a) Pt metal will not react with sodium nitrate (NaNO₃) as sodium remains at the lower position of the activity series than Pt.

Pt + NaNO₃ = No reaction.

(b) The standard reduction potential of Pt²⁺/Pt is so high (+1.188) thus there will be no reaction with HNO₃.

Pt + HNO₃ = No reaction

(c) The standard reduction potential of Pt²⁺/Pt is more than nickel Ni²⁺/Ni thus there will be no reaction between Pt and Ni(NO₃)₂.

Pt + Ni(NO₃)₂ = No reaction.

8 0

3 years ago

The rate constant for the second-order reaction: 2NOBr(g) → 2NO(g) + Br2(g) is 0.80/(M · s) at 10°C. Starting with a concentrati

a_sh-v [17]

Answer : The concentration of NOBr after 95 s is, 0.013 M

Explanation :

The integrated rate law equation for second order reaction follows:

$k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)$

where,

k = rate constant = $0.80M^{-1}s^{-1}$

t = time taken = 95 s

[A] = concentration of substance after time 't' = ?

$[A]_o$ = Initial concentration = 0.86 M

Now put all the given values in above equation, we get:

$0.80=\frac{1}{95}\left (\frac{1}{[A]}-\frac{1}{(0.86)}\right)$

[A] = 0.013 M

Hence, the concentration of NOBr after 95 s is, 0.013 M

4 0

3 years ago

18. What is the percent by mass sugar of a solution made by dissolving 12.45 grams of sugar in 100 grams of water?

Fantom [35]

The percent by mass sugar of a solution : 11.07%

<h3>Further explanation</h3>

Given

mass of sugar = 12.45 g

mass of water = 100 g

Required

The percent by mass

Solution

Mass of solution :

$\tt mass~sugar+mass~water=12.45+100=112.45~g$

Percent mass of sugar :

$\tt \%mass=\dfrac{mass~sugar}{mass~solution}\times 100\%\\\\\%mass=\dfrac{12.45}{112.45}\times 100\%=11.07\%$

6 0

3 years ago

Which of the following are cations? Check all that apply.

Lostsunrise [7]

Answer:

a

barium

b

calcium

e

aluminum

f

magnesium

g

copper

Forms cation

Explanation:

4 0

3 years ago

The following molecular equation represents the reaction that occurs when aqueous solutions of silver(I) nitrate and barium chlo

mrs_skeptik [129]

Answer:

Hey, I hope this helps. You gave the equation already balanced so there was no need to do so, the next thing we need to do after balancing is to split the strong electrolytes into ions. Once that is completed we cross off any reoccurring ions. That leaves us with our complete net ionic.

I also recommend you check out Wayne Breslyn on Yt. He is so helpful with equations like these.

5 0

2 years ago