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2 years ago

What is the molality of a solution in which 3.0 moles of NaCl is dissolved in 1.5 Kg of water?

1 answer:

5 0

Density H2O = 1g/cm³
1,5 kg H2O = 1500g = 1500cm³ (1dm³ = 1000cm³)

3moles of NaCl-----in---------1500cm³ H2O
x moles of NaCl ----in--------1000cm³ H2O
x = 2moles of NaCl

answer: 2 mol/dm³

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A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at

Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0

2 years ago

(ii) When shale gas is burned, the hydrogen sulfide reacts with oxygen.

Elenna [48]

Answer:

See explanation

Explanation:

Now , we have the equation of the reaction as;

2H2S(g) + 302(g)------->2SO2(g) + 2H2O(g)

This equation shows that SO2 gas is produced in the process. Let us recall that this same SO2 gas is the anhydride of H2SO4. This means that it can dissolve in water to form H2SO4

So, when SO2 dissolve in rain droplets, then H2SO4 is formed thereby lowering the pH of rain water. This is acid rain.

4 0

2 years ago

Suppose a soap manufacturer starts with a triglyceride that has the fatty acid chains arachidic acid, palmitic acid and palmitic

DIA [1.3K]

Answer:

Sodium arachidate; Sodium palmitate and Sodium palmitate

Explanation:

Triglycerides are esters of fatty acids with glycerol. In triglycerides, three fatty acid molecules are linked by ester bonds to each of the three carbon atoms in a glycerol molecule. The fatty acids may be same or different fatty acid molecules. Hydrolysis of triglycerides yields the three fatty acid molecules and glycerol.

Saponification is the process by which a base is used to catalyst the hydrolysis of the ester bonds in glycerides. The products of this base-catalyzed hydrolysis of triglycerides are the metallic salts of the three fatty acids and glycerol. The salts of the fatty acids are known as soaps.

For a triglyceride that has the fatty acid chains arachidic acid, palmitic acid and palmitic acid attached to the three backbone carbons glycerol, the saponification of the triglyceride with NaOH will yield the sodium salts or soaps of the three fatty acids as well as glycerol.

Arachidic acid will react with NaOH to yield sodium arachidate.

The two palmitic acid molecules will each react with NaOH to yield sodium palmitate.

8 0

1 year ago

Which description best characterizes the motion of particles in a solid?

Pavlova-9 [17]

Vibrating around fixed positions

7 0

2 years ago

Read 2 more answers

The ΔG°f of atomic oxygen is 230.1 kJ/mol. Find ΔG° for the following dissociation reactionO2 (g) <--> 2O (g)then calculat

marusya05 [52]

Answer:

Kc = 2.145 × 10⁻⁸¹

Explanation:

Let's consider the following reaction:

O₂(g) ⇄ 2O(g)

The standard Gibbs free energy for the reaction (ΔG°) can be calculated using the following expression:

ΔG° = Σnp. ΔG°f(p) - Σnp. ΔG°f(p)

where,

ni are the moles of products and reactants

ΔG°f(p) are the standard Gibbs free energy of formation of products and reactants

In this case,

ΔG° = 2 × ΔG°f(O) - 1 × ΔG°f(O₂)

ΔG° = 2 × 230.1 kJ/mol - 1 × 0 kJ/mol

ΔG° = 460.2 kJ/mol

With this information, we can calculate the equilibrium constant (Kc) using the following expression:

$Kc=e^{-\Delta G \°/R.T } = e^{-460.2 kJ/mol/(8.314 \times 10^{-3}kJ/mol.K) \times 298K }=2.145 \times 10^{-81}$

3 0

2 years ago