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sergiy2304 [10]
3 years ago
13

Determine the number of molecules present in 1.53 mol of carbon monoxide

Chemistry
1 answer:
ASHA 777 [7]3 years ago
8 0

Answer:

hi

Explanation:

this is my first time on this app so what do i do

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Which set of numbers gives the correct possible values of l for n = 2
Paha777 [63]
HEY THERE!!

For a given principal quantum number or n, the corresponding angular quantum number or l is equivalent to a range between 0 and (n-1) .

This means that the angular quantum number for a principal quantum number of 2 is equivalent to:

l = 0 -> (n-1) = 0 -> (2-1) = 0 -> 1

So the answer is 0, 1

HOPE IT HELPED YOU.
3 0
3 years ago
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Pentaoxygen monochloride formula​
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Answer:

Explanation:

O5Cl

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Choose the number of significant figures indicated. 90
kvv77 [185]
There are 2 significant figures. All numbers in a whole number are significant.
3 0
3 years ago
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Q.2. Which of the following statements is true of second ionization energies?
Ratling [72]

Answer:

a) That of Al is higher than that of Mg because Mg wants to lose the second electron, so it is easier to take the second electron away

5 0
2 years ago
Given that Delta.G for the reaction below is –957.9 kJ, what is Delta.Gf of H2O?
Alexxandr [17]

Answer:

6ΔG°(f) H₂O = -229 Kj/mol

Explanation:

                    4NH₃(g)          +      5O₂(g)       =>        4NO(g)           +     6H₂O(g)

ΔG°(f) 4mol(-16.66Kj/mol) | 5mol(0Kj/mol) || 4mol(+86.71Kj/mol) | 6ΔG°(f) H₂O

Hess's Law

ΔG°(Rxn) = ∑ΔG°(f) Products - ∑ΔG°(f) Reactants

-957.9 Kj = [(4mol(+86.71Kj/mol)) + 6ΔG°(f) H₂O(g)] - [4mol(-16.66Kj/mol) + 5mol(0Kj/mol)]

-957.9 Kj = [4(86.7)Kj + 6ΔG°(f) H₂O] - [4(-16.66)Kj] = 346.84Kj + 6ΔG°(f) H₂O + 66.64Kj

ΔG°(f) H₂O = ((-957.9 - 346.84 -66.64)/6)Kj =  -228.56 Kj ≅ -228.6 Kj*

*Verified with Standard Heat of Formation Table

8 0
3 years ago
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