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gulaghasi [49]
2 years ago
5

When a pot of water has been boiling vigorously for several minutes, the bubbles that emerge from the liquid water are composed

of
1. air
2. water vapor
3. H2 gas and O2 gas in a 1:1 ratio
4. H2 gas and O2 gas in a 2:1 ratio
Chemistry
1 answer:
kumpel [21]2 years ago
7 0
Well, water is composed H2 gas and O2 gas in a 2:1 ratio, and the bubbles that form at the bottom of the pan is just the water changing from a liquid to a gas, and im pretty sure its water vapor that comes out of the pan. I hope that helped??? :)
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The reaction N2 + 3 H2 → 2 NH3 is used to produce ammonia. When 450. g of hydrogen was reacted with nitrogen, 1575 g of ammonia
tiny-mole [99]

Answer:

The percent yield of this reaction is 70%

Explanation:

The reaction is: N₂ + 3H₂ → 2NH₃

We only have the mass of H₂, so we assume that N₂ is in excess

We convert the mass to moles, to work with the reaction:

450 g . 1mol / 2 g = 225 moles

Ratio is 2:3. 3 moles of H₂ can produce 2 moles of ammonia

Therefore 225 moles of H₂ will produce (225 .2)/ 3 = 150 moles

This is the 100% yield reaction → We convert the moles of NH₃ to mass

150 mol . 17g /1mol = 2550 g

Percent yield = (Produced yield/Theoretical yield) .100

Percent yield = (1575g/2550g) . 100 = 70%

7 0
2 years ago
During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.
Monica [59]

Answer:

0.74 grams of methane

Explanation:

The balanced equation of the combustion reaction of methane with oxygen is:

  • CH₄ + 2 O₂ → CO₂ + 2 H₂O

it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.

firstly, we need to calculate the number of moles of both

for CH₄:

number of moles = mass / molar mass = (3.00 g) /  (16.00 g/mol) = 0.1875 mol.

for O₂:

number of moles = mass / molar mass = (9.00 g) /  (32.00 g/mol) = 0.2812 mol.

  • it is clear that O₂ is the limiting reactant and methane will leftover.

using cross multiplication

1 mol of  CH₄ needs → 2 mol of O₂

???  mol of  CH₄  needs → 0.2812 mol of O₂

∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol

so 0.14 mol will react and the remaining CH₄

mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol

now we convert moles into grams

mass of CH₄ left over = no. of mol of CH₄ left over *  molar mass

                                    = 0.0469 mol * 16 g/mol = 0.7504 g

So, the right choice is 0.74 grams of methane

3 0
2 years ago
A 55.0 L steel tank at 20.0 ∘C contains acetylene gas, C2H2, at a pressure of 1.39 atm. Assuming ideal behavior, how many grams
JulijaS [17]

Answer:

PV=nRT

n = PV/RT

n = m/Mm

m/Mm = PV/RT

m = MmPV/RT

T in kelvin = T Celsius + 273.15 = 293.15 K

m = (26.04 x 1.39 x 55)/(0.08206 x 293.15)

mass in grams = 82.8 grams  

Explanation:

Ideal gases formula is PV=nRT, where:

P is the pressure (1.39 atm in this case)

V is the volume (55.0 L in this case)

R is the gas constant (0.08206 L.atm/K.mole)

T is the temperature (20.0C) should be converted to Kelvin

all the unit should correspond to the one in the R.

we also know that to find the mass, we can use number mole with the formula number of mole(n) = mass (m) divided by the molar mass (Mm). therefore we substituted that in the formula and make (m) the subject of the formula.

we found the mass to be 82.8 grams

7 0
2 years ago
Which runner has greater kinetic energy: a 46-kilogram runner moving at a speed of 8 meters per second or a 92-kilogram runner m
Lena [83]
Runner a has a greater kinetic energy
8 0
2 years ago
You can identify a metal by carefully determining its density. A 8.44 g piece of an unknown metal is 1.25 cm long, 2.50 cm wide,
Dafna1 [17]

Answer: Aluminum, 2.70 g/cm^3

Explanation:

Density is defined as the mass contained per unit volume.  It is characteristic of a substance.

Density=\frac{mass}{Volume}

Given : Mass of object = 8.44 grams

Volume of object=length\times breadth\times height=1.25cm\times 2.50cm\times 1.0cm=3.125cm^3

Putting in the values we get:

Density=\frac{8.44g}{3.125cm^3}=2.70g/cm^3

Thus density of the object will be 2.70 g/cm^3  which matches that of aluminium.

4 0
2 years ago
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