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2 points

erica [24]

9 grams of hydrogen gas (H2) will SC Johnson need to react in order to make 1 bottle of Windex.

Explanation:

Balance equation for the formation of ammonia from H2 gas.

N2 + 3H2 ⇒ 2 $NH_{3}$

Given

mass of ammonia in 1 bottle of windex = 51 gram

atomic mass of ammonia 17.01 gram/mole

number of moles = $\frac{mass}{atomic mass of 1 mole}$

number of moles = $\frac{51}{17.01}$

= 3 moles of ammonia is formed.

in 1 bottle of windex there are 3 moles of ammonia 0r 51 grams of ammonia.

From the equation it can be found that:

3 moles of hydrogen reacted to form 2 moles of ammonia

so, x moles of hydrogen will react to form 3 moles of ammonia.

$\frac{2}{3}$ = $\frac{3}{x}$

x = 4.5 moles of hydrogen will be required.

to convert moles into gram formula used:

mass = atomic mass x number of moles (atomic mass of H2 is 2grams/mole)

= 2 x 4.5

= 9 grams of hydrogen.

7 0

3 years ago

Consider the following mechanism for the oxidation of bromide ions by hydrogen peroxide in aqueous acid solution. H+ + H2O2 ? H3

Margarita [4]

Answer: The rate law for the reaction is $\text{Rate}=k'[H+][H_2O_2][Br^-]$

Explanation:

Rate law is the expression which is used to express the rate of the reaction in terms of the molar concentration of reactants where each term is raised to the power their stoichiometric coefficient respectively from a balanced chemical equation.

In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.

The chemical equation for the oxidation of bromide ions by hydrogen peroxide in aqueous acid solution follows:

$2H^++2Br^-+H_2O_2\rightarrow Br_2+2H_2O$

The intermediate reaction of the mechanism follows:

Step 1: $H^++H_2O_2\rightleftharpoons H_3O_2^+;\text{ (fast)}$

Step 2: $H_3O_2^++Br^-\rightarrow HOBr+H_2O;\text{(slow)}$

Step 3: $HOBr+H^++Br^-\rightarrow Br_2+H_2O;\text{(fast)}$

As, step 2 is the slow step. It is the rate determining step

Rate law for the reaction follows:

$\text{Rate}=k[H_3O_2^+][Br^-]$ ......(1)

As, $[H_3O_2^+]$ is not appearing as a reactant in the overall reaction. So, we apply steady state approximation in it.

Applying steady state approximation for $[H_3O_2^+]$ from step 1, we get:

$K=\frac{[H_3O_2^+]}{[H^+][H_2O_2]}$

$[H_3O_2^+]=K[H^+][H_2O_2]$

Putting the value of $[H_3O_2^+]$ in equation 1, we get:

$\text{Rate}=k.K[H^+][H_2O_2][Br^-]\\\\\text{Rate}=k'[H+][H_2O_2][Br^-]$

Hence, the rate law for the reaction is $\text{Rate}=k'[H+][H_2O_2][Br^-]$

4 0

3 years ago

Mrs. Rushing fills a balloon with hydrogen gas to demonstrate its ability to burn. Which combination could she use to produce th

ale4655 [162]

Answer:

Mg and HCI

Explanation:

3 0

3 years ago

Read 2 more answers

For the reaction: 2 A (g) + B (s)= 2 C(s) + D (g)

jolli1 [7]

Answer:

The value of the equilibrium constant = 5.213

Explanation:

Here $K_p$ (equilibrium constant) is referred to as the partial pressure of product divided by the partial pressure of reactant with each pressure term raised to power that is equal to its stoichiometric coefficient in balanced equation .

As such only gas appear in $K_p$ expression as solids takes a value of 1;

SO ; in the given equation from the question:

2 A (g) + B (s) ----> 2 C(s) + D (g)

$K_p = \dfrac{[D]}{[A]^2}$

$K_p = \dfrac{2.71}{0.721^2}$

$K_p = 5.213$

The value of the equilibrium constant = 5.213

8 0

3 years ago

The proper name for a carbohydrate polymer with 2 subunits is

valentina_108 [34]

Carbohydrates are classified in three major categories depending upon the number of sub units joining to form them. These are,

Monosaccharides

Oligosaccharides

Polysachharides

The simplest single units are monosaccharides, if units are ranging between two and twenty they are called oligosaccharides and above twenty joining units they are called polysachharides.

Result:
The proper name for a carbohydrate polymer with 2 subunits is called Oligosachharide in general and Disaccharide in specific.

8 0

4 years ago