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Olin [163]
3 years ago
9

The molecule hydrogen fluoride (HF) contains a polar bond H - F, where fluorine is more electronegitivexpensive than Hydrogen. W

hich is the appropriate representation of the H - F bond?
Chemistry
1 answer:
irina1246 [14]3 years ago
7 0
<span>Hydrogen's one electron will see that fluoride has seven electrons on its valence shell and will want to fill that eight slot to create the more stable compound Hydrogen fluoride. The flouride atom will have eight electrons orbiting its valence shell while hydrogen will have two electrons.</span>
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What two properties of a gas depend on its container
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6 0
3 years ago
What reaction will take place if h2o is added to a mixture of nanh2/nh3? draw the products of the reaction (without sodium ion)?
Maru [420]

When sodium amide i.e.NaNH_{2} reacts with water i.e. H_{2}O results in the formation of sodium hydroxide i.e. NaOH and ammonia NH_{3}.

The chemical reaction is given by:

NaNH_{2}+H_{2}O\rightarrow NH_{3}+NaOH

Now, when ammonia i.e.NH_{3} reacts with water results in the formation of ammonium hydroxide i.e. NH_{4}OH

The chemical reaction is given by:

NH_{3}+H_{2}O\rightarrow NH_4OH

Thus, the products of the above reactions are ammonia and ammonium hydroxide (without sodium ion).

The structures of the products are shown in figure (1): ammonium hydroxide and figure (2) ammonia.


8 0
3 years ago
ASAP I AM GIVING BRAINLIEST PLEASEEEEEEEEE HELPPPPPPPPPPPPPPPPPP
zepelin [54]

Answer:

B?

Explanation:

In the example, the amount of hydrogen is 202,650 x 0.025 / 293.15 x 8.314472 = 2.078 moles. Use the mass of the hydrogen gas to calculate the gas moles directly; divide the hydrogen weight by its molar mass of 2 g/mole. For example, 250 grams (g) of the hydrogen gas corresponds to 250 g / 2 g/mole = 125 moles.

7 0
3 years ago
3. How much energy is needed to raise 45 grams of water from 40°C to 115 °C?
Dafna1 [17]

Answer:

Q = 114349.5 J

Explanation:

Hello there!

In this case, since this a problem in which we need to calculate the total heat of the described process, it turns out convenient to calculate it in three steps; the first one, associated to the heating of the liquid water from 40 °C to 100 °C, next the vaporization of liquid water to steam at constant 100 °C and finally the heating of steam from 100 °C to 115 °C. In such a way, we calculate each heat as shown below:

Q_1=45g*4.18\frac{J}{g\°C}*(100\°C-40\°C)=11286J\\\\Q_2=45g* 2260 \frac{J}{g} =101700J\\\\Q_3=45*2.02\frac{J}{g\°C}*(115\°C-100\°C)=1363.5J

Thus, the total energy turns out to be:

Q_T=11286J+101700J+1363.5J\\\\Q_T=114349.5J

Best regards!

5 0
3 years ago
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