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Anestetic [448]
3 years ago
14

Which of the following does not serve as a way to neutralize the charge in a body? Question 3 options: A) Bringing the charged b

ody into contact with another body having an equal but opposite charge B) Adding more protons to a positively charged body until the number of protons matches the number of electrons C) Adding free electrons to a positively charged body D) Allowing free electrons to escape from a negatively charged body
Chemistry
1 answer:
Lelechka [254]3 years ago
8 0

Answer: B. Adding more protons to a positively charged body until the number of protons matches the number of electrons

Explanation:

took test got it right

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In which pair do both compounds exhibit predominantly ionic bonding? A) KCl and CO2 B) SO2 and BaF2 C) F2 and N2O D) N2O3 and Rb
gtnhenbr [62]

Answer:

E) NaF and SrO

Explanation:

The ionic bonding occurs between atoms with a great difference in electronegativity. This usually happens between a metal and a non-metal.

<em>In which pair do both compounds exhibit predominantly ionic bonding? </em>

A) KCl and CO₂. NO. C and O are non-metals and present covalent bonding.

B) SO₂ and BaF₂. NO. S and O are non-metals and present covalent bonding.

C) F₂ and N₂O. NO. Both compounds contain non-metals and present covalent bonding.

D) N₂O₃ and Rb₂O. NO. N and O are non-metals and present covalent bonding.

E) NaF and SrO. YES. Na and Sr are metals while F and O are non-metals.

5 0
3 years ago
All of the following show a periodic pattern except
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4 0
3 years ago
Biphenyl and benzene are common side products in this gringard reaction. explain at which points in the purification procedure t
mafiozo [28]

In Grignard reaction, Biphenyl and benzene are common side products which are removed during trituration.

In organic chemistry, a reaction in which the Grignard reagents or organometallic substances are added to organic compounds such as aldehydes and ketones to form alcohol is known as Grignard reaction.

These Grignard reagents are magnesium halides of alkyl, vinyl or allyl, which react with a carbonyl group to form alcohols.

During this reaction, primary, secondary and tertiary alcohols are formed.

While Biphenyl and benzene are common side products.

These are removed during trituration process in which cold petroleum ether is added to dissolve the biphenyl and benzene side products

If you need to learn more about Biphenyl and benzene click here:

brainly.com/question/4336669

4 0
1 year ago
Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water
madreJ [45]

Explanation:

The given data is as follows.

         Temperature of dry bulb of air = 25^{o}C

          Dew point = 10^{o}C = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

        Partial pressure of water vapor (P_{a}) = vapor pressure of water at a given temperature (P^{s}_{a})

Using Antoine's equation we get the following.

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}

                               = 0.17079

                   P^{s}_{a} = 1.18624 kPa

As total pressure (P_{t}) = atmospheric pressure = 760 mm Hg

                                   = 101..325 kPa

The absolute humidity of inlet air = \frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                  \frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                 = 0.00735 kg H_{2}O/ kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg H_{2}O/ kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

                 0.07 kg - 0.00735 kg

              = 0.06265 kg H_{2}O for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

                0.06265 kg \times 100

                  = 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

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