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Bess [88]
2 years ago
13

Given the following UNBALANCED reaction: NH3 (g) <--> N2 (g) + H2 (g) If 1

Chemistry
1 answer:
Yakvenalex [24]2 years ago
8 0

Answer:

C. 1.35

Explanation:

                                                     2NH3 (g) <-->          N2 (g) +             3H2 (g)

Initial concentration                2.2 mol/0.95L       1.1 mol/0.95L           0

change in concentration        2x                             x                           3x

                                                 -0.84 M                  +0.42M                +1.26M

Equilibrium                       1.4 mol/0.95L=1.47M        1.58 M                   1.26 M

concentration

Change in concentration(NH3) = (2.2-1.4)mol/0.95 L = 0.84M

Equilibrium concentration (N2) = 1.1/0.95 +0.42=1.58 M

Equilibrium concentration(NH3) = 1.4/0.95 = 1.47M

K = [N2]*{H2]/[NH3] = 1.58M*1.26M/1.47M = 1.35 M

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How many moles are in 68.5g of KOH?
Dafna11 [192]

Answer: It has to be 56.10564

Explanation:1 grams KOH to mol = 0.01782 mol

10 grams KOH to mol = 0.17824 mol

20 grams KOH to mol = 0.35647 mol

30 grams KOH to mol = 0.53471 mol

40 grams KOH to mol = 0.71294 mol

50 grams KOH to mol = 0.89118 mol

100 grams KOH to mol = 1.78235 mol

200 grams KOH to mol = 3.5647

8 0
3 years ago
Read 2 more answers
How many protons are in nitrogen-15?
valentina_108 [34]
7 protons. As well as in nitrogen-14. The difference between nitrogen-14 and nitrogen-15 is in a number of neutrons. In nitrogen-14 - 7 neutrons, in nitrogen-15 - 8 neutrons. 
4 0
3 years ago
A flask contains 0.340 mol of liquid bromine, br2. determine the number of bromine molecules present in the flask.
marshall27 [118]
<span>Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles. We calculate as follows:

0.340 mol Br2 ( </span>6.022 x 10^23 molecules / mol ) = 2.05 x 10^23 molecules
7 0
3 years ago
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A horizontal cylinder equipped with a frictionless piston contains 785 cm3 of steam at 400 K and 125 kPa pressure. A total of 83
guapka [62]

Answer:

a. 478.69 K

b. 939.43 cm^{3}

c. 19.30 J

d. 64.5J

Explanation:

From the question, we can identify the following;

V_{o} = 785cm^{3} = 0.000785 m^{3}

T_{o} = 400K

P_{o} = 125 Kpa =  125 000 Pa

Using the ideal gas equation,

PV = nRT

where R is the molar gas constant = 8.314 m^{3}⋅Pa⋅K^{-1}⋅mol^{-1}

Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol

a. Steam temperature in K

To calculate this, we use the constant pressure process;

q = nΔH

Where q is 83.8J according to the question

Thus;

83.8 = 0.03 × [34980 + 35.5T_{1} - (34980 + 35.5T_{o})]

83.8 = (0.03 × 35.5) (T_{1} - 400K)

83.8 = 1.065 (T_{1}  - 400)

78.69 = (T_{1}  - 400)

T_{1} = 400 + 78.69

T_{1}  = 478.69 K

b. Final cylinder volume

To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)

V_{1}/T_{1} = V_{o}/T_{o}

V_{1}  =  V_{o}T_{1}/T_{o}

V_{1}   = (785 × 478.69)/400

V_{1}   = 939.43 cm^{3}

c. Work done by the system

Mathematically, the work done by the system is calculated as follows;

w = P(V_{1}- V_{o}) = 125 KPa ( 939.43 - 785) = 19.30 J

d. Change in internal energy of the steam in J

ΔU = q - w = 83.8 - 19.3 = 64.5J

6 0
3 years ago
2000 minutes to hours dimensional analysis
elena-14-01-66 [18.8K]

Answer:

33.33 h

Explanation:

You know that

1 h = 60 min

If you divide both sides by 60 min, you get the conversion factor: 1 h/60 min = 1.

If you divide both sides by 1 h, you get the conversion factor: 1 = 60 min/1 h.

Both are conversion factors because they both equal one and multiplying a measurement by one does not change its value.

You choose the conversion factor that gives you the correct dimensions for your answer. It must have the correct dimensions on top (in the numerator),

Thus, to convert 2000 min to hours, you use the conversion factor with “h” on the top.

\text{Time} = \text{2000 min} \times \dfrac{\text{1 h}}{\text{60 min}} = \textbf{33.33 h}

5 0
2 years ago
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