All categories

3 years ago

How would you prepare a saturated solution of potassium nitrate?

1 answer:

5 0

Using titration method,

Pipette 25 cm3 of KOH into a conical flask. Add a few drops of methyl orange(indicator).
Then pour a beaker of Dilute nitric acid (HNO3) into a burette. Record the readings of the burette before starting.

1)While releasing some nitric acid into the conical flask, swirl the conical flask.
2)Stop swirlimg or adding nitric acid when color of the solution in the conical flask turns light pink. That will be end point. Record the reading of the burette.
3)Repeat the experiment without adding methyl orange(indicator)
4)Add the same vol of nitric acid used previously.
5)Solution in conical flask transfer over to a dish.
6)Heat the solution till it becomes saturated.

You might be interested in

Determine the AMOUNT OF NO2, LIMITING REACTANT, AND THE AMOUNT AND NAME OF EXCESS REACTANT.

zepelin [54]

Answer:

balanced equation mole ratio 5 2 mol NO/1 mol O2

10.00 g O2 3 1 mol O2/32.00 g O2 5 0.3125 mol O2

20.00 g NO 3 1 mol NO/30.01 g NO 5 0.6664 mol NO

actual mole ratio 5 0.6664 mol NO/0.3125 mol O2 5 2.132 mol NO/1.000 mol O2

Because the actual mole ratio of NO:O2 is larger than the balanced equation mole

ratio of NO:O2, there is an excess of NO; O2 is the limiting reactant.

Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO

0.6250 mol NO 3 30.01 g NO/1 mol NO 5 18.76 g NO

Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2

Excess NO 5 20.00 g NO 2 18.76 g NO 5 1.24 g N

Explanation:

3 0

3 years ago

2. if 0.20 m fe3 had been used instead of 0.020 m fe3 , how would the numerical value of the rate constant and the activation en

dezoksy [38]

the calculated value is Ea is 18.2 KJ and A is 12.27.

According to the exponential part in the Arrhenius equation, a reaction's rate constant rises exponentially as the activation energy falls. The rate also grows exponentially because the rate of a reaction is precisely proportional to its rate constant.

At 500K, K=0.02s−1

At 700K, k=0.07s −1

The Arrhenius equation can be used to calculate Ea and A.

RT=k=Ae Ea

lnk=lnA+(RT−Ea)

At 500 K,

ln0.02=lnA+500R−Ea

500R Ea (1) At 700K lnA=ln (0.02) + 500R

lnA = ln (0.07) + 700REa (2)

Adding (1) to (2)

700REa100R1[5Ea-7Ea] = 0.02) +500REa=0.07) +700REa.

=ln [0.02/0 .07]

Ea= 2/35×100×8.314×1.2528

Ea =18227.6J

Ea =18.2KJ

Changing the value of E an in (1),

lnA=0.02) + 500×8.314/18227.6

= (−3.9120) +4.3848

lnA=0.4728

logA=1.0889

A=antilog (1.0889)

A=12.27

Consequently, Ea is 18.2 KJ and A is 12.27.

Learn more about Arrhenius equation here-

brainly.com/question/12907018

#SPJ4

5 0

1 year ago

Magnesium ribbon reacts vigorously with hydrochloric acid. The test tube gets hot and bubbles are produced. When a metal reacts

oee [108]

The answer is C, hydrogen gas. This is because in single replacement reactions, the single element (here Magnesium) replaces whichever element in the compound it corresponds to. Because Mg loses electrons since it’s a metal, it will replace the element which also loses electrons, which is Hydrogen here. So when they switch places, MgCl2 and H2 are made— and H2 is the hydrogen gas.

6 0

3 years ago

Read 2 more answers

Salt water

lidiya [134]

Answer:

salt water heterogeneous mixture , pure substance compound,element solut8on

5 0

3 years ago

Al añadir 1g de urea a 100 ml de agua a 20ºC, la presión de vapor pasa de 17,54 mmHg a 17,49 mmHg. Determina la masa molecular d

never [62]

Answer: Los estudios teóricos y experimentales han permitido establecer, que los líquidos poseen propiedades físicas características. Entre ellas cabe mencionar: la densidad, la propiedad de ebullir, congelar y evaporar, la viscosidad y la capacidad de conducir la corriente eléctrica, etc. Cada líquido presenta valores característicos (es decir, constantes) para cada una de estas propiedades. Cuando un soluto y un solvente dan origen a una solución, la presencia del soluto determina una modificación de estas propiedades con relación a su estado normal en forma aislada, es decir, líquido puro. Estas modificaciones se conocen como PROPIEDADES DE UNA SOLUCIÓN.

3 0

3 years ago