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Deffense [45]
4 years ago
9

How would you prepare a saturated solution of potassium nitrate?

Chemistry
1 answer:
ValentinkaMS [17]4 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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Sort these elements into pairs<br> Mg<br> Cl<br> As<br> Sr<br> I<br> N
LenKa [72]

Metals:-

Mg(Magnesium)

Sr(Strontium)

Non-metals are:-

Cl(Chlorine)

I(Iodine)

N(Nitrogen)

Metalloids are:-

As(Arsenic)

4 0
3 years ago
Read 2 more answers
If there are 40 mol of NBr3 and 48 mol of NaOH, what is the excess reactant?
Nata [24]

Answer:

The correct answer is option B.

Explanation:

3NaOH+2NBr_3\rightarrow 3HOBr+3NaBr+N_2

Moles of NBr_3 = 40 mol

Moles of NaOH = 48 mol

According to reaction, 3 moles of NaOH reacts with 2 moles NBr_3

Then ,48 moles of NaOH will reacts with:

\frac{2}{3}\times 48 mol=32 mol of NBr_3

Then ,40 moles of NaBr_3 will reacts with:

\frac{3}{2}\times 40 mol=60 mol of NaOH

As we can see that 48 moles of sodium will completey react with 32 moles of nitrogen tribromide.

Moles left after reaction = 40 mol - 32 mol = 8 mol

Hence, the NBr_3 is an excessive reagent.

6 0
3 years ago
Read 2 more answers
How much heat is required, in calories, to raise the temperature of 57.8 g of silver from
Stella [2.4K]

Answer:

87.3 calories of heat is required.

Explanation:

Heat = mcΔT

m= mass, c = specific heat of silver, T = temperature

H= 57.8 g * 0.057 cal/g°C * ( 43.5 - 17 °C)

H = 57.8 * 0.057 * 26.5

H = 87.3069 cal.

The heat required to raise the temperature of 57.8 g of silver from 17 °C to 43.5 °C is 87.3 calories.

8 0
3 years ago
A sample of an element has a volume of 78.0 mL and a density of 1.85 g/mL. What is the mass in grams of the sample?
abruzzese [7]

Hey there!:

density = 1.85 g/mL

volume = 78.0 mL

Mass = ??

therefore:

D = m / V

1.85 = m / 78.0

m = 1.85 x 78.0

m = 144.3 grams

Hope this helps!

7 0
3 years ago
100 points! Please help, will name brainliest. Thank you :)
Ludmilka [50]

Answer and Explanation:

We need to remember the formulas:

- pOH = -log[OH-]

- pH = -log[H+]

- pH + pOH = 14

When given the concentration of OH-, denoted as [OH-], we can plug that value into the equation pOH = -log[OH-] to find the pOH. Once we have found the pOH, we can then subtract that value from 14 to determine the value of pH.

4 0
3 years ago
Read 2 more answers
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