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3 years ago

Express 9.39 x 109 seconds in terms of days

Physics

2 answers:

vladimir2022 [97]3 years ago

5 0

Answer:

$t = 9.39 \times 10^9 s = 108680.6 days$

Explanation:

Given time is

$t = 9.39 \times 10^9 s$

as we know that time in seconds for one day

$1 day = 24 hr\times \frac{3600 s}{1 hr}$

$1 day = 86400 s$

now we will have

$t = 9.39 \times 10^9 s\times \frac{1 day}{84600 s}$

$t = 108680.6 days$

So the given times is equivalent to 108680.6 days

Allisa [31]3 years ago

4 0

   (9.39 x 10⁹ sec) x (1 day / 8.64 x 10⁴ sec)

   = (9.39 / 8.64) x 10⁵ days

   = 108,680.56 days

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Complete Question

The complete question is shown on the first and second uploaded image

Answer:

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at

the surface is $\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = 2.5$

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Explanation:

We are told from the question that the pressure at the surface is 1 atm and for each depth of 10m below the surface the pressure increase by 1 atm

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making n the subject we have

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$n_s = \frac{P_sV}{RT}$

Substituting $1 atm = 101325 N/m^2$ for $P_s$ , $6L = 6*10^{-3}m^3$ for volume , 8.314 J/mol. K for R , (37° +273) K for T into the equation

$n_s = \frac{(1atm)(6*10^{-3} m^3)}{(8.314J/mol \cdot K)(37 +273)K}$

$= 0.2359 mol$

To obtain the number of moles at the depth of the water we use

$n_d = \frac{P_d V}{RT}$

Where $P_d \ and \ n_d \$ are pressure and no of moles at the depth of the water

Substituting values we have

$n_d = \frac{(2.5)(101325 N/m^2)(6*10^{-3}m^3)}{(8.314 J/mol \cdot K)(37 + 273)K}$

$= 0.5897 mol$

Now to obtain the number of moles released we have

$n = n_d - n_s$

$= 0.5897mol - 0.2359mol$

$=0.3538 \ mol$

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$=39.31mol/m^3$

The molar concentration at the depth of water is

$[\frac{n}{V} ]_{15m} = \frac{0.5897}{6*10^{-3}}$

$= 98.28 mol/m^3$

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at the surface is

$\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = \frac{98.28}{39.31} =2.5$

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