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2 years ago

Urgent : Hydrogen and oxygen react under a specific set of conditions to produce water according to the equation:

Chemistry

1 answer:

ziro4ka [17]2 years ago

8 0

Hydrogen and oxygen react under a specific set of conditions to produce water according to the equation: 2H2(g)+O2(g) -> 2H2O(g)

I think, but i hope this helps :)

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8 gm of pure calcium is treated with 50 gram of pure Hcl to give cacl2 and h2 .which is limiting reactant?

elena-14-01-66 [18.8K]

Ca as a limiting reactant

<h3>Further explanation</h3>

Given

8 g Calcium

50 g HCl

<h3>Required</h3>

Limiting reactant

Solution

Reaction

Ca + 2HCl → CaCl₂ + H₂

mol Ca (Ar = 40 g/mol) :

= mass : Ar

= 8 g : 40 g/mol

= 0.2

mol HCl (MW= 36.5 g/mol) :

= mass : MW

= 50 g : 36.5 g/mol

= 1.37

Mol : coefficient reactants :

Ca = 0.2/1 = 0.2

HCl = 1.37/2 = 0.685

Ca as a limiting reactant(smaller ratio)

3 0

2 years ago

What mass of potassium hypochlorite (FW-90.6 g/mol) must be added to 4.50 x 10 mL of water to give a solution with pH 10.20? [Ka

marta [7]

Answer : The mass of potassium hypochlorite is, 4.1 grams.

Explanation : Given,

pH = 10.20

Volume of water = $4.50\times 10^2ml=0.45L$

The decomposition of KClO will be :

$KClO\rightarrow K^++ClO^-$

Now the further reaction with water $(H_2O)$ to give,

$ClO^-+H_2O\rightarrow HClO+OH^-$

First we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-10.20=3.8$

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

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

$3.8=-\log [OH^-]$

$[OH^-]=1.58\times 10^{-4}M$

Now we have to calculate the base dissociation constant.

Formula used : $K_b=\frac{K_w}{K_a}$

Now put all the given values in this formula, we get :

$K_b=\frac{1.0\times 10^{-14}}{4.0\times 10^{-8}}=2.5\times 10^{-7}$

Now we have to calculate the concentration of $ClO^-$ .

The equilibrium constant expression of the reaction is:

$K_b=\frac{[OH^-][HClO]}{[ClO^-]}$

As we know that, $[OH^-]=[HClO]=1.58\times 10^{-4}M$

$2.5\times 10^{-7}=\frac{(1.58\times 10^{-4})^2}{[ClO^-]}$

$[ClO^-]=0.0999M$

Now we have to calculate the moles of $ClO^-$ .

$\text{Moles of }ClO^-=\text{Molarity of }ClO^-\times \text{Volume of solution}$

$\text{Moles of }ClO^-=0.0999mole/L\times 0.45L=0.0449mole$

As we know that, the number of moles of $ClO^-$ are equal to the number of moles of KClO.

So, the number of moles of KClO = 0.0449 mole

Now we have to calculate the mass of KClO.

$\text{Mass of }KClO=\text{Moles of }KClO\times \text{Molar mass of }KClO$

$\text{Mass of }KClO=0.0449mole\times 90.6g/mole=4.07g\approx 4.1g$

Therefore, the mass of potassium hypochlorite is, 4.1 grams.

6 0

2 years ago

2. Which of the following expressions is generally used for solubility? (1 point)

miskamm [114]

<u>Answer:</u>

For 2: The correct answer is grams of solute per 100 grams of solvent.

For 3: The correct answer is supersaturated.

For 4: the correct answer is the solubility decreases.

<u>Explanation:</u>

<u>For 2:</u>

Solubility is defined as the property which refers to the ability of the solute that can be dissolved in a solvent. It is defined as the number of grams of solute per 100 grams of solvent.

<u>For 3:</u>

Unsaturated solution is defined as the solution in which amount of solute that is dissolved in the solvent is less.

Saturated solution is defined as the solution in which no more solute can be dissolved in the given amount of solvent.

Emulsion is defined as the dispersion of one liquid in another liquid in which it is not soluble.

Supersaturated solution is defined as the solution in which solvent contains more amount of solute than the required amount. These solutions help in the process of crystallization.

When a crystal is added to a <u>supersaturated solution</u>, more and more particles come out of the solution and this process is known as crystallization.