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4 years ago

When water reaches its boiling point and turns into water vapor what happens to its molecular structure

1 answer:

8 0

When water reaches its boiling point and turns into water vapor, the molecular structure of water remains the same. It is only the state of the substance that is changed in this process. I hope this helps you

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10 In a bond between an atom of carbon and an atom of fluorine, the fluorine atom has a

Oksanka [162]

The best answer is (2) stronger attraction for electrons, for the fluorine atom has a higher electronegativity than the carbon one, if not highest of all nonmetals.
Hope this helps~

4 0

3 years ago

Calculate the number of pounds of CO2CO2 released into the atmosphere when a 22.0 gallon22.0 gallon tank of gasoline is burned i

Gnesinka [82]

Answer:

391.28771 pounds of carbon-dioxide was released into the atmosphere when 22.0 gallon tank of gasoline is burned in an automobile engine.

Explanation:

Density of the gasoline ,d= 0.692 g/mL

Volume of gasoline in an tanks,V = 22.0 gallons = 83,279.02 mL

Let mass of the gasoline be M

$Density= \frac{Mass}{Volume}$

M = V × d = 83,279.02 mL × 0.692 g/mL=57,629.081 g

Given that gasoline is primarily octane.

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

Mass of octane burnt in the tank = M = 57,629.081 g

Moles of octane = $\frac{57,629.081 g}{114.08g/mol}=505.1637 mol$

According to reaction, 2 moles of octane gives 16 moles of carbon-dioxide.

Then 505.1637 mol of octane will give:

$\frac{16}{2}\times 505.1637 mol=4,041.3100 mol$ of carbon-dioxide

Mass of 4,041.3100 mol of carbon-dioxide:

4,041.3100 mol × 44.01 g/mol = 177,858.05 g

Mass of carbon-dioxide produced in pounds = 391.28771 pounds

391.28771 pounds of carbon-dioxide was released into the atmosphere when 22.0 gallon tank of gasoline is burned in an automobile engine.

3 0

3 years ago

A vessel contained N2, Ar, He, and Ne. The total pressure in the vessel was 987 torr. The partial pressures of nitrogen, argon,

xz_007 [3.2K]

Answer:

The partial pressure of neon in the vessel was 239 torr.

Explanation:

In all cases involving gas mixtures, the total gas pressure is related to the partial pressures, that is, the pressures of the individual gaseous components of the mixture. Put simply, the partial pressure of a gas is the pressure it exerts on a mixture of gases.

Dalton's law states that the total pressure of a mixture of gases is equal to the sum of the pressures that each gas would exert if it were alone. Then:

PT= P1 + P2 + P3 + P4…+ Pn

where n is the amount of gases present in the mixture.

In this case:

PT=PN₂ + PAr + PHe + PNe

where:

PT= 987 torr
PN₂= 44 torr
PAr= 486 torr
PHe= 218 torr
PNe= ?

Replacing:

987 torr= 44 torr + 486 torr + 218 torr + PNe

Solving:

987 torr= 748 torr + PNe

PNe= 987 torr - 748 torr

PNe= 239 torr

The partial pressure of neon in the vessel was 239 torr.

4 0

3 years ago

When 125.0 g of ethylene (C2H4) burns in 60.0 grams of oxygen to give carbon dioxide and water, how many grams of CO2 are formed

gizmo_the_mogwai [7]

Answer:

66 g of CO₂

Solution:

The Balance Chemical Reaction is as follow,

C₂H₂ + 5/2 O₂ → 2 CO₂ + H₂O

Or,

2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O ------- (1)

Step 1: Find out the limiting reagent as;

According to Equation 1,

56.1 g (2 mole) C₂H₂ reacts with = 160 g (5 moles) of O₂

So,

125 g of C₂H₂ will react with = X g of O₂

Solving for X,

X = (125 g × 160 g) ÷ 56.1 g

X = 356.5 g of O₂

It means for total combustion of Ethylene we require 356.5 g of O₂, but we are only provided with 60.0 g of O₂. Therefore, O₂ is the limiting reagent and will control the yield.

Step 2: Calculate Amount of CO₂ produced as;

According to Equation 1,

160 g (5 mole) O₂ produces = 176 g (4 moles) of CO₂

So,

60.0 g of O₂ will produce = X g of CO₂

Solving for X,

X = (60.0 g × 176 g) ÷ 160 g

X = 66 g of CO₂

8 0

3 years ago

Calcium carbonate decomposes according to the following equation:

Bad White [126]

Answer:

CaCO3----------CaO+CO2

(40g+12g+16g×3) of CaCO3 produce (40g+16g) of CaO

100g of CaCO3 -----------------56g of CaO

20g of CaCO3-------------------x grams of CaO

x=(20g×56g)÷100g

x=1120gg÷100g

x=11.2g of CaO

Therefore, 11.2grams of CaO from 20grams of CaCO3

HOPE IT HELPED, PLEASE MARK BRAINLIEST

8 0

2 years ago