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alexgriva [62]
2 years ago
9

Whats the difference between a bonding electron domain and a nonbonding electron domain?​

Chemistry
1 answer:
Alika [10]2 years ago
3 0

Answer:  lone pair of electrons

Explanation: A bonding electron domain is a pair of electrons shared between two atoms and a nonbonding is a lone pair of electrons

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assuming nitrogen behaves like an ideal gas, what volume would 14.0 g of nitrogen gas (N2) occupy at STP? the gas constant is 0.
dimaraw [331]

Answer:

V = 22.41 L

Explanation:

Given data:

Mass of nitrogen = 14.0 g

Volume of gas at STP = ?

Gas constant = 0.0821 atm.L/mol.K

Solution:

Number of moles of gas:

Number of moles = mass/molar mass

Number of moles= 14 g/ 14 g/mol

Number of moles = 1 mol

Volume of gas:

PV = nRT

1 atm × V = 1 mol × 0.0821 atm.L/mol.K  × 273 K

V = 22.41 atm.L / 1 atm

V = 22.41 L

4 0
3 years ago
How many molecules of XeF6 are formed from 12.9 L of F2 (at 298 K and 2.6 atm) according to 11) the following reaction? Assume t
ddd [48]

Answer:

#Molecules XeF₆ = 2.75 x 10²³ molecules XeF₆.

Explanation:

Given … Excess Xe + 12.9L F₂ @298K & 2.6Atm => ? molecules XeF₆

1. Convert 12.9L 298K & 2.6Atm to STP conditions so 22.4L/mole can be used to determine moles of F₂ used.

=> V(F₂ @ STP) = 12.6L(273K/298K)(2.6Atm/1.0Atm) = 30.7L F₂ @ STP

2. Calculate moles of F₂ used

=> moles F₂ = 30.7L/22.4L/mole = 1.372 mole F₂ used

3. Calculate moles of XeF₆ produced from reaction ratios …

Xe + 3F₂ => XeF₆ => moles of XeF₆ = ⅓(moles F₂) = ⅓(1.372) moles XeF₆ = 0.4572 mole XeF₆

4. Calculate number molecules XeF₆ by multiplying by Avogadro’s Number  (6.02 x 10²³ molecules/mole)

=> #Molecules XeF₆ = 0.4572mole(6.02 x 10²³ molecules/mole)

                                  = 2.75 x 10²³ molecules XeF₆.

8 0
2 years ago
(4) Calculate the % of a compound that can be removed from liquid phase 1 by using ONE to FOUR extractions with a liquid phase 2
maksim [4K]

Answer:

One extraction: 50%

Two extractions: 75%

Three extractions: 87.5%

Four extractions: 93.75%

Explanation:

The following equation relates the fraction q of the compound left in volume V₁ of phase 1 that is extracted n times with volume V₂.

qⁿ = (V₁/(V₁ + KV₂))ⁿ

We also know that V₂ = 1/2(V₁) and K = 2, so these expressions can be substituted into the above equation:

qⁿ = (V₁/(V₁ + 2(1/2V₁))ⁿ = (V₁/(V₁ + V₁))ⁿ =  (V₁/(2V₁))ⁿ = (1/2)ⁿ

When n = 1, q = 1/2, so the fraction removed from phase 1 is also 1/2, or 50%.

When n = 2, q = (1/2)² = 1/4, so the fraction removed from phase 1 is (1 - 1/4) = 3/4 or 75%.

When n = 3, q = (1/2)³ = 1/8, so the fraction removed from phase 1 is (1 - 1/8) = 7/8 or 87.5%.

When n = 4, q = (1/2)⁴ = 1/16, so the fraction removed from phase 1 is (1 - 1/16) = 15/16 or 93.75%.

5 0
3 years ago
Using aluminium as an example, describe the key properties of p-block metals.
olga nikolaevna [1]

Answer:

P-block metals have classic metal characteristics like they are shiny, they are good conductors of heat and electricity, and they lose electrons easily. These metals have high melting points and readily react with nonmetals to form ionic compounds.

Explanation:

6 0
2 years ago
Read 2 more answers
Predict the whether the following reactions are possible or not.
vova2212 [387]

Answer:

1. No

2. No

3. Yes

4. No

5. Yes

6. No

Explanation:

Use the reactivity series. If the element is above it, it is more reactive. More reactive elements can displace less reactive elements from their compounds.

5 0
3 years ago
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