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rjkz [21]
3 years ago
12

It takes 486.0 kJ/mol to convert sodium atoms to Na+ ions. Sodium atoms absorb and emit light of wavelengths 589.6 and 590.0 nm.

Calculate the energy of the 590.0 nm light in kilojoules per mole.
Chemistry
1 answer:
solmaris [256]3 years ago
8 0
We first calculate the energy contained in one photon of this light using Planck's equation:

E = hc/λ

E = 6.63 x 10⁻³⁴ x 3 x 10⁸ / 590 x 10⁻⁹
E = 3.37 x 10⁻²² kJ/photon

Now, one mole of atoms will excite one mole of photons. This means that 6.02 x 10²³ photons will be excited

(3.37 x 10⁻²² kJ/photon) x (6.02 x 10²³ photons / mol)

The energy released will be 202.87 kJ/mol
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The lattice energy of a salt is 350 kJ/mol and the solvation energies of its ions add up to 320 kJ/mol for the preparation of a
4vir4ik [10]

Answer:

It would get <u>colder</u>

Explanation:

The lattice energy is the energy involved in the disruption of interactions between the ions of the salt. In this case, we have: ΔHlat  = 350 kJ/mol > 0, so it is an endothermic process (the energy is absorbed).

The solvation energy is the energy involved in forming interactions between water molecules and the ions of the salt. In this case, we have: ΔHsolv  = 320 kJ/mol > 0, so it is an endothermic process (the energy is absorbed).

The dissolution process involve both processes: the disruption of ion-ion interactions of the salt and the solvation process. Thus, the enthalphy change (ΔHsol) in the preparation of the solution is calculated as the addition of the lattice energy and solvation energy:

ΔHsol= ΔHlat + ΔHsolv = 350 kJ/mol + 320 kJ/mol = 370 kJ/mol

370 kJ/mol > 0 ⇒ endothermic process

Since the preparation of the solution is an <u>endothermic</u> process, it will absorb energy from the surroundings, so <u>the solution would get colder</u>.

3 0
3 years ago
A chemical combination of two or more atoms is a(n) .
irinina [24]

Answer:

Molecule

Explanation:

Its simple definition

6 0
3 years ago
Read 2 more answers
Which is the basis of thin-layer chromatography?
solmaris [256]

Answer:

B.) The drug gets carried through a stationary phase by a mobile phase and the retention time identifies the drug.

Explanation:

Chromatography is used in purifying complex mixtures of organic compounds. It uses the adsorption tendencies of compounds to seperate and identify them.

Chromatography is made up of two phases in contact, the stationary phase or non-mobile phase and the mobile phase. The movement of the mobile phase over the stationary phase causes the separation of a mixture into its constituents.

The stationary phase is made up of silica-gel or alumina in a solvent (an adsorbent) and the mobile phase or carrier is the organic solvent which is the drug.

8 0
4 years ago
Calculate the mass of 1.0 L of helium (He), 1.0 L of chlorine gas (Cl2), and 1.0 L of air (79% N2, 21% O2 by volume) at 25°C and
Zigmanuir [339]

To calculate the mass we use the following formulas:

PV=nRT     (1)

and

n = m / M    (2)

where:

P - pressure (atm)

V - volume (L)

n - moles

R - gas constant = 0.082 (L × atm) / (mol × K)

T - temperature (°K) (25°C + 273 = 298°K)

m - mass (g)

M - molecular mass (g/mole)

Now we rewrite equation (1):

n = PV / RT

And replace n with m / M from equation (2):

m / M = PV / RT

m = (P × V × M) / (R ×T)

1 L of He will have a mass of:

m = (1 × 1 × 4) / (0.082 × 298) = 0.1637 g

1 L of Cl₂ will have a mass of:

m = (1 × 1 × 71) / (0.082 × 298) = 2.9055 g

1.0 L of air will contain 0.79 L of N₂ and 0.21 L of O₂

0.79 L of N₂ will have a mass of:

m = (1 × 0.79 × 28) / (0.082 × 298) = 0.9052 g

0.21 L of O₂ will have a mass of:

m = (1 × 0.21 × 32) / (0.082 × 298) = 0.2750 g

mass of air = mass of N₂ + mass of O₂

mass of air = 0.9052 + 0.2750 = 1.1802 g

A balloon filed with helium will rise because as you see 1 L of helium is lighter than 1 L of air.

Chlorine gas is dangerous because chlorine is very toxic for human life and more of that is heavier than the air so will diffuse very hard from the area where the leak appeared.

7 0
4 years ago
Chemistry help please!
BARSIC [14]

A and then C

<em>Dimensional analysis</em>

\begin{array}{lll}n & = & m / M \\ & = & 8.33 \; \text{g H} / (1.008 \; \text{g H} \cdot\text{mol H}^{-1})\\ & = & 8.26 \; \text{mol H}\end{array}

\begin{array}{lll} N & = & n \cdot A_r \\ & = & 8.26 \; \text{mol H} \times 6.023 \times 10^{23} \; \text{mol}^{-1} \\ &= & 4.93 \times 10^{24} \; \text{H atoms}\end{array}

\begin{array}{lll}N & = & n \cdot A_r \\ & = & 0.417 \; \text{mol O}\times 6.02 \times 10^{23} \; \text{mol}^{-1} \\ & = &2.51 \times 10^{23} \; \text{O atoms}\end{array}

6 0
4 years ago
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