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Rasek [7]
2 years ago
13

The molecular mass of air, at standard pressure and temperature, is approximately 28.97 g/mol. Calculate the mass of 3.33 moles

of air.
First, complete the unit conversion using dimensional analysis:

A ×
B
C
= ? g Air

A:
B:
Chemistry
2 answers:
erica [24]2 years ago
8 0

A: 3.33 mol Air

B: 28.97 g Air

C: 1 mol Air

poizon [28]2 years ago
6 0
<h3>Answer:</h3>

               Mass = 96.47 g

<h3>Solution:</h3>

Data Given:

                  M.Mass  =  28.97 g.mol⁻¹

                  Moles  =  3.33 mol

                  Mass  =  ??

Formula Used:

                  Moles  =  Mass ÷ M.Mass

Solving for Mass,

                  Mass  =  Moles × M.Mass

Putting values,

                  Mass  =  3.33 mol × 28.97 g.mol⁻¹

                  Mass  =  96.4701 g

Rounding to four significant numbers,

                 Mass = 96.47 g


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Consider a saturated solution formed when 17.5 g of a solute dissolve in 28.3 g of a solvent, giving a total solution volume of
STALIN [3.7K]

Answer:

a) 38.2 % mass

b) 61.8 g solute/100 g solvent

c) 1.65 g/mL

Explanation:

Given the data:

mass of solute = 17.5 g

mass of solvent= 28.3 g

total solution volume= 27.8 mL

a)- mass percent= mass of solute/mass of solution x 100

mass of solution = mass solute + mass solvent = 17.5 g + 28.3 g = 45.8 g

mass % = 17.5 g/45.8 g x 100 = 38.2 % mass

b)- solubility = grams of solute/ 100 g solvent

                    = 17.5 g x (100 g /28.3 g solvent) = 61.8 g solute/100 g solvent  

c)- density = massof solution/total volumesolution  = 45.8 g/27.8 mL = 1.65 g/mL

7 0
2 years ago
When the reaction CO2(g) + H2(g) ⇄ H2O(g) + CO(g) is at equilibrium at 1800◦C, the equilibrium concentrations are found to be [C
UNO [17]

Answer:

The new molar concentration of CO at equilibrium will be  :[CO]=1.16 M.

Explanation:

Equilibrium concentration of all reactant and product:

[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M

Equilibrium constant of the reaction :

K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}

K = 4

CO_2(g) + H_2(g) \rightleftharpoons H_2O(g) + CO(g)

Concentration at eq'm:

0.24 M          0.24 M                 0.48 M            0.48 M

After addition of 0.34 moles per liter of CO_2 and H_2 are added.

(0.24+0.34) M    (0.24+0.34) M  (0.48+x)M         (0.48+x)M

Equilibrium constant of the reaction after addition of more carbon dioxide and water:

K=4=\frac{(0.48+x)M\times (0.48+x)M}{(0.24+0.34)\times (0.24+0.34) M}

4=\frac{(0.48+x)^2}{(0.24+0.34)^2}

Solving for x: x = 0.68

The new molar concentration of CO at equilibrium will be:

[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M

3 0
2 years ago
Which of the following is correctly balanced redox half reaction? Group of answer choices A. 14H+ + 9e- + Cr2O72- ⟶ Cr3+ + 7H2O
oee [108]

Answer:

The correct option is: B. 14 H⁺ + 6 e⁻ + Cr₂O₇²⁻ ⟶ 2 Cr³⁺ + 7 H₂O

Explanation:

Redox reactions is an reaction in which the oxidation and reduction reactions occur simultaneously due to the simultaneous movement of electrons from one chemical species to another.

The reduction of a chemical species is represented in a reduction half- reaction and the oxidation of a chemical species is represented in a oxidation half- reaction.

<u>To balance the reduction half-reaction for the reduction of Cr₂O₇²⁻ to Cr³⁺</u>:

Cr₂O₇²⁻ ⟶ Cr³⁺

First the <u>number of Cr atoms</u> on the reactant and product side is balanced

Cr₂O₇²⁻ ⟶ 2 Cr³⁺

Now, Cr is preset in +6 oxidation state in Cr₂O₇²⁻ and +3 oxidation state in Cr³⁺. So each Cr gains 3 electrons to get reduced.

