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11111nata11111 [884]
3 years ago
5

Two identical pots, one black and one silver, containing exactly the same amounts of water, are heated to the same temperature.

They are left uncovered and placed in the same room. After 15 minutes the water in the black pot is significantly cooler than the water in the silver pot. Which of these statements best explains this difference?
A)The black pot emits heat at a faster rate than the silver pot.
B)The black pot allows more heat to be lost through evaporation.
C)The silver pot absorbs heat from the surroundings in the room.
D)The silver pot reflects heat into the water and keeps it warm longer.
Chemistry
2 answers:
I am Lyosha [343]3 years ago
7 0
The most probable answer is: (a) <span>The black pot emits heat at a faster rate than the silver pot. It means the black pot conducts heat at a higher rate than the silver pot, that's why it cooled off faster. It also means that the silver pot retains heat better and is more of an insulator.</span>
WINSTONCH [101]3 years ago
4 0

Answer: Option (A) is the correct answer.

Explanation:

Black color absorbs heat more quickly and it also emits heat at a faster rate.

Whereas silver is one of the best conductor of heat and electricity. Therefore, it absorbs heat more quickly but it does not emit heat at a faster rate.

Therefore, we can conclude that the statement black pot emits heat at a faster rate than the silver pot best explains that black pot is significantly cooler than the water in the silver pot.

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Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ 1) H2(g)+12O2(g)⟶H2O
Bond [772]

Answer:

ΔH = -793,6 kJ

Explanation:

It is possible to obtain ΔH of this reaction using Hess's law that says you can sum the half-reactions ΔH to obtain the ΔH of the global reaction:

If half-reactions are:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ  

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ

The sum of (4) + 4×(3) - (2) - 2×(1) - 2×(5) is:

(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ

+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369,2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356,9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483,6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= <em>XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)</em>

Where ΔH is:

ΔH = -639,1 kJ -369,2 kJ -356,9 kJ +483,6 kJ +88,0 kJ

<em>ΔH = -793,6 kJ</em>

I hope it helps!

5 0
3 years ago
What should you do during a titration when you notice the indicator start to indicate the approach of the equilibrium point? Add
Leni [432]

B. Add the second reactant slower.

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Two moles of an ideal gas are placed in a container whose volume is 2.3 x 10^-3 m3. The absolute pressure of the gas is 6.9 x 10
PtichkaEL [24]

Answer:

K.E.=1.97\times 10^{-21}\ J

Explanation:

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The expression for the conversion of pressure in Pascal to pressure in atm is shown below:

P (Pa) = \frac {1}{101325} P (atm)

Given the value of pressure = 43,836 Pa

So,  

6.9\times 10^5\ Pa = \frac{6.9\times 10^5}{101325} atm

Pressure = 6.80977 atm

Volume = 2.3\times 10^{-3}\ m^3 = 2.3 L ( 1 m³ = 1000 L)

n = 2 mol

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

6.80977 atm × 2.3 L = 2 mol × 0.0821 L.atm/K.mol × T

⇒T = 95.39 K

The expression for the kinetic energy is:-

K.E.=\frac{3}{2}\times K\times T

k is Boltzmann's constant = 1.38\times 10^{-23}\ J/K

T is the temperature

So, K.E.=\frac{3}{2}\times 1.38\times 10^{-23}\times 95.39\ J

K.E.=1.97\times 10^{-21}\ J

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