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3 years ago

8

An elementary reaction _____. converts reactants to products in a single step never needs a catalyst has only elements as produc

ts has only elements as reactants description

1 answer:

IgorLugansk [536]3 years ago

8 0

Answer:

chain

Explanation:

I say that its chain because of the porreqwue

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HELP!!!!!! DUE IN 1 HOUR

Gnesinka [82]

Answer:

Group 17, 2, 7, 7 electrons

Explanation:

5 0

3 years ago

When aluminum is placed in concentrated hydrochloric acid, hydrogen gas is produced. 2 Al ( s ) + 6 HCl ( aq ) ⟶ 2AlCl3 ( aq ) +

Rufina [12.5K]

Answer:

The answer to your question is V = 19.9 L

Explanation:

Data

mass of Al = 8.60 g

volume of H₂ = ?

Balanced chemical reaction

2Al + 6HCl ⇒ 2AlCl₃ + 3H₂

1.- Use proportions to calculate the mass of H₂ produced

Atomic mass of Al = 2 x 27 = 54 g

Atomic mass of H₂ = 6 x 1 = 6 g

54 g of Al -------------------- 6 g of H₂

8 g of Al ------------------- x

x = (8 x 6) / 54

x = 48 / 54

x = 0.89 g of H₂

2.- Calculate the volume of H₂

- Convert the H₂ to moles

1 g of H₂ --------------- 1 mol

0.89 g ---------------- x

x = 0.89 moles

1 mol of H₂ -------------- 22.4 l

0.89 moles ------------- x

x = (0.89 x 22.4) / 1

x = 19.9 L

6 0

3 years ago

Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$

<u>Part (a) :</u>

<u>Given:</u>

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}$

and,

$\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

<u>Part (b) :</u>

<u>Given:</u>

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}$

and,

$-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

5 0

4 years ago

After reading the list of physical properties above, you realize that they are describing a substance that is a

Alecsey [184]

The answer is B

Hope this helps!

5 0

3 years ago

What is the number of atoms per unit cell for each metal?

Sav [38]

In the question, the number of atoms per unit cell is required for:
A) Polonium (Po)
In polonium, the structure is simple cubic, meaning there are 8 corner atoms, which add up to one atom per unit cell.

B) Manganese (Mn)
The structure of the Mn can be considered to be a body centered cubic (BCC) and the number of atoms for this is 8 corner atoms and 1 central atoms, making a total of 2 atoms per unit cell.

C) Silver (Ag)
Silver has a face centered cubic (FCC) unit cell structure, where there are 8 corner atoms and 6 atoms on the faces, so there are a total of 4 atoms per unit cell.

4 0

3 years ago