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gayaneshka [121]
1 year ago
7

Which statement describes the transfer of heat energy that occurs when an ice cube is added to an insulated container with 100 m

illiliters of
water at 25°C?
(1) Both the ice cube and the water lose heat energy.
(2) Both the ice cube and the water gain heat energy.
(3) The ice cube gains heat energy and the water loses heat energy.
(4) The ice cube loses heat energy and the water gains heat energy.
Chemistry
1 answer:
ICE Princess25 [194]1 year ago
4 0
<span>The answer to this question would be: (3) The ice cube gains heat energy and the water loses heat energy.

Based on the law of conservation of energy, the energy in an isolated system should be constant. If something receives energy, other must be losing energy. The option 1 and 2  definitely false because the total energy is not constant.
In this case, the ice should have lower heat energy, so the ice should be the one who receives energy from the water</span>
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At high temperature, 2.00 mol of HBr was placed in a 4.00 L container where it decomposed in the reaction: 2HBr(g) H2(g) Br2(g)
viktelen [127]

Answer: K_c for this reaction at this temperature is 0.029

Explanation:

Moles of  HBr = 2.00 mole

Volume of solution = 4.00 L

Initial concentration of HBr=\frac{moles}{Volume}=\frac{2.00}{4.00L}=0.500M

The given balanced equilibrium reaction is,

                            2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial conc.              0.500 M              0  M        0 M  

At eqm. conc.            (0.500-2x) M   (x) M   (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[H_2\times [Br_2]}{[HBr]^2}

Equilibrium concentration of [Br_2] = x =  0.0955 M

Now put all the given values in this expression, we get :

K_c=\frac{0.0955\times 0.0955}{0.500-2\times 0.0955}

K_c=0.029

Thus K_c for this reaction at this temperature is 0.029

7 0
2 years ago
Like all equilibrium constants, Kw varies somewhat with temperature. Given that Kw is 3.31 × 10−13 at some temperature, compute
Artyom0805 [142]
Since Kw= [H⁺][OH⁻], and the concentration of both substances are the same, the equation is now Kw=[H⁺]²
So,
3.31x10⁻¹³ = [H⁺]²
Take the square root= 5.75x10⁻⁷
Then take the negative log to find the pH:
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5 0
1 year ago
In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for
joja [24]

Answer:

ΔG° = -118x10³ J/mol

Explanation:

The two half-reactions in the cell are:

Oxidation half-reaction:

Co(s) → Co²⁺(aq) + 2e⁻; E° = -0,28V

Reduction half-reaction:

Cu²⁺(aq)+2e⁻ → Cu(s); E° = 0,34V

The E° of the cell is defined as:

E_{cell} = E_{red} - E_{ox}

Replacing:

0,34V - (-0,28V) = 0,62V

It is possible to obtain the keq from E°cell with Nernst equation thus:

nE°cell/0,0592 = log (keq)

Where:

E°cell is standard electrode potential (0.62 V)

n is number of electrons transferred (2 electrons, from the half-reactions)

Replacing:

0,62V×2/0,0592 = log (keq)

20,946 = log keq

keq = 8,83x10²⁰≈ 5,88x10²⁰

ΔG° is defined as:

ΔG° = -RT ln Keq

Where R is gas constant (8,314472 J/molK) and T is temperature (298K):

ΔG° = -8,314472 J/molK×298K ln5,88x10²⁰

<em>ΔG° = -118x10³ J/mol</em>

<em />

I hope it helps!

8 0
2 years ago
(f) what is the observed rotation of 100 ml of a solution that contains 0.01 mole of d and 0.005 mole of l? (assume a 1-dm path
never [62]
<span>Answer: .01 moles of D to .005 moles of L ~ so, .01+.005 = .015 total; using this total value, divide the portions of D and L. so .01/.015 to .005/.015 ~ 67% D to 33% L. And thus, the enantiomer excess will be 34%.</span>
4 0
2 years ago
Read 2 more answers
Which species does not have a noble gas electron
lara31 [8.8K]
S, sulfur does not have a noble gas electron.
6 0
2 years ago
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