Therefore, <u>6 electrons are gained</u> by 2 Cr atoms of Cr₂O₇²⁻ to get reduced.

Cr₂O₇²⁻ + 6 e⁻ ⟶ 2 Cr³⁺

Now the total charge on the reactant side is (-8) and the total charge on the product side is (+6).

From the given options it is evident that the reaction must be balanced in acidic conditions.

Therefore, to <u>balance the total charge</u> on the reactant and product side,<u> 14 H⁺ is added on the reactant side.</u>

Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺

Now to <u>balance the number of hydrogen and oxygen atoms, 7 H₂O is added on the product side.</u>

Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺ + 7 H₂O

<u>Therefore, the correct balanced reduction half-reaction is:</u>

Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺ + 7 H₂O

3 0
2 years ago
How much heat is lost when changing 65 g of water vapor (H2O) at 421 K to ice at 139 K?
poizon [28]

The heat change will be

Moles of water = mass / Molar mass = 65/ 18 = 3.61 mol

specific heat of ice =2.09J /g C

specific heat of water = 4.184 J/g C

Specific heat of vapour= 2.01 /g C

Heat of fusion = 3.33X10⁵ J /kg = 333 J /g

Heat of vaporization = 2.26 X10⁶J/kg = 2260J/g

Q1 = heat change when vapours get cooled to 373.15 K

Q2 = heat change when vapours get converted to liquid water

Q3 = heat change when liquid water cools to 273.15 K

Q4= heat change when liquid water freezes to ice

Q5= heat change when ice cools from 273.15K to 139 K

Q1= mass of water X specific heat of vapours X change in temperature

Q1 = 65 X 2.01 /g C X (421-373.15) = 6251.60 J = 6.252 kJ

Q2 = heat of vaporization X mass = 2260 X 65 = 146900 = 146.9 kJ

Q3 = mass X specific heat of water X change in temperature =

Q3 = 65 X 4.184 X (373.15-273.15) = 65 X 4.184 X 100 = 27196 J = 27.196kJ

Q4 = heat of fusion X mass =333X65 = 21645 J = 21.645 kJ

Q5 =  mass X specific heat of ice X change in temperature

Q5 = 65 X 2.09 X (273.15-139) = 18224.3 J = 18.224 kJ

Total energy = 6.252 +146.9+27.196+ 21.645+ 18.224 = 220.217

As this is energy released so it will be expressed in negative

-220.217

from the given options the correct answer will be -219.4 kJ

The answer is little different as the reference values of specific heats or enthalpy may vary.

3 0
2 years ago
Describe the hybrid orbitals used by the central atom and the types of bonds formed in o3
Tems11 [23]
Hybridization in ozone, O3...... 

<span>...O = O ........ 1 lone pair on central O, 2 lone pairs on terminal O </span>
<span>../ </span>
<span>O .................. 3 lone pairs on terminal O </span>

<span>I didn't show the second of two resonance structures in which the single and double bonds are reversed. In reality, both bonds are identical have a bond order of 1.5 due to delocalized pi-bonding. </span>

<span>The central atom exhibits sp2 hybridization since there is trigonal planar electron pair geometry. The notion of hybrid orbitals was "invented" by Linus Pauling in the 1930's as a way of explaining the geometry of molecules, primarily the geometry of carbon compounds. </span>

<span>If the electron pair geometry is linear, the hybridization is sp. </span>
<span>If the electron pair geometry is trigonal planar, the hybridization is sp2. </span>
<span>If the electron pair geometry is tetrahedral, the hybridization is sp3. </span>

<span>The notion that there is sp3d and sp3d2 because of d-orbital participation has been debunked. Chemists know today that there is no d-orbital involvement in hypervalent molecules regardless of what some out-of-date textbooks and some teachers' dusty old notes may say. Instead, the best explanation involves 3-center, 4-electron bonding.</span>
8 0
2 years ago
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