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lara31
2 months ago
14

A 126-gram sample of titanium metal is heated from 20.0°C to 45.4°C while absorbing 1.68 kJ of heat. What is the specific heat o

f titanium?
Chemistry
1 answer:
lorasvet [2.7K]2 months ago
3 0
The specific heat of titanium metal is 0.524 J/g°C. Given that Q = 1.68 kJ, which equates to 1680 Joules, with a mass of 126 grams and initial and final temperatures of 20°C and 45.4°C respectively, the specific heat is computed using the formula Q = (mass)(ΔT)(Cp), where ΔT = T₂ - T₁ = 25.4°C. Plugging in the numbers leads us to Cp = 0.524 J/g°C.
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Primordial swamps decomposing under ancient seas and tons of rock layers gave rise to an important fuel used today. That fuel is
lorasvet [2795]
I think the right choice is C. Coal, as it's utilized in making a multitude of products across the globe. I trust this information is useful to you.:)
8 0
2 months ago
Read 2 more answers
PART A: Use the following glycolytic reaction to answer the question. If the concentration of DHAP is 0.125 M and the concentrat
alisha [2963]

Answer:

For A: The change in free energy for the reaction is -5339.76 J/mol

For B: Free energy change is expressed in kJ/mol

For C: The forward reaction favors progression, while the reverse reaction does not.

Explanation:

Regarding the specified chemical reaction:

DHAP\rightleftharpoons G_3P

  • For A:

The relationship between standard Gibbs free energy and equilibrium constant is as follows:

\Delta G^o=-RT\ln K_{eq}

The free energy change can be calculated using the following equation:

\Delta G=\Delta G^o+RT\ln Q

Or,

\Delta G=-RT^o\ln K_{eq}+RT\ln Q

where,

\Delta G = Change in free energy

R = Gas constant = 8.314J/K mol

T^o = standard temperature = 25^oC=[273+25]K=298K

T = temperature of the cell = 37^oC=[273+37]K=310K

K_[eq} = equilibrium constant = 5.4\times 10^{-2}

Q = reaction quotient = \frac{[G_3P]}{[DHAP]}

[G_3P] = 0.06 M

[DHAP] = 0.125 M

Substituting the values into the equation yields:

\Delta G=[-(8.314J/mol.K\times 298K\times \ln (5.4\times 10^{-2}))]+[(8.314J/mol.K\times 310K\times \ln (\frac{0.06}{0.125}))]\\\\\Delta G=-[-7231.46]+[-1891.7]=-5339.76J/mol

Thus, the change in free energy for the reaction is -5339.76 J/mol

  • For B:

To convert the free energy change to kilojoules, we apply the conversion factor:

1 kJ = 1000 J

So, -5339.76J/mol\times \frac{1kJ}{1000J}=-5.34kJ/mol

Consequently, the free energy change's units are kJ/mol

  • For C:

For spontaneity in the reaction, the Gibbs free energy must be negative. However, the calculations indicate a positive Gibbs free energy, leading to the conclusion that the reaction is not spontaneous.

The free energy change of the reaction is negative.

Consequently, the forward reaction is favored and the reverse reaction is not favored.

8 0
2 months ago
Los automóviles actuales tienen “parachoques de 5 mi/h (8 km/h)” diseñados para comprimirse y rebotar elásticamente sin ningún d
KiRa [2933]

Response: k = 23045 N/m

Clarification:

To determine the spring constant, one must consider the maximum elastic potential energy that the spring can withstand. The kinetic energy of the vehicle should equal at minimum the elastic potential energy of the spring when it is fully compressed. Hence, we express it as:

K=U\\\\\frac{1}{2}Mv^2=\frac{1}{2}kx^2    (1)

M: mass of the vehicle = 1050 kg

k: spring constant =?

v: car speed = 8 km/h

x: maximum spring compression = 1.5 cm = 0.015m

You need to resolve equation (1) for k. Beforehand, convert the speed v to meters per second:

v=8\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=2.222\frac{m}{s}

k=\frac{Mv^2}{x^2}=\frac{(1050kg)(2.222m/s)^2}{(0.015m)^2}=23045\frac{N}{m}

The spring constant calculates to 23045 N/m

3 0
2 months ago
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