All categories

2 years ago

I neeeeeeeeed help you within 30 min

Chemistry

1 answer:

sladkih [1.3K]2 years ago

8 0

Answer:

The answers to your question are given below

Explanation:

The equation for the reaction is given below:

2KClO₃ —> 2KCl + 3O₂

The total number of each atoms present in the reactants and products can be obtained by doing a head count as illustrated below:

Reactants >>>>>> Number of atoms

Potassium (K) >>> 2 × 1 = 2

Chlorine (Cl) >>>> 2 × 1 = 2

Oxygen (O) >>>>> 2 × 3 = 6

Products >>>>>>> Number of atoms

Potassium (K) >>> 2 × 1 = 2

Chlorine (Cl) >>>> 2 × 1 = 2

Oxygen (O) >>>>> 3 × 2 = 6

You might be interested in

Seans formula did not have any atmos of oxygen

Dafna1 [17]

What are you asking???? If the formula had no atoms of oxygen then......

8 0

3 years ago

How would you prepare 250 mL of 0.125 M HCl from concentrated HCl (aq) that is 38.0% by mass with a density of 1.19 g/mL

hoa [83]

<h2>Step 1 : Identify the given </h2>

Volume = 250mL

Density = 1.19 g/ML

<h2>Step 2 . Calculate the mass of HCL </h2>

Density = mass/volume

∴Mass = Density * Volume

= 1.19g/mL* 250mL

= 297,5g

<h2>Step 3 : Calculate the total mass of the solution, given that concentration HCL is 38% </h2>

Mass of the total solution can be calculated by the following :

38% = Mc /297.5 * 100

Mc = 38/100 *297.5

= 113.05grams

• Finally, this means that mass of the total solution of 0.125M HCL i,s 113grams, ,you would use this mass to prepare 250 mL of 0.125 M HCl from concentrated HCl (aq) that is 38.0%

6 0

1 year ago

How many nanometers are in 0.0006245101 km?

maw [93]

Answer:

624510100

Explanation:

Doing a conversion factor:

$0,0006245101[km]*\frac{1000[m]}{1 km} *\frac{1x10^{9} nanometer}{1 m} =624510100 [nanometer]$

5 0

3 years ago

Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:

Phantasy [73]

What is Chemical Reaction?

A chemical reaction is the chemical transformation of one set of chemical components into another.

Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: $Al_{2}S_{3} +H_{2}O _\to Al(OH)_{3} + H_{2}S$

First, balance the chemical equation

$Al_{2}S_{3} + 6H_{2}O \to 2Al(OH)_{3} + 3H_{2}S$

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

$n_{Al_{2}S_{3} } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3} } = \frac{158g}{150.16g/mol} \\n_{Al_{2}S_{3} }=1.05 mol$

From the chemical reaction , the ratio of molar is 3mol $H_{2}S/1 mol Al_{2}S_{3}.$ So, the moles of hydrogen sulfide are:

$n_{H_{2} O} =\frac{131g}{18.02g/mol}$

= 7.26mol

From the chemical reaction, the molar ratio is 3 mol $H_{2}S/6$ mol $H_{2}O.$ So, the moles of hydrogen sulfide are:

Moles of $H_{2}S$ formed = 7.26 mol $H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }$

Th liming reactant is $Al_{2}S_{3}$ beacuse the mass of $Al_{2}S_{3}$ forms less product than water. Therefore, the maximum number of moles of $H_{2}S$ is 3.15 mol. We know that molar mass of $H_{2}S$ is 34.10g/mol. So, the maximum mass of $H_{2}S$ formed is,

$m_{H_{2}S } = n_{H_{2}S } \times$ Molar mass of $H_{2}S$

= 3.15 mol $\times$ 34.10g/mol

= 107.4g

Now, multiplying the number of moles of $Al_{2}S_{3}$ by the molar ratio between $Al_2S_3$ and $H_2O$ which is 6mol $H_2O/1mol$ $Al_2S_3$ we get the number of moles of $H_2O$ reacted.

Moles of $H_2O$ reacted = 1.05 mol $Al_{2}S_3 \times \frac{6 mol H_2O}{1 mol Al_2S_3}$

= 6.31 mol $H_2O$

The mass of $H_2O$ is,

$m_{H_{2} O}$ = 6.31 mol $\times$ 18.02g/ mol

= 114g

On subtracting, the mass of $H_2O$ reacted from the given mass of $H_2O$ is,

$m_{H_2O}$ = (131-114)g

= 17g

Hence, the excess remaining reactant is 17g

To learn more about Chemical Reaction

brainly.com/question/11231920

#SPJ4

8 0

2 years ago

effect. RbClO(s) → Rb+(aq) + ClO−(aq) HClO(aq) + H2O(l) equilibrium reaction arrow H3O+(aq) + ClO−(aq) The degree of dissociatio

Lemur [1.5K]

Explanation:

Common ion effect is defined as the effect which occurs on equilibrium when a common ion (an ion which is already present in the solution) is added to a solution. This effect generally decreases the solubility of a solute.

Equilibrium reaction of strontium sulfate and sodium sulfate follows the equation:

$RbClO(s)\rightarrow Rb^{+}(aq.)+ClO^{-}(aq.)$

$HClO(aq)+H_2O\rightleftharpoons H_3O^+(aq.)+ClO^{-}(aq.)$

According to Le-Chateliers principle: If there is any change in the variables of the reaction, the equilibrium will shift in the direction in order to minimize the effect.

In the equilibrium reactions, hypochlorite ion is getting increased on the product side, so the equilibrium will shift in the direction to minimize this effect, which is in the direction of hydrogen hypochlorite.

Thus, the addition hypochlorite ions will shift the equilibrium in the left direction.

The dissociation of hydrogen hypochlorite is suppressed due to the common ion effect.

8 0

3 years ago