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3 years ago

Consider Nal → Na+ + - and the following information.

Chemistry

1 answer:

Jobisdone [24]3 years ago

6 0

Answer:

47 kJ·mol⁻¹

Explanation:

NaI(s) ⟶ Na⁺(aq) + I⁻(aq); ΔH = ?

We can carry out this process in a series of steps.

1. Convert the solid to its gaseous ions.

NaI(s) ⟶ Na⁺(g) + I⁻(g); -ΔH(lat) = 704 kJ

2. Hydrate the Na⁺ ions

Na⁺(g) ⟶ Na⁺(aq); ΔH(hyd) = - 410.0 kJ

3. Hydrate the I⁻ ions

I⁻(g) ⟶ I⁻(aq); ΔH(hyd) = -247

4. Add the equations

Cancel ions that occur on opposite sides of the reaction arrow

ΔH/kJ·mol⁻¹

NaI(s) ⟶ Na⁺(g) + I⁻(g); +704

Na⁺(g) ⟶ Na⁺(aq); - 410.0

I⁻(g) ⟶ I⁻(aq); -247

NaI(s) ⟶ Na⁺(g) + I⁻(g) 47

The heat of solution of NaI is 47 kJ·mol⁻¹.

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An unknown gaseous substance has a density of 1.06 g/L at 31 °C and 371 torr. If the substance has the following percent composi

Anarel [89]

Answer:

C) C4H6 - Right answer

Explanation:

Let's combine the Ideal Gases Law with density to get the molecular formula for the unknown gas.

Density = mass / volume

1.06 g /L means that 1.06 grams of compound occupy 1 liter of volume.

P . V = n . R . T

Pressure in Torr must be converted to atm

760 Torr are 1 atm

371 Torr are __ (371 .1)/760 = 0.488 atm

0.488 atm . 1L = 1.06g/MM . 0.082 . 304K

(0.488 atm . 1L) / 0.082 . 304K = 1.06g/ MM

Mass / Molar mass = Moles → That's why the 1.06 g / MM

0.0195 mol = 1.06g / MM

1.06g/0.0195 mol = MM → 54.3 g/m

Now, let's use the composition

100 g of compound have 88.8 g of C

54.3 g of compound have ___ (54.3 . 88.8) /100 = 48 g of C

100 g of compound have 11.2 g of H

54.3 g of compound have __ (54.3 . 11.2)/100 = 6 g of H

48 g of C are included un 4 atoms

6 g of H are included in 6 atoms

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3 years ago

Who was the first to state the concept of an atom

marin [14]

Democritus, a Greek philosopher, first developed the idea of atoms (around 460 B.C., I believe).
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4 years ago

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3 years ago

The equilibrium constant for the gas phase reaction

stellarik [79]

The equilibrium constant is found by [product]/[reactant]

If the equilibrium constant is very small, such as 4.20 * 10^-31, then that means at equilibrium there is very little product and a lot of reactant.
And likewise, if there is a lot of product formed, and very little reactant, then the K value will be very large, which tells us that it is predominantly product.

At equilibrium, for any reaction, there will always be some reactant and some product present. There cannot be zero reactant or zero product. Also keep in mind that the equilibrium constant is dependent on temperature.

At equilibrium, for your reaction, it is predominantly reactants.

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4 years ago

What mass of Sodium Chloride is required to make 100.0 mL of 3.0 M solution?

Julli [10]

Answer:

17.55 g of NaCl

Explanation:

The following data were obtained from the question:

Molarity = 3 M

Volume = 100.0 mL

Mass of NaCl =..?

Next, we shall convert 100.0 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

100 mL = 100/1000

100 mL = 0.1 L

Therefore, 100 mL is equivalent to 0.1 L.

Next, we shall determine the number of mole NaCl in the solution. This can be obtained as follow:

Molarity = 3 M

Volume = 0.1 L

Mole of NaCl =?

Molarity = mole /Volume

3 = mole of NaCl /0.1

Cross multiply

Mole of NaCl = 3 × 0.1

Mole of NaCl = 0.3 mole

Finally, we determine the mass of NaCl required to prepare the solution as follow:

Mole of NaCl = 0.3 mole

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl =?

Mole = mass /Molar mass

0.3 = mass of NaCl /58.5

Cross multiply

Mass of NaCl = 0.3 × 58.5

Mass of NaCl = 17.55 g

Therefore, 17.55 g of NaCl is needed to prepare the solution.

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3 years ago