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12 days ago

compare and contrast melting 10 kg of ice freezing 1 kg of water. be sure to address temperature,heat flow, and thermal energy

1 answer:

6 0

<span>Melting is a process that absorbs heat (endothermic), while freezing is one that releases heat (exothermic), or it can be wrongly perceived in thermodynamic terms as "absorbing cold." The standard enthalpy of fusion for water applies to both processes, but remember standard enthalpy is measured in energy per mass, so melting 10 kg of ice will require 10 times the energy compared to freezing 1 kg of water. For both cases, assuming the water is pure and the pressure remains at standard atmospheric levels, the temperature of the solid and the liquid will stabilize at exactly 0 degrees Celsius (273 K) during the entire phase transition.</span>

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A laboratory analysis of an unknown sample yields 74.0% carbon, 7.4% hydrogen, 8.6% nitrogen, and 10.0% oxygen. What is the empi

Tems11 [2624]

Assuming we have a 100g sample, the mass of each element is as follows:
C: 74 g
H: 7.4 g
N: 8.6 g
O: 10 g
Next, we calculate the moles of each by dividing the mass of each element by its molar mass:
C: (74 / 12) = 6.17
H: (7.4 / 1) = 7.4
N: (8.6 / 14) = 0.61
O: (10 / 16) = 0.625
Now, we take the smallest value to determine the ratio:
C: 10
H: 12
N: 1
O: 1
Thus, the empirical formula can be expressed as
C10H12NO

3 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 [2782]

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

Given the connection between Aw and K (Aw=2k) could you use the ideal gas law and derive the Boltzmann constant. Water freezes a

lions [2782]

Answer:

Explanation:

The relationship between the new temperature scale and the absolute temperature scale is defined as follows

Aw = 2 K

for K = 273.15 (the freezing point of water on the absolute scale)

Aw = 2 x 273.15 = 546.3 K

Each division of the new scale is equivalent to half that of each division on the absolute scale

each division of the new scale is minimal.

The value of R = 8.314 J per mole per K

Here, per K corresponds to 2Aw

Hence, the value of R in the new scale = 8.314/2 J per mole per Aw

= 4.157 J per mole per Aw

k = R / N

= 4.157 / 6.02 x 10²³

= .69 x 10⁻²³

= 6.9 x 10⁻²⁴ J per molecule per Aw .

7 0

1 month ago

Read 2 more answers

eduard [2645]

1) Calcium carbonate comprises 40.0% calcium by weight.
M(CaCO₃)=100.1 g/mol
M(Ca)=40.1 g/mol
w(Ca)=40.1/100.1=0.400 (which is 40.0%)!

2) The mass fraction mentioned is superfluous information.

3) The resulting solution is:

m(Ca)=1.2 g

m(CaCO₃)=M(CaCO₃)*m(Ca)/M(Ca)

m(CaCO₃)=100.1g/mol*1.2g/40.1g/mol=3.0 g

4 0

1 month ago

Read 2 more answers

What are the respective central-metal oxidation state, coordination number, and overall charge on the complex ion in Na2[Cr(NH3)

lions [2782]

Answer:

Central metal oxidation state: +2

Coordination number: 6

Overall charge: -2

Explanation:

For the ion complex:

Na₂[Cr(NH₃)₂(NCS)₄]

The central metal is chromium, with NH₃ and NCS as the ligands.

NH₃ acts as a neutral ligand, while NCS carries a negative charge.

The entire complex has a charge of:

2Na⁺ + [Cr(NH₃)₂(NCS)₄]⁻² → -2

Since each NCS contributes -1 and there are four NCS, the Cr must possess an oxidation state of +2 to achieve an overall charge of -2.

With 2 NH₃ and 4 NCS attached, the coordination number sums to 2+4 = 6

I trust this clarifies the matter!

6 0

1 month ago