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Softa [21]
3 years ago
13

Please hep its my last question

Chemistry
1 answer:
atroni [7]3 years ago
7 0

Answer:Noble gases:

 are highly reactive.

 react only with other gases.

 do not appear in the periodic table.

 are not very reactive with other elements.

Explanation:Noble gases:

 are highly reactive.

 react only with other gases.

 do not appear in the periodic table.

 are not very reactive with other elements.

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PLEASE HELP ME
pishuonlain [190]

Answer:

Explanation:

Li(s)+1/2 F₂(g)→LiF(s)

In this reaction one mole of product is formed from gaseous state of reactants . So this reaction will represent the reaction for which ΔH∘rxn equals to ΔH∘f .

In other reaction one mole is not produced if we balance the reaction. The reactant must be in gaseous state .  ΔH∘f . represents heat of formation .  ΔH∘rxn represents heat of reaction .

3 0
3 years ago
How to find the oxidation number of an element
Virty [35]
The thing you MUST do FIRST is look for any H's, O's, or F's in the equation

1)any element just by itself not in a compound, their oxidation number is 0
ex: H2's oxidation number is 0
ex: Ag: oxidation number is 0 if its just something like Ag + BLA = LALA

2) the oxidation number of H is always +1, unless its just by itself (see #1)
3) the oxidation number of O is always -2, unless its just by itself (see #1)
4) the oxidation number of F is always -1, unless its just by itself (see#1)


ok so after you have written those oxidation numbers in rules 1-4 over each H, F, or O atom in the compound, you can look at the elements that we havent talked about yet

for example::::
N2O4

the oxidation number of O is -2.

since there are 4 O's, the charge is -8. now remember that N2O4 has to be neutral so the N2 must have a charge of +8
+8 divided by 2 = +4

N has an oxidation number of +4.

more rules:
5) the sum of oxidation numbers in a compound add up to 0 (when multiplied by the subscripts!!!) (see above example)
6) the sum of oxidation numbers in a polyatomic ion is the charge (for example, PO4 has a charge of (-3) so

oxidation # of O = -2. (there are 4 O's = -8 charge on that side ) P must have an oxidation number of 5. (-8+5= -3), and -3 is the total charge of the polyatomic ion
3 0
4 years ago
If 21.42g of KMnO4 is actually produced what is the percent yield
mrs_skeptik [129]

Answer:

Percentage yield  = 6.776%

Explanation:

Data Given:

Actual yield of KMnO₄ = 21.42g

Percentage Yield = ?

Formula Used to find Percent yield

                 Percentage yield = Actual yield/ theoretical yield x 100        (1)

For this Pupose First step is

We have to know the theoretical yield KMnO₄

Potasium permagnate form from MnO₄ and KOH in the presence of Oxygen by heating, in 1st step in second 2K₂MnO₄ react with HCl and give KMnO₄ .

The Reactions of formation of KMnO₄

1st Step                      

2MnO₄ + 4KOH + O₂  ------------> 2K₂MnO₄ + 2H₂O

2nd Step

2K₂MnO₄ + 4HCl ------------->2 KMnO₄ + H₂O + 4KCl

So form the above equation we come to know it produced 2 Mole of KMnO₄

Now we will calculate the mass of KMnO₄ by mass formulae

     mass of KMnO₄ = Number of mole of KMnO₄ x Molar Mass of KMnO₄

Molar Mass of KMnO₄ = 158.034g /mol

Put value in Mass Formula

 mass of KMnO₄ = 2 mol x  158.034g/mol

mass of KMnO₄ = 316.1 g

So the theoratical yield per standard reaction = 316.1 g

Now put all values in equation 1

           Percentage yield = Actual yield/ theoretical yield x 100  

           Percentage yield  = 21.42g / 316.1 g x 100

          Percentage yield =  0.0678 x 100

          Percentage yield  = 6.776%

4 0
3 years ago
The formation of nitryl fluoride, 2no2(g)+f2(g)⇌2no2f(g) is first order in no2 and first order in f2. if the concentration of no
ra1l [238]
The reaction is:

2 NO₂ (g) + F₂ (g) ⇆  2 NO₂F (g)

The stoichiometric coefficients of the substances balance out each other to obey the Law of Definite Proportions. Now, you have to note that determining the reaction rate expression is specific to a certain type of reaction. So, this are determined empirically through doing experiments. But in chemical reaction engineering, to make things simple, you assume that the reaction is elementary. This means that the order of a reaction with respect to a certain substance follows their individual stoichiometric coefficients. What I'm saying is, the stoichiometric coefficients are the basis of our reaction rate orders. For this reaction, the rate order is 2 for NO₂, 1 for F₂ and 2 for NO₂F. When the forward and reverse reactions are in equilibrium, then it applies that:

Reaction rate of disappearance of reactants = Reaction rate of formation of products.

Therefore, we can have two reaction rate constants for this. But since the conditions manipulated are the reactant side, let's find the expression for reaction rate of disappearance of reactants.

-r = k[NO₂]²[F₂]

The negative sign before r signifies the rate of disappearance. If it were in terms of the product, that would have been positive. The term k denotes for the reaction rate constant. That is also empirical. As you can notice the stoichiometric coefficients are exponents of the concentrations of the reactants. Let's say initially, there are 1 M of NO₂ and 1 M of F₂. Then,

-r = k(1)²(1)
-r = k

Now, if we change 1 M of NO₂ by increasing it to its half, it would now be 1.5 M NO₂. Then, if we quadruple the concentration of F₂, that would be 4 M F₂. Substituting the values:

-r = k(1.5)²(4)
-r = 9k

So, as you can see the reaction rate increase by a factor of 9.
4 0
4 years ago
Read the descriptions below of two substances and an experiment on each. Decide whether the result of the experiment tells you t
Lelu [443]

Answer:

Sample A - mixture

Sample B- Mixture

Explanation:

Looking at sample A, we can see that as water was poured over sample A, the sample was separated into its components as the powder disappeared leaving behind a gritty black sand-like material on the surface of the paper. A separation of the mixture has taken place.

In sample B, we can clearly see that it is a mixture because the amount of solid recovered is much less than the total mass of the solid put into the beaker. The sample must have been separated into its components.

8 0
3 years ago
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