Ask question

Ask question

All categories

3 years ago

9

When an atom gives off energy in the form of light, a. it becomes excited. b. it becomes more stable. c. it returns to ground st

ate. d. both b & c e. none of the above.

1 answer:

egoroff_w [7]3 years ago

6 0

Answer is only B.
<span>it becomes more stable</span>

You might be interested in

The reaction of hydrogen(g) with fluorine(g) to form hydrogen fluoride(g) proceeds as follows: H2(g) + F2(g) 2 HF(g) When 0.647

Nata [24]

<u>Answer:</u> The value of $\Delta H$ for the reaction will be -537 kJ

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of hydrogen gas = 0.647 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in above equation, we get:

$\text{Moles of hydrogen gas}=\frac{0.647g}{2g/mol}=0.324mol$

We are given:

Energy released for 0.324 moles of hydrogen reacted is 174 kJ

For the given chemical reaction:

$H_2(g)+F_2(g)\rightarrow 2HF(g)$

By Stoichiometry of the reaction:

When 0.324 moles of hydrogen gas is reacted, the energy evolved is 174 kJ

So, when 1 mole of hydrogen gas will react, the energy evolved will be = $\frac{174kJ}{0.324mol}\times 1mol=537kJ$

<u>Sign convention of heat:</u>

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

Hence, the value of $\Delta H$ for the reaction will be -537 kJ

6 0

3 years ago

A gas has an initial volume of 455 mL at 105ºC and a final volume of 235 mL. What is its final temperature in Celsius degrees?

Oksana_A [137]

Hello!

To solve this problem we're going to use the Charles' Law. This Law describes the relationship between Volume and Temperature in an ideal gas. Applying this law we have the following equation:

$\frac{V1}{T1} = \frac{V2}{T2} \\ \\ T2= \frac{V2*T1}{V1}= \frac{235 mL * 105 ^{\circ}C }{455 mL}=54,23 ^{\circ}C$

So, the final temperature is 54,23 °C

Have a nice day!

5 0

3 years ago

Read 2 more answers

CAN SOMEONE PLEASE EXPLAIN

Kipish [7]

The answer will be b

5 0

3 years ago

what is the mass of water vapor produced when 3.2 liters reacts with 8.7 liters of oxygen gas at STP?

uranmaximum [27]

Answer:

2.57g

Explanation:

First, let us write a balanced equation for the reaction. This is illustrated below:

2H2 + O2 —> 2H2O

Next let us determine the limiting reactant. This is achieved as follows:

From the equation,

2L H2 required 1L of O2.

Therefore, 3.2L of H will require = 3.2/2 = 1.6L of O2

From the calculation above, O2 is excess because the volume of O2 given from the question is far greater than the volume of O2 obtained from our calculation. Therefore, H2 is the limiting reactant.

Now let us covert 3.2L of H2 to mole. This is illustrated below:

1mole of a gas occupy 22.4L at stp

Therefore, Xmol of H2 will occupy 3.2L i.e

Xmol of H2 = 3.2/22.4 = 0.143mol

From the equation,

2moles of H2 produced 2moles of H2O.

Therefore, 0.143mol of H2 will also produce 0.143moles of H2O.

Now, we can obtain the mass of the water vapour produced by convert 0.143mol of H2O to gram. This is illustrated below:

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Number of mole of H2O = 0.143mol

Mass of H2O =?

Mass = mole x Molar Mass

Mass of H2O = 0.143 x 18 = 2.57g

The mass of water vapour produce is 2.57g

8 0

3 years ago

Ammonia NH3 may react with oxygen to form nitrogen gas and water.4NH3 (aq) + 3O2 (g) \rightarrow 2 N2 (g) + 6H2O (l)If 2.15g of

bagirrra123 [75]

Answer:

NH3 is the limiting reactant

The % yield is 36.1 %

Explanation:

<u>Step 1: </u>Data given

Mass of NH3 = 2.15 grams

Mass of O2 = 3.23 grams

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

volume of N2 produced = 0.550 L

Temperature = 295 K

Pressure = 1.00 atm

<u>Step 2:</u> The balanced equation:

4NH3 (aq) + 3O2 (g) → 2 N2 (g) + 6H2O (l)

<u>Step 3:</u> Calculate moles of NH3

Moles NH3 = Mass NH3 / Molar Mass NH3

Moles NH3 = 2.15 grams / 17.03 g/mol

Moles NH3 = 0.126 moles

<u>Step 4:</u> Calculate moles of O2

Moles O2 = 3.23 grams / 32 g/mol

Moles O2 = 0.101 moles

<u>Step 5: </u>Calculate the limiting reactant

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

NH3 is the limiting reactant. It will completely be consumed ( 0.126 moles).

O2 is in excess, there will be 3/4 * 0.126 = 0.0945 moles consumed

There will remain 0.101 - 0.945 = 0.0065 moles of O2

<u>Step 6:</u> Calculate moles of N2

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

For 4 moles NH3 , we'll have 2 moles of N2 produced

For 0.126 moles NH3 consumed, we'll have 0.063 moles of N2 produced.

<u>Step 7</u>: Calculate volume of N2 produced

p*V = n*R*T

⇒ with p = the pressure of the gas = 1.00 atm

⇒ with V = the volume = TO BE DETERMINED

⇒ with n = the number of moles N2 = 0.063 moles

⇒ with R = the gasconstant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 295

V = (nRT)/p

V = (0.063*0.08206*295)/1

V = 1.525 L = theoretical yield

<u>Step 8:</u> Calculate the % yield

% yield = actual yield / theoretical yield

% yield = (0.550 L / 1.525 L)*100%

% yield = 36.1 %

4 0

3 years ago