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1 year ago

How many molecules of O₂ will be required to produce 28.8 g of water? 2H₂ + O₂ ---> 2H₂O

Chemistry

1 answer:

juin [17]1 year ago

4 0

To produce 28.8 g of water 4.81 x 10²³ molecules of O₂ will be required

<h3>What is limiting reagent ?</h3>

The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.

According to given equation;

2H₂ + O₂ ---> 2H₂O

If 36 g of H₂O (2 mole) is produced by 1 mole O₂ i.e, 32 gram.

Therefore, 28.8 g of H₂O is produced by ;

32/36 x 28.8 = 25.6 g of O₂

Number of molecules of O₂ = given weight / molecular weight x 6.023 x 10²³

= 25.6 / 32 x 6.023 x 10²³

= 4.81 x 10²³ molecules of O₂

Therefore,4.81 x 10²³ molecules of O₂ will be required to produce 28.8 g of water.

Learn more about Limiting reagent here ;

brainly.com/question/20070272

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Chemists have developed insulation.

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3 years ago

Which metals are called noble metals

Diano4ka-milaya [45]

I don't know if this is multiple choice but if it isn't I can name of few.

Fun Fact: Noble Metals are chemical elements that have outstanding tolerance and resistance to oxidation.

Answer: Palladium, Silver, Platinum, and Gold. Those are all examples of Noble Metals.

Other: Ruthenium, osmium, and rhodium.

Hope this helps!

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2 years ago

1s22s²2p63s23p64s²3d104p5 Which element is this?

bearhunter [10]

Answer: bromine

Explanation:

There are a total of 2+2+6+2+6+2+10+5=35 electrons, meaning there are 35 protons. The element with atomic number 35 is bromine

3 0

1 year ago

At 85°C, the vapor pressure of A is 566 torr and that of B is 250 torr. Calculate the composition of a mixture of A and B that b

Phantasy [73]

Answer:

Composition of the mixture:

$x_{A} =0.652=65.2$ %

$x_{B} =0.348=34.8$ %

Composition of the vapor mixture:

$y_{A} =0.809=80.9$ %

$y_{B} =0.191=19.1$ %

Explanation:

If the ideal solution model is assumed, and the vapor phase is modeled as an ideal gas, the vapor pressure of a binary mixture with $x_{A}$ and $x_{B}$ molar fractions can be calculated as:

$P_{vap}=x_{A}P_{A}+x_{B}P_{B}$

Where $P_{A}$ and $P_{B}$ are the vapor pressures of the pure compounds. A substance boils when its vapor pressure is equal to the pressure under it is; so it boils when $P_{vap}=P$ . When the pressure is 0.60 atm, the vapor pressure has to be the same if the mixture is boiling, so:

$0.60*760=P_{vap}=x_{A}P_{A}+x_{B}P_{B}\\456=x_{A}P_{A}+(1-x_{A})P_{B}\\456=x_{A}*(P_{A}-P_{B})+P_{B}\\\frac{456-P_{B}}{P_{A}-P_{B}}=x_{A}\\\\\frac{456-250}{566-250}=x_{A}=0.652$

With the same assumptions, the vapor mixture may obey to the equation:

$x_{A}P_{A}=y_{A}P$ , where P is the total pressure and y is the fraction in the vapor phase, so:

$y_{A} =\frac{x_{A}P_{A}}{P}=\frac{0.652*566}{456} =0.809=80.9$ %

The fractions of B can be calculated according to the fact that the sum of the molar fractions is equal to 1.

7 0

2 years ago

Consider the aldol condensation between dibenzyl ketone and benzil in strong base. Calculate the molar masses of the reactants a

Sati [7]

Answer:

a) 210.3 g/mol

b) 210.2 g/mol

c) 384.5 g/mol

Explanation:

First step we will calculate the molar masses of ; carbon atom, hydrogen atom and oxygen atom in each .

Molar mass of dibenzyl ketone

Molar mass of dibenzyl ketone = ∑ molar masses of atoms in dibenzyl ketone

= carbon( 15 ) = 15 ( 12.0107 ) + oxygen ( 14 ) = 1 ( 15.999 ) + hydrogen(14) =14(1.00784)

= 210.26926 ≈ 210.3 g/mol

Molar mass of benzil

Molar mass of Benzil = ∑ molar masses of atoms in Benzil

= carbon( 14) = 14(12.0107) + oxygen(2) = 2 ( 15.999) + hydrogen(10) =10(1.00784)

= 210.2262 ≈ 210.2 g/mol

Molar mass of 2,3,4,5-tetraphenylcyclopentadienone

Molar mass = ∑ molar masses of atoms

= carbon ( 29) = 29(12.0107) + oxygen (1) = 1( 15.999 ) + hydrogen(20) = 20(1.00784 )

≈ 384.5 g/mol

6 0

2 years ago