Examples of exponential decay in the real-world.

How many electrons are in O2.

Liula [17]

Number of electron in O = 8
So, in O2 it would be: 8 * 2 = 16

So, your final answer is 16

Hope this helps!

6 0

2 years ago

Chemitey half equations

shepuryov [24]

Isn't it a because in b at the start of the equation the E in Fe just disappeared

8 0

3 years ago

How many grams of Ni are formed from 55.3 g of Ni2O3? 2Ni2O3(s)⟶4Ni(s)+3O2(g)

Neko [114]

Answer:

39.2 g

Explanation:

2Ni₂O₃(s) ⟶ 4Ni(s) + 3O₂(g)

First we convert 55.3 grams of Ni₂O₃ into moles of Ni₂O₃, using its molar mass:

55.3 g ÷ 165.39 g/mol = 0.334 mol Ni₂O₃

Then we convert 0.334 moles of Ni₂O₃ into moles of Ni, using the stoichiometric coefficients of the balanced reaction:

0.334 mol Ni₂O₃ * $\frac{4molNi}{2molNi_2O_3}$ = 0.668 mol Ni

Finally we calculate how much do 0.668 Ni moles weigh, using the molar mass of Ni :

0.668 mol Ni * 58.69 g/mol = 39.2 g

7 0

3 years ago

The polymer formed from the monomer ch2=ch–cn is select one:

podryga [215]

Answer:
None of the given options show polymer made up of H₂C=CH-CN (Acrylonitrile).

Explanation:
Acrylonitrile (H₂C=CH-CN) which is a monomer on self linkage results in a large chain polymer called as Polyacrylonitrile.
The structure of Polyacrylonitrile is as follow,

--(H₂C-CHCN-)n--

Where n shows the number of Acrylonitrile units joined together in the formation of Polyacrylonitrile. This polymerization reaction can take place by different mechanisms including free radical mechanism, acid catalyzed addition or base catalyzed addition reaction.

The polymerization is shown below,

5 0

3 years ago

What is the percent composition of NaHCO3?

zhuklara [117]

Answer:

Option-C (27.36% Na, 1.20% H, 14.30% C, and 57.14% O)

Explanation:

Percent Composition is defined as the %age by mass of each element present in a compound. Therefore, it is a relative amount of each element present in a compound.

Calculating Percent Composition of NaHCO₃:

1: Calculating Molar Masses of all elements present in NaHCO₃:

a) Na = 22.99 g/mol

b) H = 1.01 g/mol

c) C = 12.01 g/mol

d) O₃ = 16.0 × 3 = 48 g/mol

2: Calculating Molecular Mass of NaHCO₃:

Na = 22.99 g/mol

H = 1.01 g/mol

C = 12.01 g/mol

O₃ = 48 g/mol

----------------------------------

Total 84.01 g/mol

3: Divide each element's molar mass by molar mass of NaHCO₃ and multiply it by 100:

For Na:

= 22.99 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

= 27.36 %

For H:

= 1.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

= 1.20 %

For C:

= 12.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

= 14.29 % ≈ 14.30 %

For O:

= 48.0 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

= 57.13 % ≈ 57.14 %

8 0

3 years ago