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12 days ago

Miley partridge if you see this I am CJ and these hearts are for you ❤❤❤❤❤❤

2 answers:

6 0

Response: I don't mean to be impolite, but who is Miley Partridge? If she's a friend, then you must have a generous spirit

Clarification:

KiRa [2.9K]12 days ago

6 0

How adorable that is!

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What is the activation energy (in kJ/mol) of a reaction whose rate constant increases by a factor of 89 upon increasing the temp

lorasvet [2795]

15.9 KJ/mol Explanation: Given data: Temperature = T1 = 307 K, Temperature = T2 = 343 K, Gas constant R = 8.314 J/(mol • K), rate constant = k2/K1 = 89. To determine: Activation energy (in kJ/mol) = Ea =? Formula: The Arrhenius equation establishes the relationship between temperature and reaction rates. Here, in this equation, k = the rate constant, Ea = the activation energy, R = the Universal Gas Constant, T = the temperature. Solution: ln 89 = Ea / 8.314 J/mol.K * (0.0325 - 0.00291). then ln 89 = Ea / 8.314 J/mol.K * (2.95 x 10^2). Resulting in 4.488 = Ea / 8.314 J/mol.K * (2.95 x 10^2). Therefore, Ea = 4.488 * (2.95 x 10^2) / 8.314 J/mol.K which simplifies to Ea = 0.1324 / 8.314. Thus, Ea = 0.0159 and finally, Ea = 1.59 x 10^2 J/mol or 15.9 KJ/mol.

3 0

1 month ago

Parker has a toy car he has made out of plastic building blocks. He breaks it apart so he can build something different with the

Tems11 [2777]

Response:

Reasoning:

3 0

23 days ago

Read 2 more answers

In 2021, you were given a 100. g wine sample to verify its age. Using tritium dating you observe that the sample has 0.688 decay

lorasvet [2795]

Answer:

1984

Explanation:

Utilizing the equation;

0.693/t1/2 = 2.303/t log (Ao/A)

Where;

t1/2 = half-life of the radioactive isotope

t= age of the wine

Ao= initial activity of the wine

A= activity at time = t

Substituting values, we have 0.693/12.3 = 2.303/t log (5.5/0.688)

0.693/12.3 = 2.079/t

0.056 = 2.079/t

t= 2.079/0.056

t= 37 years

7 0

1 month ago

Approximately 220 million tires are discarded in the U.S. each year. These tires present a disposal problem because they take up

lions [2927]

Answer:

A total of 2667 tires are required to satisfy the annual power needs of ten homes.

Explanation:

According to the Second Law of Thermodynamics, not all energy produced when tires are incinerated can be effectively used due to losses associated with finite temperature differences. The energy obtainable from a tire when burned, measured in kilowatt-hours ( $E_{out}$ ), can be calculated using the efficiency definition:

$E_{out} = \eta \cdot E_{in}$

Where:

$\eta$ - Efficiency, which is dimensionless.

$E_{in}$ - Energy released from burning, measured in kilowatt-hours.

Taking into account $\eta = 0.5$ and $E_{in} = 75\,kWh$ , the yearly energy yield from a tire amounts to:

$E_{out} = 0.5\cdot (75\,kWh)$

$E_{out} = 37.5\,kWh$

Thus, the number of tires necessary to meet the electricity demand of ten homes for one year is:

$n = \frac{(10\,homes)\cdot \left(10000\,\frac{kWh}{home} \right)}{37.5\,\frac{kWh}{tire} }$

$n = 2666.667\,tires$

A total of 2667 tires are necessary to satisfy the annual power needs of ten homes.

8 0

1 month ago

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

1 month ago