All categories

3 years ago

Which of the following explains how all compounds are created?

Chemistry

2 answers:

Vinvika [58]3 years ago

8 0

Answer. i think that the answer is D hope it helps :)

sweet [91]3 years ago

5 0

Compounds are formed by bonding atoms of two or more different elements through chemical reaction. Atoms are bonded by losing or gaining electrons in ionic bonding and by sharing electrons in covalent bonding between two atoms participating in the bond formation. For example, one molecule of the compound water is formed by bonding two atoms of hydrogen and one atom of oxygen by sharing electrons between the atoms and forming covalent bonds. Therefore, compounds are created by losing, gaining or sharing electrons between other atoms.

(Answer) (D) By losing, gaining, or sharing electrons between other atoms.

You might be interested in

A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0

3 years ago

Read 2 more answers

Show your calculation by uploading a picture. Calculate the molar mass of ammonia, NH3

Cloud [144]

Answer:

17.04 g/mol

Explanation:

Molar Mass of NH₃

we know that

Nitrogen has 14.01 gram/mol

And Hydrogen has 1.01 gram/mol

but we have 3 Hydrogens So we multiply

1.01 by 3 i.e., 3.03

Now, add

14.01

+ 3.03

17.04

So, The molar mass of ammonia, NH₃ is

17.04 g/mol

-TheUnknownScientist

5 0

3 years ago

Elements with atomic numbers 7 and 83 have how many valence electrons? *

IRINA_888 [86]

Explanation:

element with atomic numbers 7 and 83 have 5 and 5 valence electrons

5 0

2 years ago

What is the volume of 0.1 mole of methane (CH4) ? (One mole of any gas occupies 22.4 L under certain conditions of temperature a

Veronika [31]

Answer:

Option A = 2.2 L

Explanation:

Given data:

volume of one mole of gas = 22.4 L

Volume of 0.1 mole of gas at same condition = ?

Solution:

It is known that one mole of gas at STP occupy 22.4 L volume. The standard temperature is 273.15 K and standard pressure is 1 atm.

For 0.1 mole of methane.

0.1/1 × 22.4 = 2.24 L

0.1 mole of methane occupy 2.24 L volume.

8 0

3 years ago

Read 2 more answers

2Al + 6HCl → 2AlCl3 + 3H2

Nadya [2.5K]

Answer: The reaction produces 2.93 g H₂.

M_r: 133.34 2.016

2Al + 6HCl → 2AlCl₃ + 3H₂

Moles of AlCl₃ = 129 g AlCl₃ × (1 mol AlCl₃/133.34 g AlCl₃) = 0.9675 mol AlCl₃

Moles of H₂ = 0.9675 mol AlCl₃ × (3 mol H₂/2 mol AlCl₃) = 1.451 mol H₂

Mass of H₂ = 1.451 mol H₂ × (2.016 g H₂/1 mol H₂) = 2.93 g H₂

Explanation:

6 0

3 years ago