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zalisa [80]
3 years ago
10

View Available Hint(s) Check all that apply. Hydrogen bonding occurs when a hydrogen atom is covalently bonded to an NN, OO, or

FF atom. The CH4CH4 molecule exhibits hydrogen bonding. A hydrogen atom acquires a partial positive charge when it is covalently bonded to an FF atom. A hydrogen bond is possible with only certain hydrogen-containing compounds. A hydrogen bond is equivalent to a covalent bond.
Chemistry
1 answer:
Sauron [17]3 years ago
7 0

Answer:

Hydrogen bonding occurs when a hydrogen atom is covalently bonded to an NN, OO, or FF atom.

A hydrogen atom acquires a partial positive charge when it is covalently bonded to an FF atom.

A hydrogen bond is possible with only certain hydrogen-containing compounds.

Explanation:

A hydrogen bond does not occur in all hydrogen containing compounds. Hydrogen bonds only occur in those compounds where hydrogen is bonded to a highly electronegative element such as fluorine, oxygen or nitrogen.

In a hydrogen bonded specie, hydrogen acquires a partial positive charge and the electronegative element acquires a partial negative charge which extends throughout the molecule.

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Why is the chemical symbol for calcium ca
Assoli18 [71]
You are right, it's CA Calcium, 40.08, Group 2 and Row 4.
6 0
3 years ago
A 40.2 g sample of a metal heated to 99.3°C is placed into a calorimeter containing 120 g of water at 21.8°C. The final temper
aliina [53]

Answer:

B) Iron (c=0.45 J/g°C)

Explanation:

Given that:-

Heat gain by water = Heat lost by metal

Thus,  

m_{water}\times C_{water}\times (T_f-T_i)=-m_{metal}\times C_{metal}\times (T_f-T_i)

Where, negative sign signifies heat loss

Or,  

m_{water}\times C_{water}\times (T_f-T_i)=m_{metal}\times C_{metal}\times (T_i-T_f)

For water:

Mass = 120 g

Initial temperature = 21.8 °C

Final temperature = 24.5 °C

Specific heat of water = 4.184 J/g°C

For metal:

Mass = 40.2 g

Initial temperature = 99.3 °C

Final temperature = 24.5 °C

Specific heat of metal = ?

So,  

120\times 4.184\times (24.5-21.8)=40.2\times C_{metal}\times (99.3-24.5)

40.2C_{metal}\left(99.3-24.5\right)=120\times \:2.7\times \:4.184

40.2C_{metal}\left(99.3-24.5\right)=1355.616

C_{metal}=0.45\ J/g^0C

<u>This value corresponds to iron. Thus answer is B.</u>

3 0
3 years ago
What is the de Broglie wavelength (in meters) of a 45-g golf ball traveling at 72 m/s?
Cerrena [4.2K]

Answer: The de broglie wavelength is 2.037 \times 10^{-34} m.

Explanation:

Calculate  \lambda = \frac{h}{p}as follows.

          \lambda = \frac{h}{p}

where,

          h = plank's constant = 6.6 \times 10^{-34} m^{2} kg/s

         p = momentum = mass \times velocity

Putting the values in the formula as follows.

        \lambda = \frac{h}{mass \times velocity}

                               =  \frac {6.6 \times 10^{-34} m^{2} kg/s}{0.045 kg \times 72 m/s}                        

                               =  2.037 \times 10^{-34} m

Thus, the de broglie wavelength is 2.037 \times 10^{-34} m.

                               

6 0
3 years ago
Read 2 more answers
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zloy xaker [14]

Answer:

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Explanation:

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4 0
3 years ago
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A sample of ammonia gas at 75°c and 445 mm hg has a volume of 16.0 l. what volume will it occupy if the pressure rises to 1225 m
ioda
In this kind of exercises, you should  use the "ideal gas" rules: PV = nRT
P should be in Pascal: 
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V should be in cubic meter:
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==> P1 * V1 = P2 * V2
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V2 = 0.00581 m3 = 5.81 L


7 0
3 years ago
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