What is the mass of the Iron (Fe) involved in this reaction? (4pts) 1

Why is argon used in a light bulb? using the terms of the electronic structure of an argon atom.

AVprozaik [17]

The property of a filament of argon which lasts for a long time than air because argon is inert and will not oxidize the filament even at high temperatures is the key reason why argon is used in light bulbs compared to air which reacts quickly in the atmosphere.

3 0

3 years ago

The element chlorine has two stable isotopes, chlorine-35 with a mass of 34.97 amu and chlorine-37 with a mass of 36.95 amu. Fro

sergejj [24]

Answer: One isotope has a percentage abundance of 75.75 % and the percentage abundance of another isotope is 24.24%.

Explanation:

We are given the two stable isotopes of chlorine with their respective masses. The average atomic mass of chlorine is also given.

Average atomic mass of chlorine = 35.45 amu.

Let us assume the fractional abundance of one isotope be 'x' and the fractional abundance for another isotope will be (1 - x) because the total fractional abundance is always equal to 1.

For Chlorine-35 isotope:

Fractional abundance = x

Mass = 34.97 amu

For Chlorine-37 isotope:

Fractional abundance = 1 - x

Mass = 36.95 amu

The formula for the calculation of average atomic mass is given by:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

Putting values in above equation, we get:

$35.45=34.97(x)+36.95(1-x)\\\\35.45=34.97x+36.95-36.95\\\\-1.5=-1.98x\\\\x=\frac{-1.5}{-1.98}=0.7575$

$1-x=1-0.7575=0.2424$

Converting these two fractional abundances into percentage abundances by multiplying it with 100.

$x=0.7575\times 100=75.75\%\\\\1-x=0.2424\times 100=24.24\%$

Hence, one isotope has a percentage abundance of 75.75 % and the percentage abundance of another isotope is 24.24%.

7 0

3 years ago

What would happen if the Earth was tilted more than its current tilt of 23.5°? HELP ASAP!!

Oduvanchick [21]

Answer:

Wouldn't the Earth's atmosphere be moving too fast that it eventually breaks out?

Explanation:

Do NOT trust me.

7 0

3 years ago

Gaseous indium dihydride is formed from the elements at elevated temperature:

3241004551 [841]

Answer:

For 1: The value of $Q_p$ for above reaction is 36.83

For 2: The value of $Q_p$ for above reaction is 36.83

For 3: The equilibrium partial pressure of Indium is 0.126 atm

For 4: The equilibrium partial pressure of hydrogen gas is 0.094 atm

For 5: The equilibrium partial pressure of Indium dihydrogen is 0.018 atm

Explanation:

We are given:

Partial pressure of Indium gas = 0.0650 atm

Partial pressure of hydrogen gas = 0.0330 atm

Partial pressure of Indium dihydride = 0.0790 atm

The given chemical equation follows:

$ln(g)+H_2(g)\rightleftharpoons InH_2(g)$

Initial: 0.065 0.033 0.079

At eqllm: 0.065-x 0.033-x 0.079+x

For 1:

The expression of $Q_p$ for above reaction follows:

$Q_p=\frac{p_{InH_2}}{p_{In}\times p_{H_2}}$

Putting values in above equation, we get:

$Q_p=\frac{0.079}{0.065\times 0.033}=36.83$

Hence, the value of $Q_p$ for above reaction is 36.83

For 2:

We are given:

$K_p$ of the reaction = 1.48

There are 3 conditions:

When $K_{p}>Q_p$ ; the reaction is product favored.
When $K_{p}$ ; the reaction is reactant favored.
When $K_{p}=Q_p$ ; the reaction is in equilibrium.

As, $Q_{p}>K_p$ for the given reaction, the reaction is reactant favored.

Hence, the reaction proceed in the backward direction to attain equilibrium

For 3:

The expression of $K_p$ for above reaction follows:

$K_p=\frac{p_{InH_2}}{p_{In}\times p_{H_2}}$

Putting values in above equation, we get:

$1.48=\frac{(0.079+x)}{(0.065-x)\times (0.033-x)}\\\\x=-0.061,0.835$

Neglecting the value of x = 0.835 because the reaction is going backwards. So, by taking this value, the pressure of the reactants will decrease

So, equilibrium partial pressure of Indium = (0.065 - x) = [0.065 - (-0.061)] = 0.126 atm

For 4:

The equilibrium partial pressure of hydrogen gas = (0.033 - x) = [0.033 - (-0.061)] = 0.094 atm

For 5:

The equilibrium partial pressure of Indium dihydrogen = (0.079 + x) = [0.079 + (-0.061)] = 0.018 atm

6 0

3 years ago

Ethanoic acid and 1-butanol can react to produce water and a compound classified as an 1) aldehyde 2) amine 3) ester 4) halide

kvv77 [185]

Answer:

3) ester

Explanation:

Esterification is the process in which alkanol and alkanoic acids reacts in the presence of a catalyst and heat. The product is usually an ester(alkanoate) and water.

For the reaction between ethanoic acid and 1-butanol, the product is butylethanoate and water as shown below:

CH₃COOH + C₄H₈OH → CH₃COOC₄H₈ + H₂O

4 0

3 years ago