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Kitty [74]
3 years ago
11

A 100g sample of a metal was heated to 100oC and then quickly transferred to an insulated container holding 100g of water at 22o

C. The temperature of the water rose to reach a final temperature of 35oC. Which of the following can be concluded?
A) The metal temperature changed more than the water temperature did; therefore the metal
lost more thermal energy than the water gained.
B) The metal temperature changed more than the water temperature did, but the metal lost
the same amount of thermal energy as the water gained.
C) The metal temperature changed more than the water temperature did; therefore the heat
capacity of the metal must be greater than the heat capacity of the water.
D) The final temperature is less than the average starting temperature of the metal and the
water; therefore the total energy of the metal and water decreased.
Chemistry
1 answer:
Lady_Fox [76]3 years ago
4 0

Answer:

B) The metal temperature changed more than the water temperature did, but the metal lost

the same amount of thermal energy as the water gained.

Explanation:

Heat capacity or thermal capacity is defined as the amount of heat required by a given mass of a material to raise its temperature by one unit which means that the heat capacity of the water, that is the quantity of heat required to cause a rise from 22°C to 35°C that is a rise of 13°C is the quantity of heat that caused the drop in temperature of the metal from 100°C to 35°C a change of 65°C

The water has more capacity to absorb heat or a higher heat capacity than the metal

However, the first law of thermodynamics states that energy is neither created nor destroyed, but it changes from one form to another. In this case, the thermal energy lost by the metal is the same as the thermal or heat energy gained by the water

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½O2(g) + H2(g) ⇌ H2O(g)
ira [324]

Answer:

-241.826 kJ·mol⁻¹;  -146.9 J·K⁻¹mol⁻¹; 664.6 J·K⁻¹mol⁻¹; spontaneous

Explanation:

                        ½O₂(g)   +  H₂(g) ⟶ H₂O(g)

ΔHf°/kJ·mol⁻¹:      0                0        -241.826

S°/J·K⁻¹mol⁻¹:   205.0         130.6       188.7

1. ΔᵣH

ΔᵣH = products -reactants = -241.826 -(0 + 0) = -241.826 kJ·mol⁻¹

2. ΔᵣS

ΔᵣS = products - reactants = 188.7 - (205.0 + 130.6) = 188.7 - 335.6 = -146.9 J·K⁻¹mol⁻¹

3. ΔS(univ)

\begin{array}{rcl}\Delta S_{\text{univ}} &=& \Delta S_{\text{sys}}  +\Delta S_{\text{surr}}\\\\ &=& \Delta S_{\text{sys}}  -\dfrac{\Delta H_{\text{sys}}}{T}\\\\& = & -146.9 - \dfrac{-241826}{298}\\\\& = & -146.9 + 811.5\\& = & \mathbf{664.6 \,\, J\cdot K^{-1}mol^{-1}}\\\end{array}

4. Spontaneity

\begin{array}{rcl}\Delta G &=& \Delta H - T\Delta S\\& = & -241.826 - 298 \times (-0.1469)\\& = & -241.826 + 43.776\\& = &  \textbf{-198.050 kJ}\cdot\textbf{mol}^{\mathbf{-1}}\\\end{array}

ΔG is negative, so the reaction is spontaneous.

4 0
3 years ago
How many significant figures<br> are in this number?<br> 107.051
Anestetic [448]

Answer:

there are 6 significant figures in 107.051

7 0
3 years ago
Read 2 more answers
Why is performing extraction with several small portions of a solvent more officient than a single extraction with the same tota
elena55 [62]

With various extractions the amount of material left in the trash will be lower, ergo the extraction will be more perfect. Various extractions with fewer amounts of solvent are more efficient than a single extraction with a huge amount of solvent.

<u>Explanation:</u>

Surely multiple extractions are better than the single large extraction. Because extraction is about maximizing outside field communication between the two solvents, and you easily get more surface area contact with fewer amounts.

You can merge two smaller portions quicker and more completely than with large portions.

6 0
3 years ago
A 0.80 L sample of gas has a temperature of 27°C and a pressure of 0.925 atm. How many moles of gas are present?
ololo11 [35]

Answer:

n= 0.03 moles

Explanation:

Using the ideal gas law:

PV=nRT

nRT=PV

n= PV/RT

 n: moles

 P: pressure in atm

 V= volume in L

 R= Avogadro's constant = 0.0821

 T= Temperature in K => ºC+273.15

n= (0.925 atm)(0.80 L) / (0.0821)(300.15 K)

n= 0.03 moles

6 0
3 years ago
Two spherical objects with a mass of 7.73 kg each are placed at a distance of 1.11 m apart. how many electrons need to leave eac
aliina [53]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Equate the gravitational force to the electrostatic force: 

<span>KC²/D² = Gm²/D² → C = m√[G/K] = 7.6√[6.67E-11/9E9] = 6.54E-10 coulombs </span>

<span>Number of electrons N = 6.24E18*C = 4.083E9 electrons</span>

4 0
3 years ago
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