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photoshop1234 [79]
2 years ago
8

What phases (solid, liquid, gas) are present in the unlighted candle? In the burning candle? Which phase appears to take part in

the chemical reaction?
Chemistry
2 answers:
PSYCHO15rus [73]2 years ago
8 0
Only the solid phase.
When lit all three phases of matter are present, solid, liquid and gas. The candle itself is still solid. But liquid wax pools under the flame. And the flame itself is releasing gases consisting mostly of water vapor and carbon dioxide. Many consider the flame is in another state of matter, the plasma state. It's a high energy state of matter exhibited by stars, flames and lightning.
Feliz [49]2 years ago
5 0

Answer:

As explained below.

Explanation:

  • In an unlighted, the solid phase can be seen when it's in solid shape and not ignited when it starts to burn it's said to have entered into the liquid phase or wax phase.
  • The liquid phase is to be part of the chemical reaction because the chemical reactions are seen when it is lit with the presence of oxygen and the last phase of gas is given out in the form of flames of fire upon burning as to give out carbon dioxide and ash.

You might be interested in
Americium-242 has a half-life of 6 hours. If you started with 24 g and you now have 3 g, how much time
kogti [31]
How many times has it halved?

24/2 = 12
12/2 = 6
6/2 = 3

It halved three times.
It halves once every 6 hours.

18 hours have passed.
7 0
2 years ago
A chemist adds 180.0 mL of a 1.42M sodium carbonate (Na CO,) solution to a reaction flask. Calculate the millimoles of sodium ca
svet-max [94.6K]

Answer: The millimoles of sodium carbonate the chemist has added to the flask are 256

Explanation:

Molarity is defined as the number of moles dissolved per liter of the solution.

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{milli moles of solute}}{\text{Volume of solution in ml}}     .....(1)

Molarity of BaCl_2 solution = 1.42 M

Volume of solution = 180.0 mL

Putting values in equation 1, we get:

1.42M=\frac{\text{milli moles of }BaCl_2}{180.0ml}\\\\\text{milli moles of }BaCl_2}={1.42M\times 180.0ml}=256milli mol

Thus the millimoles of sodium carbonate the chemist has added to the flask are 256.

6 0
2 years ago
(A) A chemical reaction takes place in a container of cross-sectional area 100 cm2. As a result of the reaction, a piston is pus
balandron [24]

Answer:

(A) The work done by the system is -101.325J

(B) The workdone by the system is -90.75J

Explanation:

(A) Workdone = -PΔV

Given that A = 100cm2 = 0.01m2

distance d = 10cm = 0.1m

ΔV= Area × distance

ΔV= 0.01 ×0.1

ΔV = 0.001m3

P= external pressure = 1atm = 101325Pa

Workdone = -0.001 × 101325

W= - 101.325Pa m3

1Pam3 = 1J

Therefore W = - 101.325J

The work done on the system is -101.325J

(B) Workdone = -PΔV

Given that A = 50cm2 = 0.005m2

distance d = 15cm = 0.15m

ΔV= Area × distance

ΔV= 0.005×0.15

ΔV = 0.00075m3

P=121kPa = 121000Pa

W= - 121000 × 0.00075

W= -90.75Pa m3

1Pam3 = 1J

W = - 90.75J

The woekdone by the system is -90.75J

5 0
2 years 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
A 100 mL reaction vessel initially contains 2.60×10^-2 moles of NO and 1.30×10^-2 moles of H2. At equilibrium the concentration
Sliva [168]

Answer:

<h2>The equilibrium constant Kc for this reaction is 19.4760</h2>

Explanation:

The volume of vessel used= 100 ml

Initial moles of NO= \frac{2.60}{10^2} moles

Initial moles of H2= \frac{1.30}{10^2} moles

Concentration of NO at equilibrium= 0.161M

Concentration(in M)=\frac{moles}{volume(in litre)}

Moles of NO at equilibrium= 0.161(\frac{100}{1000})

                                            =\frac{1.61}{10^2} moles

               

                    2H2 (g)        +    2NO(g) <—>    2H2O (g) +    N2 (g)

<u>Initial</u>          :1.3*10^-2          2.6*10^-2                0                   0        moles

<u>Equilibrium</u>:1.3*10^-2 - x     2.6*10^-2-x              x                   x/2     moles

∴\frac{2.60}{10^2}-x=\frac{1.61}{10^2}

⇒x=\frac{0.99}{10^2}

Kc=\frac{[H2O]^2[N2]}{[H2]^2[NO]^2} (volume of vesselin litre)

<u>Equilibrium</u>:0.31*10^-2      1.61*10^-2          0.99*10^-2        0.495*10^-2  moles

⇒Kc=\frac{(0.0099)^2(0.00495)}{(0.0031)^2(0.0161)^2}  (0.1)

⇒Kc=19.4760

3 0
2 years ago
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