Calculate the number of moles of iodine molecules

An excited ozone molecule, O3*, in the atmosphere can undergo one of the following reactions,O3* → O3 (1) fluorescenceO3* → O +

Maurinko [17]

Answer:

The simplified expression for the fraction is $\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$

Explanation:

From the given information:

O3* → O3 (1) fluorescence

O + O2 (2) decomposition

O3* + M → O3 + M (3) deactivation

The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)

The rate of decomposition is = k₂ × cO

The rate of deactivation = k₃ × cO × cM

where cM is the concentration of the inert molecule

The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:

$\text {X} = \dfrac{ \text {rate of deactivation} }{ \text {(rate of fluorescence) +(rate of decomposition) + (rate of deactivation) } } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{ {(k_1 \times cO) +(k_2 \times cO) + (k_3 \times cO \times cM) } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{cO (k_1 +k_2 + k_3 \times cM) }$

$\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$ since cM is the concentration of the inert molecule

7 0

3 years ago

How many grams of XeF6 are required to react with 0.579 L of hydrogen gas at 6.46 atm and 45°C in the reaction shown below?

ankoles [38]

The chemical reaction equation for this is

XeF6 + 3H2 ---> Xe + 6HF

Assuming gas behaves ideally, we use the ideal gas formula to solve for number of moles H2 with T = 318.15K (45C), P = 6.46 atm, V = 0.579L. Then we use the gas constant R = 0.08206 L atm K-1 mol-1.

we get n = 0.1433 moles H2

to get the mass of XeF6,

we divide 0.1433 moles H2 by 3 since 1 mole XeF6 needs 3 moles H2 to react then multiply by the molecular weight of XeF6 which is 245.28 g/mole XeF6.

0.1433 moles H2 x $\frac{1 mole XeF6}{3 moles H2}$ x $\frac{245.28 g XeF6}{1 mole XeF6}$ = 11.71 g XeF6

Therefore, 11.71 g of XeF6 is needed to completely react with 0.579 L of Hydrogen gas at 45 degrees Celcius and 6.46 atm.

3 0

3 years ago

The element californium (cf) sells for $1000 per µg. assuming 6.02 x 1023 atoms of cf have a mass of 251 grams, how many atoms o

saveliy_v [14]

Answer:- $2.40*10^1^0atoms$

Solution:- It is a simple unit conversion problem. We could solve this using dimensional analysis.

We know that, 1 US dollar = 100 cents

1 cent = 1 US penny

So, 1 US dollar = 100 US pennies

$1g=10^6\mu g$

Let's make the set up starting with 1 penny as:

$1penny(\frac{$1}{100pennies})(\frac{1\mu g}{$1000})(\frac{1g}{10^6\mu g})(\frac{6.02*10^2^3atoms}{251g})$

= $2.40*10^1^0atoms$

Therefore, we can bye $2.40*10^1^0atoms$ of Cf in one US penny.

3 0

3 years ago

The half-life of radon-222 is 3.8 days. If a sample currently contains 3.1 grams of radon-222, how much radon-222 did this sampl

vodka [1.7K]

The number of grams of radon 222 did it have 15.2 ago was 49.6 grams( answer C)

<u>calculation</u>

calculate the number of half life it has covered from 15.2 days to 3.8 days

that is divide 15.2/ 3.8 = 4 half life

half life is time taken for a radio activity of a specified isotope to fall to half its original mass

therefore 3.8 days ago it was 3.1 x2 = 6.2 grams

7.6 days ago it was 6.2 x2 = 12.4 grams

11.4 days ago it was 12.4 x2= 24.8 grams

15.2 days ago it was 24.8 x2=49.6 grams

5 0

3 years ago

Chose the element pairings that are likely to react with each other<br> Na and K or Na and Br

kaheart [24]

Answer:

Na k

Explanation:

because na is a metal and potassium is also a metal and both are active metal so is less likely to react as no bond is formed between them

7 0

3 years ago