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vova2212 [387]
3 years ago
15

Maya wrote if you step to describe how carbon circulates between the atmosphere and living organisms

Chemistry
1 answer:
guapka [62]3 years ago
5 0

A). Carbon enters the atmosphere as carbon dioxide from respiration and combustion.

You might be interested in
Calculating Density Warm Up
Artist 52 [7]

Answer:

2

Explanation:

33-25=8

48/8=6

6 0
3 years ago
Cu + 2AgNO3 es002-1.jpg 2Ag + Cu(NO3)2 How many moles of copper must react to form 0.854 mol Ag?
marin [14]
Balance Chemical Equation is as follow,

<span>                        Cu + 2 AgNO</span>₃     →    2 Ag + Cu(NO₃)₂

According to Balance Equation,

                   2 Moles of Ag is produced by reacting  =  1 Mole of Cu
So, 
     0.854 Moles of Ag will be produced by reacting  =  X Moles of Cu

Solving for X,
                             X  =  (0.854 mol × 1 mol) ÷ 2 mol

                             X  =  0.427 Moles of Cu
Result:
            0.854 Moles of Ag 
are produced by reacting 0.427 Moles of Cu.
4 0
3 years ago
B. For the following questions, use the reaction NO2(g) N2(g) + O2(g), with ΔH = –33.1 kJ/mol and ΔS= 63.02 J/(mol·K).
Troyanec [42]

Answer:

I. Kindly, see the attached image.

II. The reaction is exothermic.

III. - 51.88 kJ/mol.

IV. The reaction is spontaneous.

Explanation:

I. Draw a possible potential energy diagram of the reaction. Label the enthalpy of the reaction.

  • Since the sign of ΔH is negative, the reaction is exothermic reaction.

In an exothermic reaction, the energy of the reactants is higher than that of the products.

<u><em>Kindly see the attached image to show you the potential energy diagram of the reaction.</em></u>

     

<em>II. Is the reaction endothermic or exothermic? Explain your answer.</em>

  • The reaction is exothermic reaction.
  • The sign of ΔH indicates wither the reaction is endothermic or exothermic one:

If the sign is positive, the reaction is endothermic.

If the sign is negative, the reaction is exothermic.

Herein, <em>ΔH = - 33.1 kJ/mol, </em>so the reaction is exothermic.

<em>III. What is the Gibbs free energy of the reaction at 25°C? </em>

∵ ΔG = ΔH - TΔS.

Where, ΔG is the Gibbs free energy change (J/mol).

ΔH is the enthalpy change (ΔH = - 33.1 kJ/mol).

T is the temperature (T = 25°C + 273 = 298 K).

ΔS is the entorpy change (ΔS = 63.02 J/mol.K = 0.06302 J/mol.K).

<em>∴ ΔG = ΔH - TΔS</em> = (- 33.1 kJ/mol) - (298 K)(0.06302 J/mol.K) = <em>- 51.88 kJ/mol.</em>

IV. Is the reaction spontaneous or nonspontaneous at 25°C?

The sign of ΔG indicates the spontaneity of the reaction:

If ΔG < 0, the reaction is spontaneous.

If ΔG = 0, the reaction is at equilibrium.

If ΔG > 0, the reaction is nonspontaneous.

Herein, <em>ΔG = - 51.88 kJ/mol, </em>so the reaction is spontaneous.

7 0
3 years ago
The K a of propanoic acid ( C 2 H 5 COOH ) is 1.34 × 10 − 5 . Calculate the pH of the solution and the concentrations of C 2 H 5
Zigmanuir [339]

Answer:

2.62.

Explanation:

Okay let us first write the parameters in the question in question above out. We are given the ka value of propanoic acid, C2H5COOH to be equals to 1.34 × 10^- 5. Also, we are given the value for the initial concentration of propanoic acid to be 0.441 M.

So, let us delve right into the solution to the question and we will be starting by writting the equation below;

C2H5COOH <--------> H^+ + C2H5COO^-.

Please note that this Reaction is a reversible Reaction.

Therefore, the basic things about acid is its great tendency to release Hydrogen ion in an aqeous solution.

So, we will be taken equation above and correspond it with the time and Concentration.

C2H5COOH <----> H^+ C2H5COO^-.

Initial concentration of the C2H5COOH = 0.441 M and the initial concentration of H^+ and C2H5COO^- are both zero.

So, after a time, t, concentration of C2H5COOH= 0.441 - x and at that time the concentration of H^+ and C2H5COO^- are both x and x respectively.

Hence, Ka = [C2H5COO^-] [H^+]/ C2H5COOH. -----------------------(**).

Therefore, slotting in the values from above into equation (**), we have;

1.34 × 10^-5 = [x] [x]/ [0.441 - x].

1.34 × 10^-5= x^2/ [0.441 - x].

x^2 = 1.34 × 10^-5(0.441) - 1.34 × 10^-5x.

x^2 + 1.34 × 10^-5x - 5.91× 10^-6.

x = 2.4×10^-3.

Hence, the concentration of the propanoic acid at time, t= 0.441 - 2.4 ×10^-3.

==> 0.44 M.

pH = -log [H^+].

Then, we have; pH= - log[2.4× 10^-3].

pH= 2.62.

4 0
2 years ago
Trevor, Neville Longbottom’s pet toad, has a mass of 13.6 kg. Trevor jumps into a large cylinder containing water. The water lev
nasty-shy [4]

Answer:

5.44 kg/L

Explanation:

Mass = 13.6 kg

Change in water level as a result of Trevor = Volume of Trevor = 2.50 L

Density = ?

The relationship between the three parameters is given as;

Density = Mass  /  Volume

Density = 13.6 / 2.5

Density = 5.44 kg/L

8 0
3 years ago
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