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Deffense [45]
3 years ago
9

How would you prepare a saturated solution of potassium nitrate?

Chemistry
1 answer:
ValentinkaMS [17]3 years ago
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.
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A car traveling with constant speed travels 150km in 7200 s. What is the speed of the car
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75km/hr

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4.5cm wide by 5.750cm long by 1.50cm tall what is the volume of the box
Anni [7]

Answer:

volume of box is 38.81 cm³.

Explanation:

Given data:

Width of box = 4.5 cm

Height of box = 5.750 cm

Length of box = 1.50 cm

Solution:

Formula:

Volume = length × height × width

by putting values,

V = 1.50 cm × 5.750 cm× 4.5 cm

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5 0
2 years ago
What’s the answer ?
Romashka-Z-Leto [24]

Answer:

45.3°C

Explanation:

Step 1:

Data obtained from the question.

Initial pressure (P1) = 82KPa

Initial temperature (T1) = 26°C

Final pressure (P2) = 87.3KPa.

Final temperature (T2) =.?

Step 2:

Conversion of celsius temperature to Kelvin temperature.

This is illustrated below:

T(K) = T(°C) + 273

Initial temperature (T1) = 26°C

Initial temperature (T1) = 26°C + 273 = 299K.

Step 3:

Determination of the new temperature of the gas. This can be obtained as follow:

P1/T1 = P2/T2

82/299 = 87.3/T2

Cross multiply to express in linear form

82 x T2 = 299 x 87.3

Divide both side by 82

T2 = (299 x 87.3) /82

T2 = 318.3K

Step 4:

Conversion of 318.3K to celsius temperature. This is illustrated below:

T(°C) = T(K) – 273

T(K) = 318.3K

T(°C) = 318.3 – 273

T(°C) = 45.3°C.

Therefore, the new temperature of the gas in th tire is 45.3°C

6 0
3 years ago
When balancing a chemical equation can you adjust the number that is subscripted to a substance formula?
Ugo [173]
No, you can't do that
3 0
3 years ago
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