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r-ruslan [8.4K]
2 years ago
13

I can't write one song that's not about you Can't drink without thinkin' about you

Chemistry
1 answer:
Tems11 [23]2 years ago
3 0

Answer:

ok.....................

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How many fluorine atoms bond with calcium to form calcium fluoride? one two three four five
Elena-2011 [213]
Calcium fluoride.
Ca is metal, F is non-metal, so they form ionic bond.
Ca as metal can form only positive ion. Ca in the second group, so the charge of Ca ion is 2+.   Ca²⁺
F is in the 17th group, so it has 7 electrons on the last level. It is non-metal, non-metal, so it has negative charge -(8-7)=-1. "8" because on the last level cannot be more than 8 electrons. F-ion is F¹⁻.

Ca²⁺  F¹⁻
Number of positive charges should be equal to number of negative charges,
Formula of calcium fluoride
CaF2.
2 atoms Fluorine bond with Calcium.
7 0
3 years ago
How many moles of oxygen gas are required to form 55g of carbon monoxide gas
marta [7]

Answer:

8.33 moles CO2 X. 25mol O2. 16mol CO2. = 13.0 moles

6 0
3 years ago
What is a combination of two or more substances that are all solids
Sophie [7]

Answer:

Solids cant react to each other because they are the same room temperature

7 0
3 years ago
Keq for the reaction below is 2400. If the initial conditions of the reaction are a 1.0 L flask that contains 0.024 mol NO (g),
podryga [215]

Answer:

The answer to your question is it is not at equilibrium, it will move to the products.

Explanation:

Data

Keq = 2400

Volume = 1 L

moles of NO = 0.024

moles of N₂ = 2

moles of O₂ = 2.6

Process

1.- Determine the concentration of reactants and products

[NO] = 0.024 / 1 = 0.024

[N₂] = 2/1 = 2

[O₂] = 2.6/ 1= 2.6

2.- Balanced chemical reaction

                     N₂ + O₂    ⇒   2NO

3.- Write the equation for the equilibrium of this reaction

                     Keq = [NO]²/[N₂][O₂]

- Substitution

                    Keq = [0.024]² / [2][2.6]

-Simplification

                    Keq = 0.000576 / 5.2

-Result

                    Keq = 1.11 x 10⁻⁴

Conclusion

It is not at equilibrium, it will move to the products because the experimental Keq was lower than the Keq theoretical-

                         1.11 x 10⁻⁴ < 2400

7 0
3 years ago
The atomic weight of iodine is less than the atomic weight of tellurium. However, Mendeleev listed iodine after tellurium in his
QveST [7]

Explanation :

As we know that Mendeleev arranged the elements in horizontal rows and vertical columns of a table in order of their increasing relative atomic weights.

He placed the elements with similar nature in the same group.

According to the question, the atomic weight of iodine is less than the atomic weight of tellurium. So according to this, iodine should be placed before tellurium in Mendeleev's tables. But Mendeleev placed iodine after tellurium in his original periodic table.

However, iodine has similar chemical properties to chlorine and bromine. So, in order to make iodine queue up with chlorine and bromine in his periodic table, Mendeleev exchanged the positions of iodine and tellurium.

As we know that the positions of iodine and tellurium were reversed in Mendeleev's table because iodine has one naturally occurring isotope that is iodine-127  and tellurium isotopes are tellurium-128 and tellurium-130.

Due to high relative abundance of tellurium isotopes gives tellurium the greater relative atomic mass.

3 0
3 years ago
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