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azamat
3 years ago
12

How many moles of potassium nitrate, KNO3 are present in a sample with a mass of 85.2 g?

Chemistry
1 answer:
Bingel [31]3 years ago
5 0
0.843 moles would be the answer
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Predict, using Boyle’s Law, what will happen to a balloon that an ocean diver takes to a pressure of 202 kPa (normal atmospheric
kaheart [24]

Explanation:

Balloon that an ocean diver takes to a pressure of 202 k Pa will get reduced in size that is the volume of the balloon will get reduced. This is because pressure and volume of the gas are inversely related to each other.

According to Boyle's law: The pressure of the gas  is inversely proportional to the volume occupied by the gas at constant temperature(in Kelvins).

pressure\propto \frac{1}{volume} (At constant temperature)

The pressure beneath the sea is 202 kPa and the atmospheric pressure  is 101.3 kPa . This increase in pressure will result in decrease in volume occupied by the gas inside the balloon with decrease in size of a balloon. Hence, the size of the balloon will get reduced at 202 kPa (under sea).

7 0
3 years ago
Rank the solutions below in order of increasing acidity. (Drag and drop into the appropriate area)
Vladimir79 [104]

Answer:

0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HCl

Explanation:

Strong acids are more acids than weak acids. In the same way, strong bases are more basic than weak bases that are in the same concentration.

Then, the more concentrated acid or base will be more acidic or basic.

CH3COOH. Weak acid

NaOH. Strong base

H2SO4. Strong acid

NH3. Weak base.

HCl. Strong acid

The less acid (More basic):

<h3>0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HCl</h3>

Strong base, weak base, weak acid, diluted strong acid, undiluted strong acid

7 0
3 years ago
Which of the following compound names is obviously INCORRECT?
Irina18 [472]

The compound which is obviously incorrect is dihydrogen oxide

3 0
3 years ago
In an experiment, 16.8 g of k2so4 was dissolved in 1.00 kg of water to make a solution. the freezing point of the solution was m
Mashutka [201]
Answer is: V<span>an't Hoff factor (i) for this solution is 2,26.
</span>Change in freezing point from pure solvent to solution: ΔT =i · Kf · m.
<span>Kf - molal freezing-point depression constant for water is 1,86°C/m.
</span>m -  molality, moles of solute per kilogram of solvent.
n(K₂SO₄) = 16,8 g ÷ 174,25 g/mol
n(K₂SO₄) = 0,096 mol.
m(K₂SO₄) = 0,096 mol/kg.
ΔT = 0,405°C.
i = 0,405 ÷ (1,86 · 0,096)
i = 2,26.

8 0
3 years ago
If 25.0 mL of a 2.00 M CaCl2 solution is used for the reaction
RideAnS [48]

0.1 moles of chloride ions were involved  in the reaction if 25.0 mL of a 2.00 M CaCl2 solution is used for the reaction.

Explanation:

Data given:

volume of CaCl_{2} = 25 ml 0r 0.025

molarity of the calcium chloride solution = 2M

number of chloride ions =?

Balance chemical reaction:

Ca + 2Cl_{2} ⇒Ca_{} Cl_{2}

number of moles in 25 ml is calculated as:

molarity = \frac{number of moles}{volume}

number of moles of calcium chloride  = molarity x volume

putting the values in the equation:

number of moles = 2 x 0.025

                            = 0.05 moles of calcium chloride

1 mole of  CaCl_{2}  decomposes as 1 calcium ion and 2 chloride ions

so 0.05 moles will have x moles of chloride ion

\frac{2}{1} = \frac{x}{0.05}

x= 0.1

0.1 moles of chloride ions will be involved in the reaction.

5 0
3 years ago
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