All categories

3 years ago

How would adding the catalyst nitrogen monoxide (NO) affect this reaction?

Chemistry

1 answer:

RSB [31]3 years ago

8 0

Answer: A) NO increases the rate at which $SO_3$ molecules are formed.

Explanation:

A catalyst is a substance which increases the rate of a reaction by taking the reaction through a different path which involves lower activation energy and thus more reactant molecules can cross the energy barrier by undergoing collisions and convert to products.

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

Thus NO will increase the rate of reaction by lowering the activation energy and thus the colllisions among $SO_2$ and $O_2$ molecules will incraese which in turn will lead to formatioon of more $SO_3$ molecules.

You might be interested in

Which of the following possess the greatest concentration of hydroxide ions?

jek_recluse [69]

Answer : The correct option is (d) a solution of 0.10 M NaOH

Explanation :

(a) a solution of pH 3.0

First we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-3.0=11$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$11=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-11}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-11}M$

(b) a solution of 0.10 M $NH_3$

As we know that 1 mole of $NH_3$ is a weak base. So, in a solution it will not dissociates completely.

So, the $OH^-$ concentration will be less than 0.10 M

(c) a solution with a pOH of 12.

We have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$12=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-12}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-12}M$

(d) a solution of 0.10 M NaOH

As we know that $NaOH$ is a strong base. So, it dissociates to give $Na^+$ ion and $OH^-$ ion.

So, 0.10 M of $NaOH$ in a solution dissociates to give 0.10 M of $Na^+$ ion and 0.10 M of $OH^-$ ion.

Thus, the $OH^-$ concentration is, 0.10 M

(e) a $1\times 10^{-4}M$ solution of $HNO_2$

As we know that 1 mole of $HNO_2$ in a solution dissociates to give 1 mole of $H^+$ ion and 1 mole of $NO_2^-$ ion.

So, $1\times 10^{-4}M$ of $HNO_2$ in a solution dissociates to give $1\times 10^{-4}M$ of $H^+$ ion and $1\times 10^{-4}M$ of $NO_2^-$ ion.

The concentration of $H^+$ ion is $1\times 10^{-4}M$

First we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (1.0\times 10^{-4})$

$pH=4$

Now we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-4=10$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$10=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-10}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-10}M$

From this we conclude that, a solution of 0.10 M NaOH possess the greatest concentration of hydroxide ions.

Hence, the correct option is (d)

3 0

4 years ago

In an experiment, 4.14 g of phosphorus combined with chlorine to produce 27.8 g of a white solid compound. what is the empirical

Umnica [9.8K]

Grams of Phosphorus = 4.14 grams
Grams of white compound = 27.8 grams
Grams of Chlorine would be = 27.8 - 4.14 = 23.66 grams
Calculating moles which would be grams / molar mass
Molar mass of P = 30.97 grams / moles; Molar mass of Cl = 35.45 grams / moles
Moles of Phosphorus = 4.14 grams / 30.97 grams / moles = 0.1337 moles
Moles of Chlorine = 23.66 grams / 35.45 grams / moles = 0.6674 moles
Calculating the ratios by dividing with the small entity
P = 0.1337 moles / 0.1337 moles = 1
Cl = 0.6674 moles / 0.1337 moles = 5
So the empirical formula would be PCl5

3 0

3 years ago

What do all ions have in common in terms of their electrical structure?????

Free_Kalibri [48]

Answer:

The ratio between protons to electrons is not 1:1

Explanation:

A normal atom will be neutral in charge having 1 electron for each atom. An Atom that gains or looses an electron loses that perfect ratio. It is positive is electrons are loss and negative if electrons are gained.

3 0

4 years ago

The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is ________ M sodium ion and ________ M sulfate i

omeli [17]

Answer:

The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion. (option E)

Explanation:

Step 1: Data given

Volume = 500 mL = 0.500 L

The concentration sodium sulfate = 2.104 M

Step 2: The equation

Na2SO4 → 2Na+ + SO4^2-

For 1 mol Na2SO4 we have 2 moles sodium ion (Na+) and 1 mol sulfate ion (SO4^2-)

Step 3: Calculate the concentration of the ions

[Na+] = 2*2.104 M = 4.208 M

[SO4^2-] = 1*2.104 M = 2.104 M

The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion. (option E)

8 0

3 years ago

What is the atomic number of the atoms that make up carbon (C)?

Vsevolod [243]

Answer:

Explanation:

The atomic number (chemical sombol) is 6

7 0

3 years ago

Read 2 more answers