All categories

3 years ago

What is the chemical symbol for iron

2 answers:

8 0

The chemical symbol for iron is Fe.
Iron is a chemical element with symbol Fe (from Latin: ferrum) and atomic number 26. It is a metal in the first transition series. It is by mass the most common element on Earth, forming much of Earth's outer and inner core

Genrish500 [490]3 years ago

7 0

The atomic symbol for Iron is Fe, as the Latin word for Iron is ferrum.

You might be interested in

Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations (Ka is given

tekilochka [14]

Answer:

a) = 0.704%

b) = 1.30%

c) = 2.60%

Explanation:

Given that:

$K_a$ = $1.34*10^{-5$

For Part A; where Concentration of A = 0.270 M

Percentage Ionization(∝) $\alpha = \sqrt{\frac{K_a}{C} }$

$\alpha = \sqrt{\frac{1.34*10^{-5}}{0.270} }$

$\alpha = \sqrt{4.9629*10^{-5}}$

$\alpha = 7.044*10^{-3$

percentage% (∝) = $7.044*10^{-3}*100$

= 0.704%

For Part B; where Concentration of B = $7.84*10^{-2$ M

$\alpha = \sqrt{\frac{1.34*10^{-5}}{7.84*10^{-2}} }$

$\alpha = \sqrt{1.709*10^{-4} }$

$\alpha = 0.0130\\$

percentage% (∝) = 0.0130 × 100%

= 1.30%

For Part C; where Concentration of C= $1.92*10^{-2} M$

$\alpha = \sqrt{\frac{1.34*10^{-5}}{1.97*10^{-2}} }$

$\alpha = \sqrt{6.802*10^{-4}}$

$\alpha =0.02608$

percentage% (∝) = 0.02608 × 100%

= 2.60%

7 0

3 years ago

What is the metal cation in K,SO,? (cation is pronounced cat-ion and refers to a positively charged ion)

slamgirl [31]

Answer:

Potassium cation = K⁺²

Explanation:

The metal cation in K₂SO₄ is K⁺². While the anion is SO₄²⁻.

All the metals have tendency to lose the electrons and form cation. In given compound the metal is potassium so it should form the cation. The overall compound is neutral.

The charge on sulfate is -2. While the oxidation state of potassium is +1. So in order to make compound overall neutral there should be two potassium cation so that potassium becomes +2 and cancel the -2 charge on sulfate and make the charge on compound zero.

2K⁺² , SO₄²⁻

K₂SO₄

8 0

3 years ago

3. How many atoms in 3.91 moles of sodium?

Gala2k [10]

Answer:

Explanation:

The formula for sodium is Na. It does not form a molecule in some way.

1 mol Na = 6.02*10^23 atoms

3.91 mol = x Cross multiply

x = 3.91 * 6.02 * 10^23

x = 23.65 * 10^23

x = 2.365 * 10^24

Scientific notation is always expressed as a number 1 ≤ x < 10

3 0

3 years ago

An antacid tablet contains 640.0 mg of magnesium oxide per tablet.

Salsk061 [2.6K]

Answer:

317.6 mL

Explanation:

Step 1: Write the balanced neutralization equation

MgO + 2 HCl ⇒ MgCl₂ + H₂O

Step 2: Calculate the mass corresponding to 640.0 mg of MgO

The molar mass of MgO is 40.30 g/mol. The moles corresponding to 640.0 mg (0.6400 g) of MgO are:

0.6400 g × (1 mol/40.30 g) = 0.01588 mol

Step 3: Calculate the moles of HCl that react with 0.01588 moles of MgO

The molar ratio of MgO to HCl is 1:2. The moles of HCl are 2/1 × 0.01588 mol = 0.03176 mol

Step 4: Calculate the volume of 0.1000 M HCl that contains 0.03176 moles

0.03176 mol × (1 L/0.1000 mol) = 0.3176 L = 317.6 mL

3 0

3 years ago

Noble gas compounds like KrF, XeCl, and XeBr are used in excimer lasers. Draw an approximate molecular orbital diagram appropria

Aneli [31]

Answer:

Here's what I get.

Explanation:

The MO diagrams of KrBr, XeCl, and XeBr are shown below.

They are similar, except for the numbering of the valence shell orbitals.

Also, I have drawn the s and p orbitals at the same energy levels for both atoms in the compounds. That is obviously not the case.

However, the MO diagrams are approximately correct.

The ground state electron configuration of KrF is

$(1\sigma_{g})^{2}\, (1\sigma_{u}^{*})^{2} \, (2\sigma_{g})^{2} \, (2\sigma_{u}^{*})^{2} \, (3\sigma_{g})^{2} \, (1\pi_{u})^{4} \, (1\pi_{g}^{*})^{4} \, (3\sigma_{g}^{*})^{1}$

KrF⁺ will have one less electron than KrF.

You remove the antibonding electron from the highest energy orbital, so the bond order increases.

The KrF bond will be stronger.

6 0

4 years ago