Is NaF a weak acid/base?

HEEEEELP ASAPP IM AWARDING 30 points!!!!

alexira [117]

Answer:

$\rm S^{2-}$ .

Explanation:

Based on the electron configuration of this ion, count the number of electrons in this ion in total:

$2 + (2 + 6) + (2 + 6) = 18$ .

Each electron has a charge of $(-1)$ .

Atoms are neutral and have $0$ charge. However, when an atom gains one extra electron, it becomes an ion with a charge of $(-1)$ . Likewise, when that ion gains another electron, the charge on this ion would become $(-2)$ .

The ion in this question has a charge of $(-2)$ . In other words, this ion is formed after its corresponding atom gains two extra electrons. This ion has $18$ electrons in total. Therefore, the atom would have initially contained $18 - 2 = 16$ electrons. The atomic number of this atom would be $16$ .

Refer to a modern copy of the periodic table. The element with an atomic number of $16$ is sulphur with atomic symbol $\rm S$ . To denote the ion, place the charge written backwards (" $2-$ " for a charge of $(-2)$ ) as the superscript of the atomic symbol:

$\rm S^{2-}$ .

5 0

3 years ago

Read 2 more answers

A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The initial concentrations of Pb2+ and Cu2+ are

VARVARA [1.3K]

Answer:

a) Ecell = 0.5123 V

b) Ecell = 0.4695 V

c) [Pb2 +] = 4.75 M

Explanation:

a)

The reaction at the cathode is represented as follows:

Cu2 + + 2e- -> Cu (s) Eocathode = 0.34 V

The reaction at the anode is equal to:

Pb (s) -> Pb2 + + 2e- Eoanode = -0.13 V

The number of moles of the electrons that are involved is equal to n = 2

Standard cell potential equals Eo = Eocathode - Eoanode = 0.34 V- (-0.13 V) = 0.47 V

The initial cell potential can be calculated with the following formula:

Ecell = Eocell - - 0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V

b)

The reaction in the cell is equal to:

Cu2 + + Pb (s) -> Cu (s) + Pb2 +

The concentration of Cu2 that gives the exercise is equal 0.2 M

Therefore, the change in concentration for Cu2 + is equal to:

Cu2 + = 1.4 M - 0.2 M = 1.2 M

We use the formula from part a)

Ecell = Eocell - (0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V

c)

To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:

Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))

0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)

Clearing Pb2 +:

[Pb2 +] = 4.75 M

8 0

3 years ago

A student dissolves some solid NaNO3 in a beaker of water. Which are the solute and the solvent? Choose all answers that are cor

Paul [167]

The correct answer is D. The solute in this solution is the solid sodium nitrate (NaNO3) which is dissolved in the solvent, the water. Solute is the minor component in a solution whereas the solvent is the major component in the solution.

3 0

3 years ago

What does a student need to know about double bonds and triple bonds when predicting molecular geometry of molecules?

zhuklara [117]

This problem is asking for an explanation of what we need to know about double and triple bonds to successfully predict molecular geometries in molecules. At the end, one comes to the conclusion that double and triple bonds contribute to the degree in which an atom is bonded and they also determine the lone pairs, which, at the same time, define the molecular geometry.

<h3>Molecular geometry:</h3>

In chemistry, molecules are not necessarily flat arrangements of atoms, yet they have specific bond angles, orientations and shapes, which define the molecular geometry. In such a way, we can use the VSEPR theory in order to know the molecular geometry of a molecule; however, we first need its Lewis structure or at least the number and type of bonds to do so.

Consider water and carbon dioxide; the former has two hydrogen to oxygen bonds (O-H) and 2 lone pairs because O has six valence electrons but just 2 are bonded to complete the octet, so 4 unpaired electrons lead to two lone pairs. On the other hand, the latter has two double bonds (C=O) and 0 lone pairs because carbon has four valence electrons and they are all bonded to complete the octet.

In such a way, one can see how the double bond affected the bonding in CO2 in contrast to the H2O; situation that also applies to triple bonds, because CO2 has a linear molecular geometry whereas water has a bent one (see attached picture)

Hence, one comes to the conclusion that double and triple bonds contribute to the degree in which an atom is bonded and they also determine the lone pairs, which, at the same time, define the molecular geometry.

Learn more about molecular geometry: brainly.com/question/7558603

Learn more about the VSEPR theory: brainly.com/question/14225705

5 0

2 years ago

A student is given the question: "what is the mass of a gold bar that is 7.379 × 10–4 m3 in volume? the density of gold is 19.3

IRINA_888 [86]

<h3><em>Further Explanation</em></h3>

Density is a quantity derived from the mass and volume

density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than type

The unit of density can be expressed in g / cm3 or kg / m3

Density formula:

$\large{\boxed{{\bold{\rho~=~\frac{m}{V}}}}$

ρ = density

m = mass

v = volume

A common example is the water density of 1 gr / cm3

The mass itself is often equated with weight, even though it is different. Mass is the amount of matter in the matter while weight is related to the gravitational force

<em>Known variable</em>

volume gold bar 7.379 × 10⁻⁴ m³

a density of 19.3 g/cm³

<em>Asked</em>

the mass of a gold bar

<em>Answer</em>

volume is known to be 7.379 × 10⁻⁴ m³, then we change it first to units of cm³ adjusting to units of density

7.379 × 10⁻⁴ m³ = 7.379 × 10² cm³

then:

mass = volume x density

mass = 7.379 × 10² cm³ x 19.3 g/cm³

mass = 1.424 x 10⁴ gram