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algol13
6 days ago
8

An old Magi cube camera flash bulb (1960s) used Mg metal sealed in bulb with oxygen. Calculate ∆G for its reaction Mg + 1/2 O2=

MgO. where S° Mg= 32.7, 1/2 O2= 205.0, MgO= 26.9 J/mol/K, ΔΗf° -601.2 kJ/mol
​
Chemistry
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Sea water's density can be calculated as a function of the compressibility, B, where p = po exp[(p - Patm)/B]. Calculate the pre
lions [2927]

Answer:

A pressure of 137.14 MPa exists 10,000 m beneath the ocean surface.

At this same depth, the density measures 2039 kg/m3.

Explanation:

P0 and ρ0 symbolize the pressure and density at sea level (indicative of atmospheric conditions). With an increase in ocean depth, both pressure and density likewise rise.

The relationship between pressure and density can be expressed as:

\frac{dP}{dy}=\rho*g=\rho_0*g*e^{(P-P_0)/\beta\\\\

By rearranging

\frac{dP}{e^{(P-P_0)/\beta}}= \rho_0*g*dy\\\\\int\limits^{P}_{P_0} {e^{-(P-P_0)/\beta}}dP =\int\limits^y_0 {\rho_0*g*dy}\\\\(-\beta*e^{-(P-P_0)/\beta})-(\beta*e^0)=\rho_0*g*(y-0)\\\\-\beta*(e^{-(P-P_0)/\beta}-1)=\rho_0*g*y\\\\e^{-(P-P_0)/\beta}=1-\frac{\rho_0*g*y}{\beta}\\\\-\frac{P-P_0}{\beta} =ln(1-\frac{\rho_0*g*y}{\beta})\\\\P-P_0=-\beta*ln(1-\frac{\rho_0*g*y}{\beta})\\

This equation allows for computation of P at 10,000 m beneath the ocean's surface:

P-P_0=-\beta*ln(1-\frac{\rho_0*g*y}{\beta})\\\\P-P_0=-200MPa*ln(1-\frac{1027kg/m^3*9.81m/s^2*10,000m}{200MPa})\\\\P-P_0=-200MPa*ln(1-\frac{1027*9.81*10,000Pa}{200*10^6Pa})\\\\P-P_0=-200MPa*ln(1-0.5037)\\\\P-P_0=-200MPa*(-0.6857)=137.14MPa

The density found at a depth of 10,000 m in the ocean is

\rho=\rho_0*e^{(P-P_0)/\beta}\\\rho=1027kg/m^3*e^{(137.14/200)}=1027*e^{0.686}kg/m^3\\\rho=1027*1.985 kg/m^3\\\rho=2039\,kg/m^3

4 0
3 months ago
How is the periodic law demonstrated in halogens
Anarel [2989]
Halogens are characterized as a group of 7 on the periodic table. Each of these elements has 7 valence electrons, needing just one additional electron to fill their outer shell, resulting in their high reactivity. Moreover, the periodic law indicates that when the elements are arranged by increasing atomic number, similar patterns in their properties emerge. Thus, halogens exhibit comparable behaviors during chemical reactions with one another, which is indeed accurate.
3 0
3 months ago
Read 2 more answers
Assume the weight of an average adult is 70. kg, and that 420. kJ of heat are evolved per mole of oxygen consumed as a result of
castortr0y [3046]
The temperature difference after 3 hours is 5.16 K. Given that the moles of O₂ inhaled rate at 0.02 mole/min, which converts to 1.2 mole/hour, we know the average heat released during metabolism is 7.2 kJ/h·kg. Therefore, the amount of heat generated within 3 hours will be 7.2 kJ/h·kg multiplied by 3 hours, giving a result of 21.6 kJ/kg, or 21.6 x 10³ J/kg. Applying the formula Qp = Cp x ΔT, and taking the body's heat capacity to be 4.18 J/g·K, we find ΔT = 5.16 K.
6 0
3 months ago
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