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MAVERICK [17]
2 years ago
15

The wavelength of the characteristic, bright yellow-orange flame test color of sodium is 590 nm. Calculate the average energy (Δ

E) associated with this atomic emission line.
Chemistry
1 answer:
Assoli18 [71]2 years ago
7 0
<span>The energy associated with the 590nm atomic emission line is 3.369 x 10^49 Joules. This can be determined by multiplying Plank’s constant (6.626 x 10^34 Js) by the speed of light (3x 10^8 m/s), and then dividing the result by the wavelength (590 x 10^-9 m).</span>
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For the reaction of oxygen and nitrogen to form nitric oxide, consider the following thermodynamic data (Due to variations in th
CaHeK987 [17]

Answer:

a. 7278 K

b. 4.542 × 10⁻³¹

Explanation:

a.

Let´s consider the following reaction.

N₂(g) + O₂(g) ⇄ 2 NO(g)

The reaction is spontaneous when:

ΔG° < 0  [1]

Let's consider a second relation:

ΔG° = ΔH° - T × ΔS° [2]

Combining [1] and [2],

ΔH° - T × ΔS° < 0

ΔH° < T × ΔS°

T > ΔH°/ΔS°

T > (180.5 × 10³ J/mol)/(24.80 J/mol.K)

T >  7278 K

b.

First, we will calculate ΔG° at 25°C + 273.15 = 298 K

ΔG° = ΔH° - T × ΔS°

ΔG° = 180.5 kJ/mol - 298 K × 24.80 × 10⁻³ kJ/mol.K

ΔG° = 173.1 kJ/mol

We can calculate the equilibrium constant using the following expression.

ΔG° = - R × T × lnK

lnK = - ΔG° / R × T

lnK = - 173.1 × 10³ J/mol / (8.314 J/mol.K) × 298 K

K = 4.542 × 10⁻³¹

7 0
2 years ago
A 63.5 g sample of an unidentified metal absorbs 355 ) of heat when its temperature changes
insens350 [35]

0.208 is the specific heat capacity of the metal.

Explanation:

Given:

mass (m)  = 63.5 grams 0R 0.0635 kg

Heat absorbed (q) = 355 Joules

Δ T (change in temperature) = 4.56 degrees or 273.15+4.56 = 268.59 K

cp (specific heat capacity) = ?

the formula used for heat absorbed  and to calculate specific heat capacity of a substance will be calculated by using the equation:

q = mc Δ T

c = \frac{q}{mΔ T}

c = \frac{355}{63.5X 268.59}

 = 0.208 J/gm K

specific heat capacity of 0.208 J/gm K

The specific heat capacity is defined as  the heat required to raise the temperature of a substance which is 1 gram. The temperature is in Kelvin and energy required is in joules.

 

5 0
2 years ago
A collection of coins contains 14 pennies, 16 dimes, and 7 quarters. What is the percentage of pennies in the collection?A colle
marin [14]

Answer:

=37.83783784

Explanation:

Find the total sum of all coins,

which is 37, take the number of pennies and the total of all coins put in parenthesis( 14/37) like so and than * times them by 100

you equation should look like this

(14/37)* 100= and than the answer shown above should be the one you received. I have checked this with multiple calculators, it should be accurate.  

8 0
2 years ago
Read 2 more answers
If the actual yield of the reaction was 75% instead of 100%, how many molecules of no would be present after the reaction was ov
artcher [175]
To be able to answer this equations, we must set given information. Suppose the reaction to yield NO is:

N₂ + O₂ → 2 NO

Next, suppose you have 1 g of each of the reactants. Determine first which is the limiting reactant.

1 g N₂ (1 mol N₂/ 28 g)(2 mol NO/1 mol N₂)= 0.07154 mol NO present

Number of molecules = 0.07154 mol NO(6.022×10²³ molecules/mol)
<em>Number of molecules = 4.3×10²² molecules NO present</em>
8 0
2 years ago
Read 2 more answers
Determine the specific heat (in J/g C) for a 2.508 kilogram substance which increases its temperature from 4.051 C to 42.061 C w
just olya [345]

Answer: 0.036 J/g°C

Explanation:

The quantity of heat energy (Q) required to raise the temperature of a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Given that,

Q = 3.42 Kilojoules

[Convert 3.42 kilojoules to joules

If 1 kilojoule = 1000 joules

3.42 kilojoules = 3.42 x 1000 = 3420J]

Mass = 2.508Kg

[Convert 2.508 kg to grams

If 1 kg = 1000 grams

2.508kg = 2.508 x 1000 = 2508g]

C = ? (let unknown value be Z)

Φ = (Final temperature - Initial temperature)

= 42.061°C - 4.051°C

= 38.01°C

Apply the formula, Q = MCΦ

3420J = 2508g x Z x 38.01°C

3420J = 95329.08g•°C x Z

Z = (3420J / 95329.08g•°C)

Z = 0.03588 J/g°C

Round the value of Z to the nearest thousandth, hence Z = 0.036 J/g°C

Thus, the specific heat of the substance is 0.036 J/g°C

7 0
2 years ago
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