A ________ alchohol has a -OH radical bonded to a central carbon atom.

Some atoms are chemically stable and will not bond with atoms of other

tigry1 [53]

Answer:

The stability of atoms depends on whether or not their outer-most shell is filled with electrons. If the outer shell is filled, the atom is stable. Atoms with unfilled outer shells are unstable, and will usually form chemical bonds with other atoms to achieve stability.

Explanation:

7 0

3 years ago

Read 2 more answers

Someone please help me with one of them

EastWind [94]

Explanation:

Are there any values given in the question?

3 0

2 years ago

How many Sodium atoms are in 2.00 mols of Na2O

Anna71 [15]

There’s 1024 atoms of sodium in 2 moles of sodium

7 0

3 years ago

How many moles of O₂ are needed to react completely with 35.0 mol of FeCl₃? *

densk [106]

Answer:

26.3 moles of O₂ are needed to react completely with 35.0 mol of FeCl₃

Explanation:

To determine the number of moles of O₂ that are needed to react completely with 35.0 mol of FeCl₃, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), and rule of three as follows: if 4 moles of FeCl₃ react with 3 moles of O₂, 35 moles of FeCl₃ with how many moles of O₂ will it react?

$molesofO_{2} =\frac{35 moles of FeCl_{3}*3 moles of O_{2} }{4 moles of FeCl_{3}}$

moles of O₂= 26.25 ≅ 26.3

<u><em>26.3 moles of O₂ are needed to react completely with 35.0 mol of FeCl₃</em></u>

7 0

3 years ago

Consider the following reaction between calcium oxide and carbon dioxide: CaO(s)+CO2(g)→CaCO3(s) A chemist allows 14.4 g of CaO

sweet-ann [11.9K]

Answer:

Theoretical yield =26.03 g

Percent yield = 87%

Limiting reactant = CaO

Explanation:

Given data:

Mass of CaO = 14.4 g

Mass of CO₂ = 13.8 g

Actual yield of CaCO₃ = 22.6 g

Theoretical yield = ?

Percent yield = ?

Limiting reactant = ?

Solution:

Chemical equation:

CaO + CO₂ → CaCO₃

Number of moles of CaO:

Number of moles = Mass /molar mass

Number of moles = 14.4 g / 56.1 g/mol

Number of moles = 0.26 mol

Number of moles of CO₂:

Number of moles = Mass /molar mass

Number of moles = 13.8 g / 44 g/mol

Number of moles = 0.31 mol

Now we will compare the moles of CO₂ and CaO with CaCO₃ .

CO₂ : CaCO₃

1 : 1

0.31 : 0.31

CaO : CaCO₃

1 : 1

0.26 : 0.26

The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.

Mass of CaCO₃: Theoretical yield

Mass of CaCO₃ = moles × molar mass

Mass of CaCO₃ =0.26 mol × 100.1 g/mol

Mass of CaCO₃ = 26.03 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 22.6 g/ 26.03 g × 100

Percent yield = 0.87× 100

Percent yield = 87%

Limiting reactant:

The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.

7 0

3 years ago