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GarryVolchara [31]
3 years ago
7

Does the identity of gas matter when predicting its behavior why or why not?

Chemistry
1 answer:
just olya [345]3 years ago
4 0

Answer: The gas phase is unique among the three states of matter in that there are some simple models we can use to predict the physical behavior of all gases—independent of their identities. We cannot do this for the solid and liquid states. ... Gas particles do not experience any force of attraction or repulsion with each other.

Explanation:

You might be interested in
The reaction described by H2(g)+I2(g)⟶2HI(g) has an experimentally determined rate law of rate=k[H2][I2] Some proposed mechanism
MatroZZZ [7]

Answer:

Mechanism A and B are consistent with observed rate law

Mechanism A is consistent with the observation of J. H. Sullivan

Explanation:

In a mechanism of a reaction, the rate is determinated by the slow step of the mechanism.

In the proposed mechanisms:

Mechanism A

(1) H2(g)+I2(g)→2HI(g)(one-step reaction)

Mechanism B

(1) I2(g)⇄2I(g)(fast, equilibrium)

(2) H2(g)+2I(g)→2HI(g) (slow)

Mechanism C

(1) I2(g) ⇄ 2I(g)(fast, equilibrium)

(2) I(g)+H2(g) ⇄ HI(g)+H(g) (slow)

(3) H(g)+I(g)→HI(g) (fast)

The rate laws are:

A: rate = k₁ [H2] [I2]

B: rate = k₂ [H2] [I]²

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]

<em>Where K' = K1 * K2</em>

C: rate = k₁ [H2] [I]

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]^1/2

Thus, just <em>mechanism A and B are consistent with observed rate law</em>

In the equilibrium of B, you can see the I-I bond is broken in a fast equilibrium (That means the rupture of the bond is not a determinating step in the reaction), but in mechanism A, the fast rupture of I-I bond could increase in a big way the rate of the reaction. Thus, just <em>mechanism A is consistent with the observation of J. H. Sullivan</em>

5 0
3 years ago
A 25.0 mL sample of a solution of an unknown compound is titrated with a 0.115 M NaOH solution. The titration curve above was ob
Mnenie [13.5K]

Answer:

Weak acid

Explanation:

A titration curve is a graphical description of the change in pH of the solution in the conical flask as the reagent is added from the burette. A titration curve can be plotted for the different kinds of acid and base titrations. The volume of the titrant is always plotted as the independent variable and the pH of the solution as the dependent variable. The equivalence point is read off from the titration curve. A titration curve is very important because it shows the pH at various points during the titration.

A weak acid/strong base titration leads to an equivalence point above 7. From the question, we were told that the pH at equivalence point lies around 8. Hence the unknown substance must be a weak acid.

3 0
3 years ago
How many grams of oxygen are required to burn 60 grams of ethane gas, C2H6? 2 C2H6 (g) + 7 O2(g) → 4 CO2 (g) + 6 H2O (g)
timofeeve [1]
60 g C2H6 × 1 mol C2H6 x   7 mol O2     x     32 g O2   =      ~223.5 g O2
                        30.068 g      2 mol C2H6        1 mol O2
8 0
3 years ago
When 7.80 mL of 0.500 M AgNO3 is added to 6.25 mL of 0.300 M NH4Cl, how many grams of AgCl are formed?
irina1246 [14]

Answer:

The answer to your question is 0.269 grams of AgCl

Explanation:

Data

[AgNO₃] = 0.50 M

Vol AgNO₃ = 7.80 ml

[NH₄Cl] = 0.30 M

Vol NH₄Cl = 6.25 ml

mass of AgCL

Balanced reaction

                 AgNO₃(aq)  +  NH₄Cl(aq)   ⇒   AgCl (s) + NH₄NO₃ (aq)

Process

1.- Calculate the moles of AgNO₃

Molarity = moles / volume

moles = Molarity x volume

moles = 0.50 x 0.0078

moles = 0.0039

2.- Calculate the moles of NH₄Cl

moles = 0.30 x 0.0063

moles = 0.00188

3.- Calculate the limiting reactant

The proportion of     AgNO₃(aq)  to  NH₄Cl(aq) is 1 :1, then, we conclude that the limiting reactant is NH₄Cl(aq), because there are less amount of this reactant in the experiment.

4.- Calculate the moles of AgCl

                     1 mol of NH₄Cl  ---------------- 1 mol of AgCl

              0.00188 mol of NH₄Cl ------------- x

                     x = (0.00188 x 1) /1

                     x = 0.00188 moles of AgCl

5.- Calculate the grams of AgCl

molecular mass of AgCl = 108 + 35.5 = 143.5 g

                         143.5 grams of AgCl -------------- 1 mol

                         x -------------------------------------------0.00188 moles of AgCl

                          x = (0.00188 x 143.5) / 1

                          x = 0.269 grams of AgCl

8 0
3 years ago
How many milliliters of 0.0510 m edta are required to react with 50.0 ml of 0.0200 m cu2 ?
krok68 [10]
Cu  ions plus EDTA2-   ->cu(EDTA)2-  plus 2H-
   number  of  moles  of  CU  ions  used  which is   equal to  molarity  multiplied  by    volume  in  litres
that  is  50xo.o2  divided  by 1000  that  is  0.001moles
Since  ratio  is  1:1  the  moles  of EDTA  is  also0.001moles
volume  of  EDTA  is  0.001  divided  by  0.0510m  which  is  0.0196  litres in  ml is  0.0196x1000  which  is  19.61ml
4 0
3 years ago
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