Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
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
It is found that scuba diver’s can safely ascend up to 52.53 ft without Breathing out.
By what factor would a scuba diver’s lungs expand if she ascended rapidly to the surface from a depth of 125 ft without inhaling or exhaling then
=patm+
= 101325+ρ
It is given that height is 125ft. Put the value of h in above formula:
h1 =125ft=38.1m
ρ=1.04g/mL=1040kg/
g=9.81
=101325Pa+388711.44
=490036.44Pa
=p atm =101325Pa
It is known that volume and pressure can be expressed as:
V*P=const.
where, V is volume and P is pressure.
Now,
=
=
=490036.44/101325
=4.84
Assume constant temperature
d of seawater = 1.04 g/mL; d of Hg = 13.5 g/mL.
now =p atm+ =490036.44Pa
V*p=const
where, V is volume and P is pressure.
Now,
=
=
=490036.44/X
=490036.44pa/(V2/V1) =326690.96Pa
=patm +p
=101325Pa+ρ
326690.96Pa=101325Pa+ρ
ρgh1 =151987.5-101325=225365.96Pa
ρ=1,04g/mL=1040kg/m3
g=9.81
=225365.96/ρ∗g
=225365.96 / 1040∗9.81
=22.09m= 72.47ft
ΔH=
=125-72.47
=52.53ft
So she can safely ascend up to 52.53 ft without Breathing out
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Answer:
balanced
Explanation:
its balance since all the elments add up on both sides
Answer:
what would the question be?
The mass of Calcium required to complete this reaction is 4.008 g.
- Law of conservation of mass states that In a closed system, mass cannot be produced or destroyed, but it can be changed from one form to another.
- The mass of the chemical constituents before a chemical reaction is equal to the mass of the constituents after the reaction.
- In several disciplines, including chemistry, mechanics, and fluid dynamics, the idea of mass conservation is widely applied.
In the given reaction mass of product after completion of reaction is 13.614 g that means total mass of constituents before reaction should also be 13.614.
So,
mass of Ca + mass of O₂ + mass of S = mass of CaSO4
Ca + 6.400 g + 3.206 g = 13.614 g
mass of Ca = 13.614 - 9.606 = 4.008 g
Therefore, by law of conservation of mass 4.008 g of Ca is required for the completion of the reaction.
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