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azamat
3 years ago
6

Identify the intermolecular forces present in each of these substances. hcl, ch4

Chemistry
1 answer:
Amiraneli [1.4K]3 years ago
8 0
<span>Intermolecular Forces present in HCl:
                                                           The Electronegativity difference between Chlorine and Hydrogen is 0.96 showing that the bond is polar covalent in nature. The Hydrogen atom is partially positive and Chlorine is partially positive making the molecule Dipole. Hence, the Intermolecular forces present in HCl are Dipole-Dipole Interactions.

</span>Intermolecular Forces present in CH₄:
                                                                The Electronegativity difference between Chlorine and Hydrogen is 0.35 showing that the bond is non-polar covalent in nature. Hence, the Intermolecular forces present in CH₄ are London Dispersion Forces.
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N2 reacts with hydrogen gas according to the following equation:
slega [8]

Answer:

Mass = 51 g

Explanation:

Given data:

Mass of nitrogen = 41.93 g

Mass of ammonia formed = ?

Solution:

Chemical equation:

N₂ + 3H₂      →       2NH₃

Number of moles of nitrogen:

Number of moles = mass/molar mass

Number of moles = 41.93 g/ 28 g/mol

Number of moles = 1.5 mol

now we will compare the moles of nitrogen and ammonia.

                N₂          :           NH₃

                  1          :           2

                1.5         :         2/1×1.5 = 3 mol

Mass of ammonia formed:

Mass = number of moles × molar mass

Mass = 3 mol × 17 g/mol

Mass = 51 g

6 0
3 years ago
What mass of solid lead would displace exactly 234.6 liters of water?
eimsori [14]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

The mass of solid lead would displace exactly 234.6 liters of water should be <span>2,674,440</span>
6 0
3 years ago
Read 2 more answers
Determine the AMOUNT OF NO2, LIMITING REACTANT, AND THE AMOUNT AND NAME OF EXCESS REACTANT.
zepelin [54]

Answer:

balanced equation mole ratio 5 2 mol NO/1 mol O2

10.00 g O2 3 1 mol O2/32.00 g O2 5 0.3125 mol O2

20.00 g NO 3 1 mol NO/30.01 g NO 5 0.6664 mol NO

actual mole ratio 5 0.6664 mol NO/0.3125 mol O2 5 2.132 mol NO/1.000 mol O2

Because the actual mole ratio of NO:O2 is larger than the balanced equation mole

ratio of NO:O2, there is an excess of NO; O2 is the limiting reactant.

Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO

0.6250 mol NO 3 30.01 g NO/1 mol NO 5 18.76 g NO

Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2

Excess NO 5 20.00 g NO 2 18.76 g NO 5 1.24 g N

Explanation:

3 0
3 years ago
DDT:
Assoli18 [71]

Answer:

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3 years ago
Using the black numbers on the stopwatch to answer the questions.
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Answer: the top one is 5.3 s and the bottom one is tenths of seconds

Explanation:

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3 years ago
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