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3 years ago

How many lone pairs of electrons are represented in the Lewis structure of a phosphate ion (PO43-)? 10 12 21 24

1 answer:

6 0

To illustrate the Lewis structure,
P has 5 valence electrons
O has 6 valence electrons (each for 4 oxygen)
And finally, for every negative charge, there is an additional valence electron

We should add these all up = 5 + 24 + 3 = 32 valence electrons

With this, we can be guided to illustrate the lewis structure as P as central atom and the 3 oxygen each with a single bond with P and 1 oxygen with a double bond with P. We place the valence electrons until octet rule is satisfied,
we will be left with 12 lone pairs for phosphate ion.

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3 years ago

At 506 K and 2.05 atm, if 6.00 grams of H2 will reacts with excess N2 what volume in liters will be produced of NH3? (R = 0.0820

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Answer:

90.4 liter

Explanation:

You must convert the 6.00 grams of H₂ into number of moles, and then use the stoichiometry of the reaction to find the number of moles of NH₃, which can be converted into volume using the ideal gas equation.

1. Number of moles of H₂

number of moles = mass in grams / molar mass

molar mass of H₂ = 2.016g/mol

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2. Number of moles of NH₃

i) Chemical equation:

3H₂(g) + N₂(g) → 2NH₃(g)

ii) Mole ratio:

3 mol H₂ : 2 mol NH₃

iii) Proportion:

x mol NH₃ / 2.97619mol H₂ = 3 mol NH₃ / 2 mol H₂

x = 4.4642857mol NH₃

3. Volume of NH₃

pV = nRT

V = nRT/p

V = 4.4642857mol × 0.08206 atm·liter/(K·mol) × 506K / 2.05atm

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3 years ago

Determine the number of moles of Mn and the number of moles of X in compound A. Clearly distinguish which is which in your calcu

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3 years ago

In a heat engine of 1000 j of heat enters the system and the piston does 500 j of work what is the final internal energy of the

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Answer : The final energy of the system if the initial energy was 2000 J is, 3500 J

Solution :

(1) The equation used is,

$\Delta U=q+w\\\\U_{final}-U_{initial}=q+w$

where,

$U_{final}$ = final internal energy

$U_{initial}$ = initial internal energy

q = heat energy

w = work done

(2) The known variables are, q, w and $U_{initial}$

initial internal energy = $U_{initial}$ = 2000 J

heat energy = q = 1000 J

work done = w = 500 J

(3) Now plug the numbers into the equation, we get

$U_{final}-(2000J)=(1000J)+(500J)$

(4) By solving the terms, we get

$U_{final}-(2000J)=(1000J)+(500J)$

$U_{final}-(2000J)=1500J$

$U_{final}=2000J+1500J$

$U_{final}=3500J$

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4 years ago

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