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4 years ago

What is the hybridization of the nitrogen atom in ammonia?

1 answer:

6 0

The nitrogen atom in NH3 is sp3 hybridized. ... In NH3 molecule, three of the sp3 orbitals are used to formbonds to the three hydrogen atoms and the fourth sp3 orbital is used to hold the lone pair. and is sometimes called a non-bonding pair.

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I REALLY NEED THE HELP

timama [110]

Answer:4

Explanation:number 4 is the right answer :)

7 0

2 years ago

A calorimeter contains 72.0 g of water at 19.2 oC. A 141 g piece of metal is heated to 89.0 oC and dropped into the water. The e

kumpel [21]

Answer:

The specific heat of the metal is 0.212 J/(g°C).

Explanation:

We can calculate the specific heat of the metal by the following equilibrium:

$q_{a} = -q_{b}$

$m_{a}C_{a}\Delta T_{a} = -m_{b}C_{b}\Delta T_{b}$

$m_{a}C_{a}(T_{f_{a}} - T_{i_{a}}) = -m_{b}C_{b}(T_{f_{b}} - T_{i_{b}})$

In the above equation, we have that the heat loses by the metal (b) is gained by the water (a).

$m_{a}$ : is the water's mass = 72.0 g

$C_{a}$ : is the specific heat of water = 4.184 J/(g°C)

$T_{i_{a}}$ : is the initial temperature of the water = 19.2 °C

$T_{f_{a}}$ : is the final temperature of the water = 25.5 °C

$m_{b}$ : is the metal's mass = 141 g

$C_{b}$ : is the specific heat of metal =?

$T_{i_{b}}$ : is the initial temperature of the metal = 89.0 °C

$T_{f_{b}}$ : is the final temperature of the water = 25.5 °C

$m_{a}C_{a}(T_{f_{a}} - T_{i_{a}}) = -m_{b}C_{b}(T_{f_{b}} - T_{i_{b}})$

$72.0 g*4.184 J/(g^{\circ} C)*(25.5 ^{\circ} C - 19.2 ^{\circ} C) = -141 g*C_{b}*(25.5 ^{\circ} C - 89.0 ^{\circ} C)$

$C_{b} = -\frac{72.0 g*4.184 J/(g^{\circ} C)(25.5 ^{\circ} C - 19.2 ^{\circ} C)}{141 g(25.5 ^{\circ} C - 89.0 ^{\circ} C)} = 0.212 J/(g^{\circ} C)$

Therefore, the specific heat of the metal is 0.212 J/(g°C).

I hope it helps you!

7 0

3 years ago

Anastasy [175]

Answer:

I think it's letter B

Explanation:

Hope it helps correct if I'm wrong

7 0

3 years ago

I need help!!!! Plz help me with the above question

maksim [4K]

The answer will be bar graph

4 0

4 years ago

What is the mass of an object that has a total heat loss of 550. cal, a specific heat of

Galina-37 [17]

Answer:

the mass is .587 J/g°C

Explanation:

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4 0

3 years ago