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Anna11 [10]
3 years ago
9

Which is the Lewis structure for H3PO4? An upper P is single bonded above to an O, and to the left, right, and below to an O sin

gle bonded to an H. The O above the P has three pairs of dots to the left, above, and below; the O's to the sides have pairs of dots above and below, and the O below the P has pairs of dots right and left. A central upper P is single bonded left, right, above, and below to upper Os. The O above the P is single bonded to upper H on the left and the right, and has two electron dots above it. The O below the P is single bonded to an H below, and has pairs of electron dots to the left and right. A central upper P is double bonded to an O above, and single-bonded to an upper O single-bonded to an upper H to the left and the right. The O above the P has three pairs of electron dots, to the left, above, and to the right; the O's to the right and left have pairs of dots above and below. A central upper P is bonded to an upper H above, an upper O below, and upper O's bonded to upper H's to the left and the right. The O below the P has three pairs of electron dots, to the left, right, and below; the other two O's have pairs of dots above and below. A central upper P is double bonded to an O above, and single-bonded to an upper O single-bonded to an upper H to the left and the right. The O above the P has three pairs of electron dots, to the left, above, and to the right; the O's to the right and left have pairs of dots above and below.
Chemistry
2 answers:
Kamila [148]3 years ago
8 0

Answer:

1st part is B.) sigma bond and C represents two electrons

2nd part is A.) nonbonding electrons

3rd part is B.) bonding electrons and C.) sigma bond

Explanation:

Got it right on edg. 2020

Margaret [11]3 years ago
3 0

Answer:

It is A.

Explanation:

I took the test.

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A 1.00 liter container holds a mixture of 0.52 mg of He and 2.05 mg of Ne at 25oC. Determine the partial pressures of He and Ne
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Answer:

pHe = 3.2 × 10⁻³ atm

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Explanation:

Given data

Volume = 1.00 L

Temperature = 25°C + 273 = 298 K

mHe = 0.52 mg = 0.52 × 10⁻³ g

mNe = 2.05 mg = 2.05 × 10⁻³ g

The molar mass of He is 4.00 g/mol. The moles of He are:

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We can find the partial pressure of He using the ideal gas equation.

P × V = n × R × T

P × 1.00 L = 1.3 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K

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The molar mass of Ne is 20.18 g/mol. The moles of Ne are:

2.05 × 10⁻³ g × (1 mol / 20.18 g) = 1.02 × 10⁻⁴ mol

We can find the partial pressure of Ne using the ideal gas equation.

P × V = n × R × T

P × 1.00 L = 1.02 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K

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The total pressure is the sum of the partial pressures.

P = 3.2 × 10⁻³ atm + 2.5 × 10⁻³ atm = 5.7 × 10⁻³ atm

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