How many bonds exist in this molecule?

Round off the measurement to three significant figures 12.17º C

prohojiy [21]

12.2 C
It has 3 significant figures now.

6 0

3 years ago

What are molecules, and how are the properties of molecules different from the atoms they come from? Give an example.

goldenfox [79]

Molecule, a group of two or more atoms that form the smallest identifiable unit into which a pure substance can be divided and still retain the composition and chemical properties of that substance.

While Atoms are single neutral particles,

Molecules are neutral particles made of two or more atoms bonded together.

Exaplmes for molecules

H2O (water)

N2 (nitrogen)

O3 (ozone)

CaO (calcium oxide)

C6H12O6 (glucose, a type of sugar)

NaCl (table salt

And examples for atoms

Neon (Ne)

Hydrogen (H)

Argon (Ar)

Iron (Fe)

Calcium (Ca)

Deuterium, an isotope of hydrogen that has one proton and one neutron.

Plutonium (Pu)

F-, a fluorine anion.

3 0

3 years ago

Does Kinetic energy depend only on an object’s speed and velocity?

BaLLatris [955]

yes .. . . . .. . . .. ....... . ..dlnv3r;'mw,c kc;oqc,

xkdnnd

4 0

3 years ago

Read 2 more answers

Identify whether each species functions as a Brønsted-Lowry acid or a Brønsted-Lowry base in this net ionic equation. HF (aq) +

VMariaS [17]

Explanation:

A Bronsted-Lowry base is a substance that accepts a proton in the form of a hydrogen (H) atom.

On the other hand;

Bronsted-Lowry acid is the substance that donates the proton.

HF (aq) + SO32- ⇌ F- + HSO3-

In the forward reaction;

Bronsted-Lowry acid : HF

Bronsted-Lowry base: SO32-

In the backward reaction;

Bronsted-Lowry acid : HSO3-

Bronsted-Lowry base: F-

The conjugate base of HF is F-

The conjugate acid of SO32- is HSO3-

8 0

3 years ago

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

7 0

3 years ago