Oxidation states number (OSN) show how many electrons gained, lost or shared during chemical reactions.
Here are some rules of oxidation state numbers.
1. OSN of 1A elements in compounds is +1
2. OSN of Oxygen in compounds is -2
3. Sum of OSN of elements equals zero in neutral compounds.
4. OSN of free elements is zero. ex. F2, N2, O2.. all have 0 OSN.
If an element loses electrons during reaction, it is oxidized. That element is called as reducing agent.
If an element gains electrons during reaction, it is reduced. That element is called as oxidizing agent.
In the given reaction, potassium perchlorate decompose into potassium chlorate and oxygen.
a) KClO4.
K (potassium) is 1A element, OSN +1,
Oxygen in the compound has OSN -2.
Let OSN of Cl be x
1+ x+(4*-2)=0 (since compound is neutral)
Solve the equation. X is equal to +7.
b)Potassium perchlorate is the reactant, Potassium chloride and Oxygen are the products.
Reactants -----> Products
c) Check out change of OSN of elements.
in KCl, OSN number of K is +1, Cl is -1. So Cl+7---->Cl-1. It takes 8 electrons. It was reduced.
in KClO4, OSN number of O is -2, in O2 it becomes O. So O-2---->O. It gives electrons. It was oxidized.
Shortly,
LOSS OF ELECTRONS=OXIDIZED
GAIN OF ELECTRONS=REDUCED
The statements that are true regarding a catalyst are as follows:
It undergoes no permanent change during a reaction
It does not participate in a chemical reaction
What is a catalyst?
A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the process.
A catalyst increases the rate at which a chemical reaction occurs by decreasing the activation energy of the reaction.
Since a catalyst is not used up in the reaction, the following statement applies:
It undergoes no permanent change during a reaction
It does not participate in a chemical reaction
Learn more about catalyst at:
#SPJ1
Answer:
From gas laws (pressure law and Boyles law), the pressure exerted by a gas depends on Temperature of the gas and volume of the container.
Explanation:
• P → Pressure exerted by the gas.
• T → Temperature of the gas.
• V → Volume of the container.
• from the expression, pressure exerted by the gas is directly proportional to temperature of the gas and inversely proportional to the volume of the container.
Answer:
Number 1 is false
Explanation:
1.- Charged molecules are insoluble in water. The statement is false, charged molecules are soluble in water and this is the main characteristic of charged molecules.
2.- Hydrophobic molecules tend to be nonpolar and hydrophilic molecules tend to be polar. This is true, hydrophobic molecules are non polar and hydrophilic molecules are polar.
3.- Gasoline has partial positive and negative charges which allows it to dissolve in water. This is true, gasoline is soluble in water.
4.- NH3 dissolves in water because its molecules form hydrogen bonds with water. This is true, NH3 interact with water because they can form hydrogen bonds.
5.- Hydrophobic molecules prefer to interact with each other in an aqueous solution. Absolutely true, hydrophobic molecules interact among them and they isolate from the environment.
1) We need to convert 12.0 g of H2 into moles of H2, and <span> 74.5 grams of CO into moles of CO
</span><span>Molar mass of H2: M(H2) = 2*1.0= 2.0 g/mol
Molar mass of CO: M(CO) = 12.0 +16.0 = 28.0 g/mol
</span>12.0 g H2 * 1 mol/2.0 g = 6.0 mol H2
74.5 g CO * 1 mol/28.0 g = 2.66 mol CO
<span>2) Now we can use reaction to find out what substance will react completely, and what will be leftover.
CO + 2H2 -------> CH3OH
1 mol 2 mol
given 2.66 mol 6 mol (excess)
How much
we need CO? 3 mol 6 mol
We see that H2 will be leftover, because for 6 moles H2 we need 3 moles CO, but we have only 2.66 mol CO.
So, CO will react completely, and we are going to use CO to find the mass of CH3OH.
3) </span>CO + 2H2 -------> CH3OH
1 mol 1 mol
2.66 mol 2.66 mol
4) We have 2.66 mol CH3OH
Molar mass CH3OH : M(CH3OH) = 12.0 + 4*1.0 + 16.0 = 32.0 g/mol
2.66 mol CH3OH * 32.0 g CH3OH/ 1 mol CH3OH = 85.12 g CH3OH
<span>
Answer is </span>D) 85.12 grams.
