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8090 [49]
2 years ago
12

A sample of water with a mass of 587.00 kg is heated with 87 kJ of energy to a temperature of 518.4 K. The specific heat of wate

r is 1 J-1 kg K-1. What is the initial temperature of the water?

Chemistry
2 answers:
Tems11 [23]2 years ago
7 0

The correct specific heat capacity of water is <em>4.187 kJ/(kg.K)</em>.

q = mCΔT

ΔT = q/mC = 87 kJ/[587.00 kg x 4.187 kJ/(kg.K)] = 87 kJ/(2547.8 kJ/K)

= 0.035 K

Ti = Tf - ΔT = 518.4 K - 0.0354 K = 518.365 K


uysha [10]2 years ago
7 0

Answer:

370.2 K

Explanation:

The explanation is in the attached image

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Answer:

The partial pressure of carbon dioxide is 22.8 mmHg

Explanation:

Dalton's Law is a gas law that relates the partial pressures of the gases in a mixture. This law says that the pressure of a gas mixture is equal to the sum of the partial pressures of all the gases present.

In this case:

Ptotal=Pnitrogen + Poxygen + Pcarbondioxide

You know that:

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Replacing:

0.998 atm=0.770 atm + 0.198 atm + Pcarbondioxide

Solving:

Pcarbondioxide= 0.998 atm - 0.770 atm - 0.198 atm

Pcarbondioxide= 0.03 atm

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Pcarbondioxide=\frac{0.03 atm*760 mmHg}{1 atm}

Pcarbondioxide= 22.8 mmHg

<u><em>The partial pressure of carbon dioxide is 22.8 mmHg</em></u>

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2 years ago
a 0.5674 g piece of copper is added to 10.00 ml; of 16 M HNO3, producing 0.8024 g of copper (ii) nitrate. What is the precent yi
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The electron in a ground-state H atom absorbs a photon of wavelength 97.20 nm. To what energy level does it move?
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In Philip’s French class, the students are learning how to pronounce closed vowels and open vowels. The students are most likely
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A slender uniform rod 100.00 cm long is used as a meter stick. two parallel axes that are perpendicular to the rod are considere
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Refer to the diagram shown below.

The second axis is at the centroid of the rod.
The length of the rod is L = 100 cm = 1 m

The first axis is located at 20 cm = 0.2 m from the centroid.
Let m =  the mass of the rod.

The moment of inertia about the centroid (the 2nd axis) is
I_{g} =  \frac{mL^{2}}{12} = (m \, kg) \frac{(1 \, m)^{2}}{12} =  \frac{m}{12} \, kg-m^{2}

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I_{1} = I_{g} + (m \, kg)(0.2 m)^{2} \\ I_{1} =  \frac{m}{12}+ 0.04m = 0.1233m \, kg-m^{2}

The ratio of the moment of inertia through the 2nd axis (centroid) to that through the 1st axis is
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