Cách để phân biệt axit và bazo

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

.950 L of .420 M H2SO4 is mixed with .900 L of .260 M KOH. What concentration of sulfuric acid remains after neutralization?

kipiarov [429]

H₂SO₄:

V=0,95L
Cm=0,420mol/L

n = CmV = 0,42mol/L * 0,95L = 0,399mol

KOH:

V=0,9L
Cm=0,26mol/L

n = CmV = 0,26mol/L * 0,9L = 0,234mol

H₂SO₄ + 2KOH ⇒ K₂SO₄ + 2H₂O
1mol : 2mol
0,399mol : 0,234mol
limiting reagent
reamins: 0,399mol - 0,117mol = 0,282mol

n = 0,282mol
V = 0,950L + 0,900L = 1,85L

Cm = n / V = 0,282mol / 1,85L ≈ 0,152M

3 0

3 years ago

What is the empirical formula for a compound that is 83.7% carbon and 16.3% hydrogen?

zubka84 [21]

Hello!

We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100 g solution)

C: 83.7% = 83,7 g
H: 16.3% = 16.3 g

Let us use the above mentioned data (in g) and values will be converted to amount of substance (number of moles) by dividing by molecular mass (g / mol) each of the values, lets see:

$C: \dfrac{83.7\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} \approx 6.975\:mol$

$H: \dfrac{16.3\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 16.3\:mol$

We note that the values found above are not integers, so let's divide these values by the smallest of them, so that the proportion is not changed, let's see:

$C: \dfrac{6.975}{6.975} = 1$

$H: \dfrac{16.3}{6.975} \approx 2.3$

Note: So the ratio in the smallest whole numbers of carbon to hydrogen is 3:7, thus, the minimum or empirical formula found for the compound will be:

$\boxed{\boxed{C_3H_7}}\end{array}}\qquad\checkmark$

I hope this helps. =)

8 0

3 years ago

Please help! due soon!

nekit [7.7K]

Answer:

I think it's

there are the same number of molecules on each side of the equation, then a change of pressure makes no difference to the position of equilibrium

7 0

3 years ago

PH of 0,035M HCl? How to count if your calculator hasn't got -lg?

IrinaK [193]

Given information:

Concentration of HCl = 0.035 M

To determine:

pH of the solution

Explanation:

Hydrochloric acid, HCl is a strong acid. It will completely dissociate to give H+ and Cl- ions

HCl → H+ + Cl-

Hence the concentration of H+ = Cl- = 0.035M

Now, pH measures the strength of H+ ions in a given solution. It is expressed as:

pH = -log[H+]

pH (HCl) = -log(0.035) = 1.46

Ans: pH of 0.035M HCl is 1.46

6 0

2 years ago