The color of light emitted during a flame test depends on

Suppose Gabor, a scuba diver, is at a depth of 15m. Assume that: The air pressure in his air tract is the same as the net water

s2008m [1.1K]

Complete Question

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

Answer:

a

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

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

b

The number of moles of gas that must be released is $n= 0.3538\ mols$

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

This means that the pressure at the depth of the surface would be

$P_d = [\frac{15m}{10m} ] (1 atm) + 1 atm$

$= 2.5 atm$

The ideal gas equation is mathematically represented as

$PV = nRT$

Where P is pressure at the surface

V is the volume

R is the gas constant = 8.314 J/mol. K

making n the subject we have

$n = \frac{PV}{RT}$

Considering at the surface of the water the number of moles at the surface would be

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

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

$n = n_d - n_s$

$= 0.5897mol - 0.2359mol$

$=0.3538 \ mol$

The molar concentration at the surface of water is

$[\frac{n}{V} ]_{surface} = \frac{0.2359mol}{6*10^-3m^3}$

$=39.31mol/m^3$

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

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

6 0

3 years ago

A 91.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.73 r

KIM [24]

Answer:

$\omega_f = 1.08 rad/s$

Explanation:

As we know that there is no external torque on the system of platform, banana and the monkey

so the angular momentum of the system will remains conserved

so here we can say

$L_i = L_f$

$I_o\omega = (I_o + m_1r_1^2 + m_2r_2^2)\omega_f$

here we know that

$I_o = \frac{1}{2}mR^2$

$I_o = \frac{1}{2}(91.1)(1.61)^2 = 118.1 kg m^2$

$m_1 = 9.41 kg$

$r_1 = \frac{4}{5}(1.61) = 1.29 m$

$m_2 = 21.1 kg$

Now from above equation

$118.1(1.73) = (118.1 + 9.41(1.29^2) + 21.1(1.61^2))\omega_f$

$118.1(1.73) = 188.45\omega_f$

$\omega_f = 1.08 rad/s$

8 0

3 years ago

In a summer storm, the wind is blowing at a velocity of 8 m/s north. Suddenly in 3 seconds, the winds velocity is 23 m/s north.

cricket20 [7]

Answer:a=v-u/t

=23-8/3

=5m/s hope you got your answer

Explanation:

8 0

3 years ago

Kisha is trying to determine if a cell is a plant or animal cell. She observes the cell under the microscope and describes some

Ann [662]

There are organelles in the plant cell that the animal cell does not have. Plant cells have rigid cell walls and animal cells have flexible cell membrane. Plant cells contains chloroplasts, which is something animal cells don't have. Also, large, liquid filled vacuoles can only be found in plant cells, and they can occupy up to 90% of a cell's volume. By looking at the cell and see if they have any of these characteristics determines if it is plant cell or animal cell.

hope this helps.

6 0

3 years ago

* A 5 kg ball is dropped from a height of 20 m. How much kinetic energy will it have 10 m above the ground? a. 490 J b. 392 J C.

Oksi-84 [34.3K]

Answer:

A. 490

Explanation:

soln

mass = m = 5kg

Height = h = 10m

Acceleration due to gravity = g = 9.8ms²

K.E = 1/2 × mass × (velocity)²

Recall from equations of motion

v² = u² + 2gh

Therefore,

K.E = 1/2 × mass × ( u² + 2gh)

K.E = 1/2 × 5 × ( 0² + 2×10×9.8)

K.E = 1/2 × 5 × 196

K.E = 1/2 × 980

K.E = 490 Joules

6 0

3 years ago