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4 years ago

Nitric acid in urban environments is formed in the reaction of answer with answer

1 answer:

4 0

Oxidation of nitrogen to nitrogen oxide in the channel of a lightning:
N₂ + O₂ = 2NO

oxidation of natural nitrogen oxide and of nitrogen oxide of combustion gases to nitrogen dioxide:
2NO + O₂ = 2NO₂

nitrogen dioxide reaction with moisture of air or rain water:
4NO₂ + O₂ + 2H₂O = 4HNO₃

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Your task is to create a buffered solution. You are provided with 0.10 M solutions of formic acid and sodium formate. Formic aci

Anton [14]

Answer:

15.2mL of the 0.10M sodium formate solution and 4.8mL of the 0.10M formic acid solution.

Explanation:

To find the pH of a buffer based on the concentration of the acid and conjugate base we must use Henderson-Hasselbalch equation:

pH = pKa + log [A⁻] / [HA]

<em>Where [A⁻] could be taken as moles of the sodium formate and [HA] moles of the formic acid</em>

<em />

4.25 = 3.75 + log [A⁻] / [HA]

0.5 = log [A⁻] / [HA]

3.162 = [A⁻] / [HA] <em>(1)</em>

As both solutions are 0.10M and you want to create 20mL of the buffer, the moles are:

0.10M * 20x10⁻³L =

2x10⁻³moles = [A⁻] + [HA] <em>(2)</em>

Replacing (2) in (1):

3.162 = 2x10⁻³moles - [HA] / [HA]

3.162 [HA] = 2x10⁻³moles - [HA]

4.162[HA] = 2x10⁻³moles

[HA] = 4.805x10⁻⁴ moles

[A⁻] = 2x10⁻³moles - 4.805x10⁻⁴ moles = 1.5195x10⁻³moles

That means, to create the buffer you must add:

[A⁻] = 1.5195x10⁻³moles * (1L / 0.10mol) = 0.0152L =

<h3>15.2mL of the 0.10M sodium formate solution</h3>

[HA] = 4.805x10⁻⁴ moles * (1L / 0.10mol) = 0.0048L =

<h3>4.8mL of the 0.10M formic acid solution</h3>

4 0

3 years ago

An ideal gas cannot exist outside of ideal gas conditions<br><br> true or false ?

zzz [600]

Answer:

Yes

Explanation:

5 0

3 years ago

Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.85 M of reagent

elixir [45]

Answer : The initial rate for a reaction will be $3.8\times 10^{-4}Ms^{-1}$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The chemical equation will be:

$A+B+C\rightarrow P$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b[C]^c$

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

$6.1\times 10^{-5}=k(0.2)^a(0.2)^b(0.2)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-4}=k(0.2)^a(0.2)^b(0.6)^c$ ....(2)

Expression for rate law for third observation:

$2.4\times 10^{-4}=k(0.4)^a(0.2)^b(0.2)^c$ ....(3)

Expression for rate law for fourth observation:

$2.4\times 10^{-4}=k(0.4)^a(0.4)^b(0.2)^c$ ....(4)

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-4}}{6.1\times 10^{-5}}=\frac{k(0.2)^a(0.2)^b(0.6)^c}{k(0.2)^a(0.2)^b(0.2)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-4}}{6.1\times 10^{-5}}=\frac{k(0.4)^a(0.2)^b(0.2)^c}{k(0.2)^a(0.2)^b(0.2)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.4)^a(0.4)^b(0.2)^c}{k(0.4)^a(0.2)^b(0.2)^c}\\\\1=2^b\\b=0$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^0[C]^1$

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$6.1\times 10^{-5}=k(0.2)^2(0.2)^0(0.2)^1$

$k=7.6\times 10^{-3}M^{-2}s^{-1}$

Now we have to calculate the initial rate for a reaction that starts with 0.85 M of reagent A and 0.70 M of reagents B and C.

$\text{Rate}=k[A]^2[B]^0[C]^1$

$\text{Rate}=(7.6\times 10^{-3})\times (0.85)^2(0.70)^0(0.70)^1$

$\text{Rate}=3.8\times 10^{-3}Ms^{-1}$

Therefore, the initial rate for a reaction will be $3.8\times 10^{-3}Ms^{-1}$

6 0

3 years ago

In chemical kinetics, the magnitude of the rate constant depends on: I. the reaction time of the chemical reaction II. the activ

Snowcat [4.5K]

Answer:

THE TEMPERATURE AT WHICH THE CHEMICAL REACTION IS BEING CONDUCTED.

Explanation:

The reaction rate constant is dependence on the temperature in which the reaction takes place. The rate of a reaction and the activation energy relationship is given by the Arrhenius equation with the rate constant as a function of temperature.

Mathematically, the rate constant is expressed as;

k(T) = Ae^-Ea/RT

where Ea is the activation energy, R is the gas constant, Ae is the pre-exponential factor and it is the frequency at which a reactant molecule collide with each other, T is the temperature.

3 0

3 years ago

calculate the mass of calcium phosphate and the mass of sodium chloride that could be formed when a solution containing 12.00g o

Leviafan [203]

Answer : The mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

Explanation : Given,

Mass of $Na_3PO_4$ = 12.00 g

Mass of $CaCl_2$ = 10.0 g

Molar mass of $Na_3PO_4$ = 164 g/mol

Molar mass of $CaCl_2$ = 111 g/mol

Molar mass of $NaCl$ = 58.5 g/mol

Molar mass of $Ca_3(PO_4)_2$ = 310 g/mol

First we have to calculate the moles of $Na_3PO_4$ and $CaCl_2$ .

$\text{Moles of }Na_3PO_4=\frac{\text{Given mass }Na_3PO_4}{\text{Molar mass }Na_3PO_4}$

$\text{Moles of }Na_3PO_4=\frac{12.00g}{164g/mol}=0.0732mol$

and,

$\text{Moles of }CaCl_2=\frac{\text{Given mass }CaCl_2}{\text{Molar mass }CaCl_2}$

$\text{Moles of }CaCl_2=\frac{10.0g}{111g/mol}=0.0901mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is:

$2Na_3PO_4+3CaCl_2\rightarrow 6NaCl+Ca_3(PO_4)_2$

From the balanced reaction we conclude that

As, 3 mole of $CaCl_2$ react with 2 mole of $Na_3PO_4$

So, 0.0901 moles of $CaCl_2$ react with $\frac{2}{3}\times 0.0901=0.0601$ moles of $Na_3PO_4$

From this we conclude that, $Na_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $CaCl_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$ and $Ca_3(PO_4)_2$

From the reaction, we conclude that

As, 3 mole of $CaCl_2$ react to give 6 mole of $NaCl$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{6}{3}\times 0.0901=0.1802$ mole of $NaCl$