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3 years ago

11

Will you write the names for the compound feo and fe2o3

1 answer:

n200080 [17]3 years ago

3 0

FeO is Iron II Oxide. You identify that it's an ionic bond so it won't have any prefixes. By reverse criss-crossing the charges, Iron gets a 1+ charge, and Oxygen gets a 1- charge. Since you know that Oxygen has to have a 2- charge, you multiply the 1+ and 1- by 2 to fix their charges, thus giving Iron a 2+ charge and Oxygen a 2- charge.

Since Iron has a 2+ charge, it's roman numeral is II. So the answer is Iron II Oxide.

The other one is Iron III Oxide. By reverse criss-crossing the charges, Fe geta 3+ charge, and Oxygen gets a 2- charge. Since it's correct, the 3+ charge is the roman numeral. So, the answer is Iron III Oxide.

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Is the cyanide ion a strong or weak base?

11111nata11111 [884]

It's a weak base bacause H C N is weak

4 0

3 years ago

How many electrons are shared in the Lewis structure

Diano4ka-milaya [45]

Hey there,

Answer:

4 valence electrons.

Hope this helps :D

<em>~Top☺</em>

3 0

3 years ago

The freezing-point depression of a 0.100 m MgSO4 solution is 0.225°C. Determine the experimental van't Hoff factor of MgSO4 at t

Andrews [41]

<u>Answer:</u> The experimental van't Hoff factor is 1.21

<u>Explanation:</u>

The expression for the depression in freezing point is given as:

$\Delta T_f=iK_f\times m$

where,

i = van't Hoff factor = ?

$\Delta T_f$ = depression in freezing point = 0.225°C

$K_f$ = Cryoscopic constant = 1.86°C/m

m = molality of the solution = 0.100 m

Putting values in above equation, we get:

$0.225^oC=i\times 1.86^oC/m\times 0.100m\\\\i=\frac{0.225}{1.86\times 0.100}=1.21$

Hence, the experimental van't Hoff factor is 1.21

7 0

3 years ago

Convert 7.1x10^25 molecules of water to moles

ruslelena [56]

Answer:

<h2>117.94 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{7.1 \times {10}^{25} }{6.02 \times {10}^{23} } \\ = 117.940199...$

We have the final answer as

<h3>117.94 moles</h3>

Hope this helps you

8 0

3 years ago

Which answer is an isotope of 40/20x? <br><br> Question Pictured here:

Nesterboy [21]

An isotope of ₂₀⁴⁰X ⇒₂₀⁴⁴X

<h3>Further explanation </h3>

The elements in nature have several types of isotopes

Isotopes are atoms that have the same number of protons and a different number of neutrons.

So Isotopes are elements that have the same Atomic Number (Proton)

In the following element notation,

$\large {{{A} \atop {Z}} \right X}$

X = symbol of elemental atom

A = mass number

= number of protons + number of neutrons

Z = atomic number

= number of protons = number of electrons, on neutral elements

For symbol of isotope X :

₂₀X⁴⁰ : atomic number = 20, mass number = 40

A. ₄₀²⁰X : atomic number = 40, mass number = 20, neutrons=20

B. ₂₀⁴⁰X²⁺ : atomic number = 20, mass number = 40, neutrons=20

C. ₂₁⁴⁰X : atomic number = 21, mass number = 40, neutrons=19

D. ₂₀⁴⁴X : atomic number = 20, mass number = 44, neutrons=22

7 0

3 years ago