All categories

3 years ago

Why do lone pairs of electrons repel other electrons pairs farther away than bonding pairs of electrons do?

1 answer:

6 0

Because a bonding electron pair is involved in a sigma bond with another atom. Hence is at a greater distance from the nucleus of the central atom than a non bonding pair

You might be interested in

Tracy observes an organism that lives in a lake by her house. The organism has an internal skeleton, scales, fins, and gills. Th

andrey2020 [161]

I'd say B. Its most likely a fish.

8 0

3 years ago

Read 2 more answers

Which of the following statements are FALSE of models? (select all that applies)

ira [324]

Answer:

A B and C

Explanation:

Hopefully this helps

7 0

3 years ago

Which instrument is most often used to measure acid volume before a titration begins?

Darya [45]

Good Morning!

To measure the volume of an acid we use in the laboratory an instrument called burette (a). This instrument is used to calculate the volumes of chemical solutions. At first glance, these tools may be confused with syringes, but have different functions.

Hugs!

4 0

3 years ago

Read 2 more answers

How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2? Express your answer numerically in moles. View Availabl

VladimirAG [237]

Answer:

A) 8.00 mol NH₃

B) 137 g NH₃

C) 2.30 g H₂

D) 1.53 x 10²⁰ molecules NH₃

Explanation:

Let us consider the balanced equation:

N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)

Part A

3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:

$12.0molH_{2}.\frac{2molNH_{3}}{3molH_{2}} =8.00molNH_{3}$

Part B:

1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:

$4.04molN_{2}.\frac{2molNH_{3}}{1molN_{2}} .\frac{17.0gNH_{3}}{1molNH_{3}} =137gNH_{3}$

Part C:

According to the balanced equation 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:

$13.02gNH_{3}.\frac{6.00gH_{2}}{34.0gNH_{3}} =2.30gH_{2}$

Part D:

6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:

$7.62 \times 10^{-4} gH_{2}.\frac{2molNH_{3}}{6.00gH_{2}} .\frac{6.02\times 10^{23}moleculesNH_{3} }{1molNH_{3}}=1.53\times10^{20}moleculesNH_{3}$

3 0

3 years ago

When the amount of oxygen is limited, carbon and oxygen react to form carbon monoxide. How many grams of CO can be formed from 3

Alinara [238K]

61.25 grams of CO can be formed from 35 grams of oxygen.

The molecular mass of oxygen is 16 gmol⁻¹

The molecular mass of carbon monoxide is 28 gmol⁻¹

Explanation:

The molar mass of carbon monoxide is molar mass of C added to that of O;

12 + 16 = 28

= 28g/mol

The molar mass of oxygen is 16 g/mol while that of oxygen gas (O₂) is 32 g/mol

Since the ration oxygen to carbon monoxide is 1: 2 moles, we begin to find out how many moles of carbon monoxide are formed by 35 g of oxygen;

35/32 * 2

= 70/32 moles

Then multiply by the molar mass of carbon monoxide;

70/32 * 28

= 61.25 g

4 0

3 years ago