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Diano4ka-milaya [45]
2 years ago
14

5.00 mol of ammonia are introduced into a 5.00 L reactor vessel in which it partially dissociates at high temperatures. 2NH 3(g)

3H 2(g) + N 2(g) At equilibrium and a particular temperature, 1.00 mole of ammonia remains. Calculate K c for the reaction.
Chemistry
1 answer:
allochka39001 [22]2 years ago
4 0

Explanation:

system at equilibrium, will the reaction shift towards reactants ~

--?'

2. (2 Pts) Consider the reaction N2(g) + 3H2(g) =; 2NH3(g). The production of ammonia is an

exothermic reaction. Will heating the equilibrium system increase o~e amount of

ammonia produced? . .co:(

3. (2 Pts) Consider the reaction N2(g) + 3H2(g) =; 2NH3(g). Ifwe use a catalyst, which way will

the reaction shift? ':'\

.1.+- w~t s~,H (o')l r'eo.c. e~ ei~i"liht-,·u.fn\ P~~,

4. (3 Pts) ff 1ven th e o £ 11 owmg d t a a £ or th ere action: A(g) + 2B(s) =; AB2(g)

Temperature (K) Kc

300 1.5x104

600 55 k ' pr, cl l<..J~

e- ~ r fee, ct o. ~ 1<

900 3.4 X 10-3

Is the reaction endothermic or exothermic (explain your answer)?

t d- IS o.,;r-. \4\a..i~1f't~ °the te.Y'il(lf1,:J'u.r-a a•~S. j lrvdu..c,,.) +~H~to{' '\

exothe-rnh't.-- ,.. ..,. (/.., ,~.

5. (4 Pts) Consider the reaction, N2(g) + 3H2(g) =; 2NH3(g). Kc= 4.2 at 600 K.

What is the value of Kc for 4 NH3(g) =; 2N2(g) + 6H2(g)

N ... ~l + 3 H~(ri ~ ~Nli3~) kl,= ~:s;H,J3 # 4. J..

~ ;)N~~) ~ ~ H ~) ~\-_ == [A!;J:t D~~Jb

J. [,v 1+3] ~

I

4,:i.~ = 0,05

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Determine the percent yield for the reaction between 6.92 g K and 4.28g of oxygen gas if 7.36 g of potassium oxide is produced.
Illusion [34]

Answer:

percentage yield = 88.25%

Explanation:

Firstly, write the chemical reaction and balance the equation.

Potassium react with oxygen to produce potassium oxide.

K + 02 → K2O

Balance the equation

4K + 02 → 2K2O

The limiting reactant is K so the yield of potassium oxide can be calculated using grams for potassium.

atomic mass of K = 39.1g/mol

grams for 4 mole of potassium =  4(39.1) = 156.4 g

grams for 2 moles of K2O = 2( 39.1 × 2 + 16) = 188.4 g

If 156.4 g of K produces 188.4 g of K2O

6.92 g of K will produce ? gram of K2O

cross multiply

grams of K2O = 6.92 × 188.4/156.4

grams of K2O = 1303.72/156.4

grams of K2O = 8.33585677749

grams of K2O = 8.34 g

percentage yield = actual yield/theoretical yield × 100

actual yield = 7.36 g

theoretical yield = 8.34 g

percentage yield = 7.36/8.34 × 100

percentage yield =         736/8.34

percentage yield = 88.2494004796%

percentage yield = 88.25%

5 0
2 years ago
During an experiment, the percent yield of calcium chloride from a reaction was 85.22%. Theoretically, the expected amount shoul
devlian [24]

The formula to find yield is

(Actual Yield)/(Theorectical Yield) x100

Just do the math.

85.22% x 113 = 96.2986

Convert it to 3 significant figures

Ans: 96.3g

7 0
2 years ago
Read 2 more answers
In the electrochemical cell using the redox reaction below, the anode half reaction is ________. Sn4+ (aq) + Fe (s) → Sn2+ (aq)
kenny6666 [7]

Answer:

The anode half reaction is : Fe(s)\rightarrow Fe^{2+}(aq.)+2e^{-}

Explanation:

In electrochemical cell, oxidation occurs in anode and reduction occurs in cathode.

In oxidation, electrons are being released by a species. In reduction, electrons are being consumed by a species.

We can split the given cell reaction into two half-cell reaction such as-

Oxidation (anode): Fe(s)\rightarrow Fe^{2+}(aq.)+2e^{-}

Reduction (cathode): Sn^{4+}(aq.)+2e^{-}\rightarrow Sn^{2+}(aq.)

------------------------------------------------------------------------------------------------------------

overall: Fe(s)+Sn^{4+}(aq.)\rightarrow Fe^{2+}(aq.)+Sn^{2+}(aq.)

So the anode half reaction is : Fe(s)\rightarrow Fe^{2+}(aq.)+2e^{-}

8 0
2 years ago
what is the new pressure acting on a 2.5 l balloon if its original volume was 5.8 l at 3.7 ATM of pressure
Ivan

Answer : The new pressure acting on a 2.5 L balloon is, 8.6 atm.

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

or,

P_1V_1=P_2V_2

where,

P_1 = initial pressure = 3.7 atm

P_2 = final pressure = ?

V_1 = initial volume = 5.8 L

V_2 = final volume = 2.5 L

Now put all the given values in the above equation, we get:

3.7atm\times 5.8L=P_2\times 2.5L

P_2=8.6atm

Thus, the new pressure acting on a 2.5 L balloon is, 8.6 atm.

8 0
2 years ago
A student wants to draw a model of an atom. Which statement describes how to find the number of neutrons to include in the model
san4es73 [151]

c is the correct answer

4 0
2 years ago
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