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3 years ago

What characteristic bright line spectrum of an elementforced when electrons?

1 answer:

8 0

The characteristic bright-line spectrum of an element is produced when its electrons return to lower energy levels To be in the ground state all electrons must be in their lowest energy state; all excited atoms must lose energy. The lost energy appears in the form of light. Hope this helped :)

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The total pressure of a mixture of oxygen and hydrogen

Mandarinka [93]

Answer:

The composition of the original mixture in molepercent is 80% of H₂ and 20% of O₂.

Explanation:

We need to combine the ideal gas law (PV = nRT) and Dalton's law of partial pressure (Pt = Pa +Pb +Pc+...).

The total pressure of the mixture is Pt = P (H₂) + P (O₂)

The number of moles can be found by Pt = nt RT/V, in which nt = n (H₂) +n (O₂).

If Pt is 1 atm, nt is 1.0 mol.

Now we need to consider the chemical reaction below:

H₂ + 0.5O₂ → H₂O

This shows that for each mol of O₂ we need two mol of H₂.

We know that the remaining gas is pure hydrogen and that its pressure is 0.4atm. Since PV = nRT, by the end of the reaction, 0.4 mol of H₂ remains in the system.

This means that in the beginning we have n mol of H₂, and when x mol of H₂ reacts with 0,5x mol of O₂, 0.4 mol of H₂ reamains.

If we have 1 mol in the begining and 0.4 mol in the end, the total amount of gas that reacted (x + 0.5X) is equal to 0.6 mol

x + 0.5X = 0.6 mol ∴ x = 0.6 mol / 1.5 ∴ x = 0.4 mol

0.4 mol of H₂ reacted with 0.2 mol of O₂ and 0.4 mol of H₂ remained as excess.

Therefore, in the beginning we had 0.8 mol of H₂ and 0.2 mol of O₂. Thus the molepercent of the mixture is 80% of H₂ and 20% of O₂.

3 0

3 years ago

A forklift applies a force of 2,000 N to raise a box 3 m.

anygoal [31]

If the force is applied to lift the box, then there is no work done as the displacement of the box is not in the same direction of the box.

However, if the force is being applied to move the box in the direction of the force, then the work done is:

Work = force x distance

Work = 1500 x 3

Work = 4,500 Joules

6 0

2 years ago

How much heat has been lost when a 78 g piece of copper is placed into a calorimeter and cools from 120°C to

lora16 [44]

The amount of heat lost by the copper is 2402.4 J.

To Calculate the amount of heat lost, we use the formula below.

<h3>Formula:</h3>

Q = cm(t₂-t₁)................. Equation 1

<h3>Where:</h3>

Q = Amount of heat lost
c = specific heat capacity of copper
m = mass of copper
t₂ = Final temperature
t₁ = Initial temperature

From the question,

<h3>Given:</h3>

m = 78 g
c = 0.385 J/g°C
t₂ = 120°C
t₁ = 40°C

Substitute these values into equation 1.

Q = 78(0.385)(120-40)
Q = 2402.4 J

Hence, The heat lost by the copper is 2402.4 J

Learn more about heat here: brainly.com/question/13439286

3 0

2 years ago

Sammy and Marisa are asked to determine what kind of bonds are present in the

kotykmax [81]

Answer:

I think Sammy is right because I'm sure the compound contains ionic bonds

7 0

2 years ago

oth hydrogen sulfide (H2S) and ammonia (NH3) have strong, unpleasant odors. Which gas has the higher effusion rate? If you opene

Ksju [112]

Answer: Odor of ammonia would we detect first on the other side of the room.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

$\frac{Rate_{H_2S}}{Rate_{NH_3}}=\sqrt{\frac{M_{NH_3}}{M_{H_2S}}}$

$\frac{Rate_{H_2S}}{Rate_{NH_3}}=\sqrt{\frac{17.031}{34.1}}$

$\frac{Rate_{H_2S}}{Rate_{NH_3}}=0.71$

Thus the odor of ammonia would we detect first on the other side of the room as the rate of effusion of ammonia would be faster as it has low molecular weight as compared to hydrogen sulphide.

5 0

3 years ago

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