Soil covered, saturated, submerged, flooded w water, standing water
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:
The correct answer to the following question will be "4.08 × 10⁻¹⁹ Joule".
Explanation:
Given:
Wavelength, λ = 486.0 nm
As we know,
On putting the estimated values, we get
⇒ 
⇒ 
∴ 1 ev = 1.6 × 10⁻¹⁹ J
Now,
Energy, 
⇒ 
Answer:
1.) 13 g C₄H₁₀
2.) 41 g CO₂
Explanation:
To find the mass of propane (C₄H₁₀) and carbon dioxide (CO₂), you need to (1) convert mass O₂ to moles O₂ (via molar mass), then (2) convert moles O₂ to moles C₄H₁₀/CO₂ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles C₄H₁₀/CO₂ to mass C₄H₁₀/CO₂ (via molar mass). It is important to arrange the ratios in a way that allows for the cancellation of units. The final answers should have 2 sig figs to match the sig figs of the given value.
Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)
Molar Mass (C₄H₁₀): 58.124 g/mol
Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)
Molar Mass (CO₂): 44.007 g/mol
Molar Mass (O₂): 2(15.998 g/mol)
Molar Mass (O₂): 31.996 g/mol
2 C₄H₁₀ + 13 O₂ ----> 8 CO₂ + 10 H₂O
48 g O₂ 1 mole 2 moles C₄H₁₀ 58.124 g
--------------- x ----------------- x -------------------------- x ------------------ =
31.996 g 13 moles O₂ 1 mole
= 13 g C₄H₁₀
48 g O₂ 1 mole 8 moles CO₂ 44.007 g
--------------- x ----------------- x -------------------------- x ------------------ =
31.996 g 13 moles O₂ 1 mole
= 41 g CO₂
Answer:
SAMPLE A - pure substance.
SAMPLE B - homogeneous mixture.
SAMPLE C - heterogeneous mixture.
Explanation:
