HELP ASAP How many millimeters of mercury pressure does a gas exert at 3.16 atm?

Answer: , , NaOH are all ionic compounds.

Explanation:

Ionic compounds form when one atom transfers its valence electrons to another atom, resulting in partial opposite charges on the participating atoms, which creates a strong attractive force between them.

An ionic bond is typically established between a metal and a non-metal.

For instance, , , NaOH are all ionic compounds.

Conversely, a covalent compound arises through the sharing of electrons among the atoms involved, where generally a non-metal bonds with the same or another non-metal.

For example, , etc are all covalent compounds.

Therefore, we can affirm that , [tex]CH_3CH_2COONa, NaOH are indeed ionic compounds.

Response:1816.6 joules

Clarification:refer to the attached image

Let's assume that the compound formula is as follows: Experiment 1: 1.00 g of the compound yields 1.95 g of AgCl. The molar mass of AgCl is 143.32 g/mol. Thus, the moles of AgCl for 1.95g are: The moles of Cl also equal 0.0136, considering that 1 mole of AgCl corresponds to 1 mole of Cl. Experiment 2: 1.00 g of the compound results in 0.900 g of CO2 and 0.735 g of H2O. The molar mass of CO2 is 44 g/mol, and for H2O, it's 18 g/mol. Therefore, the moles of C come to 0.0205 and the moles of H stand at 0.0816 (which is 2 times the moles of H2O). Now, from the provided details, it's derived that in 1.00 g of the compound, there are 0.0136 moles of Cl, 0.0205 moles of C, and 0.0816 moles of H. In terms of mass: Mass of Cl = 0.0136 * 35.5 = 0.4828 g. Mass of C = 0.0205 * 12 = 0.246 g. Mass of H = 0.0816 * 1 = 0.0816 g. Total mass = 0.4828 + 0.246 + 0.0816 + mass of N. Given that 1.00 = 0.8104 + Mass of N, it follows that Mass of N = 0.1896. Thus, upon dividing all moles by the smallest value, we find Cl = 0.0136 / 0.0135 = 1.0007; C = 0.0205 / 0.0135 = 1.52; H = 0.0816 / 0.0135 = 6.04; N = 0.0135 / 0.0135 = 1. Multiplying by 2 allows us to reach integer values: Cl = 2, C = 3, H = 12, N = 2.

Answer:

(a) The nitrogen atom in contains 8 electrons surrounding the central nitrogen atom.

(b) The nitrogen atom in has 8 electrons around the central nitrogen atom.

(c) The nitrogen atom in has 8 electrons surrounding the central nitrogen atom.

Explanation:

The Lewis dot structure illustrates the connections between atoms within a molecule, additionally showing unpaired electrons present in the molecule.In this structure, valence electrons are represented as 'dots'.

Now we will determine the number of electrons linked to the central nitrogen atom in the specified molecule.

(a) The identified molecule is,

Recognizing that nitrogen has '5' valence electrons while hydrogen has '1', we ascertain the total valence electrons in

= 5 + 2(1) + 1 = 8

According to the Lewis dot structure, this reveals 4 bonding and 4 non-bonding electrons.

From the Lewis structure, we conclude that the nitrogen atom in

has 8 electrons surrounding it.

(b) The identified molecule is,

Knowing that nitrogen has '5' valence electrons and hydrogen has '1' valence electron, the total number of valence electrons for = 5 + 4(1) - 1 = 8

In the Lewis dot structure, there are 8 bonding electrons and 0 non-bonding electrons.

This leads us to conclude that the nitrogen atom in

contains 8 electrons surrounding it.

(c) The identified molecule is,

Recognizing nitrogen’s '5' valence electrons and hydrogen’s '1' brings the total valence electrons for

= 5 + 3(1) = 8Utilizing the Lewis dot structure indicates there are 6 bonding and 2 non-bonding electrons.

According to the Lewis dot structure, we conclude that the nitrogen atom in

has 8 electrons surrounding it.

Answer:

180.56 kilojoules of heat energy is extracted when 1.00 kg of freon-11 evaporates.

Explanation:

The molar mass of freon-11 is 137.35 g/mol

The enthalpy of vaporization for freon-11 is 24.8 kJ/mol at its normal boiling point of 24°C. Given that,

Mass of freon-11 evaporated = 1.00 kg = 1000 g

Moles of freon-11 evaporated can be calculated as

The energy removed in the form of heat when 1.00 kg of freon-11 vaporizes is: