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3 years ago

Question 6

Chemistry

1 answer:

crimeas [40]3 years ago

8 0

Answer: There are $10.5\times 10^{23}$ formula units in 1.75 mol sample of NaCl.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

1 mole of NaCl contains = $6.023\times 10^{23}$ formula units

Thus 1.75 mol of NaCl contains = $\frac{6.023\times 10^{23}}{1}\times 1.75=10.5\times 10^{23}$ formula units

Thus there are $10.5\times 10^{23}$ formula units in 1.75 mol sample of NaCl.

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Gaseous ethane (CH,CH,) will react with gaseous oxygen (0,) to produce gaseous carbon dioxide (CO) and gaseous water (H2O). Supp

notka56 [123]

Answer:

0.00 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

Mᵣ 30.07 32.00

2CH₃CH₃ + 7O₂ ⟶ 4CO₂ + 6H₂O

Mass/g: 1.50 11.

2. Calculate the moles of each reactant

$\text{moles of C$_{2}$H}_{6} = \text{1.50 g C$_{2}$H}_{6} \times \dfrac{\text{1 mol C$_{2}$H}_{6}}{\text{30.07 g C$_{2}$H}_{6}} = \text{0.04988 mol C$_{2}$H}_{6}\\\\\text{moles of O}_{2} = \text{11. g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.34 mol O}_{2}$

3. Calculate the moles of CO₂ we can obtain from each reactant

From ethane:

The molar ratio is 4 mol CO₂:2 mol C₂H₆

$\text{Moles of CO}_{2} = \text{0.04988 mol C$_{2}$H}_{6} \times \dfrac{\text{4 mol CO}_{2}}{\text{2 mol C$_{2}$H}_{6}} = \text{0.09976 mol CO}_{2}$

From oxygen:

The molar ratio is 4 mol CO₂:7 mol O₂

$\text{Moles of CO}_{2} = \text{0.34 mol O}_{2}\times \dfrac{\text{4 mol CO}_{2}}{\text{7 mol O}_{2}} = \text{0.20 mol CO}_{2}$

4. Identify the limiting and excess reactants

The limiting reactant is ethane, because it gives the smaller amount of CO₂.

The excess reactant is oxygen.

5. Mass of ethane left over.

Ethane is the limiting reactant. It will be completely used up.

The mass of ethane left over will be 0.00 g.

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3 years ago

Why can’t you perform single displacement reactions with group IA metals?

pishuonlain [190]

Answer is: because alkaline metals (group IA metals) are the strongest reducing agents and most reactive metals.
Reducing agent is an element or compound that loses an electron to another chemical species in a redox chemical reaction and they have been oxidized.
Alkaline metals tend to lose only one electron in redox reaction.

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Michela and Simone are camping in a tent during some hot weather. They

eimsori [14]

Answer:

i think that the wet cloth will keep the coolness inside the water instead of evaporating and getting warm

Explanation:

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3 years ago

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deff fn [24]

Answer:

Each oxygen atom is connected to the central O atom with 2 covalent bonds.

Explanation:

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Another explanation

A Lewis structure is also called a dot electron structure. A Lewis structure represents all the valence electrons on atoms in a molecule as dots. Lewis structures can be used to represent molecules in which the central atom obeys the octet rule as well as molecules whose central atom does not obey the octet rule.

Sometimes, one Lewis structure does not suffice in explaining the observed properties of a given chemical specie. In this case, we evoke the idea that the actual structure of the chemical specie lies somewhere between a limited number of bonding extremes called resonance or canonical structures.

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Explanation:

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Answer:

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