All categories

2 years ago

El método científico es el conjunto de pasos necesarios para obtener

Chemistry

1 answer:

Nastasia [14]2 years ago

6 0

Answer:

El método científico es una metodología para obtener nuevos conocimientos, que ha caracterizado históricamente a la ciencia, y que consiste en la observación sistemática, medición, experimentación y la formulación, análisis y modificación de hipótesis. Las principales características de un método científico válido son .

Explanation:

You might be interested in

The rate law for the reaction 3a ⟶⟶ c is rate = 4. 36 x 10–2 l • mol–1 • h–1[ a ]2 how long will it take in hours for the concen

Paha777 [63]

The hours taken for concentration to decrease from 0 to 74 min. to 0.21 m is 91.7 hours.

<h3>What is the rate law of a reaction?</h3>

Rate law depicts the rate of a chemical reaction depend on the concentration of the reactant.

The given reaction is second order reaction

Thus, the hours taken for concentration to decrease from 0 to 74 min. to 0.21 m is 91.7 hours.

Learn more about rate law of a reaction

brainly.com/question/8314253

#SPJ4

4 0

2 years ago

Be sure to answer all parts.The equilibrium constant, Kc, for the formation of nitrosyl chloride from nitric oxide and chlorine,

djverab [1.8K]

<u>Answer:</u> The reaction proceeds in the forward direction

<u>Explanation:</u>

For the given chemical equation:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

Relation of $K_p\text{ with }K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta n_g}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = ?

$K_c$ = equilibrium constant in terms of concentration = $6.5\times 10^4$

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = $35^oC=[35+273]K=308K$

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$K_p=6.5\times 10^4\times (0.0821\times 500)^{-1}\\\\K_p=1583.43$

$K_p$ is the constant of a certain reaction at equilibrium while $Q_p$ is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

The expression of $Q_p$ for above equation follows:

$Q_p=\frac{(p_{NOCl})^2}{p_{Cl_2}\times (p_{NO})^2}$

We are given:

$p_{NOCl}=1.76atm$

$p_{NO}=1.01atm$

$p_{Cl_2}=0.42atm$

Putting values in above equation, we get:

$Q_p=\frac{(1.76)^2}{0.42\times (1.01)^2}=7.23$

We are given:

$K_p=1583.43$

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, $K_p>Q_p$ , the reaction will be favoring product side.

Hence, the reaction proceeds in the forward direction

4 0

2 years ago

In this reaction, _____.

ella [17]

Answer:

AB + CD ----> AC + BD

Explanation:

If you think this reaction:

AB + CD ----> AC + BD

(Reactants) (Products)

All the statements are true.

7 0

3 years ago

4. A student is doing experiments with CO2(q) . Originally a sample of the gas is in a rigid container at 299 Kand 0.70 atm. The

marta [7]

Answer:

b. 0,99atm

c. Answer is in the explanation

d. Answer is in the explanation

Explanation:

b. Using Gay-Lussac's law:

P₁T₂ = P₂T₁

P₁: 0,70 atm; T₂: 425K; P₂: ??; T₁: 299K

0,70atm×425K / 299K = <em>0,99 atm</em>

c. Using kinetic molecular theory, the increasing of temperature increases the kinetic energy of gas particles and if kinetic energy increases, the pressure increases. That means the increasing of temperature increases the pressure in the system.

d. Now, the increases in kinetic energy of gases increase the collisions betwen particles. As these intermolecular forces that are not taken into account in ideal gas law, the observed pressure will be different to the pressure predicted by ideal gas law.

I hope it helps!

8 0

2 years ago

what is the products of the reaction between 2-methylbut-2-ene and HCL 1b. write the IUPAC names of the product of the reaction;

zzz [600]

Answer:

Explanation:

2 - chloro-2-methylbutane

8 0

2 years ago