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valina [46]
3 years ago
15

-- A scientist combines 14 grams of potassium nitrate with an unknown amount of

Chemistry
1 answer:
igor_vitrenko [27]3 years ago
5 0

Answer:

The unknown amount of potassium chloride is 13.6 grams.            

 

Explanation:

The reaction of 14 grams of KNO₃ with KCl produces a total mass of 27.6 grams of the products.

The law of conservation of mass tells us that the total mass of the reactants must be the same that the total mass of the products. So, we can find the mass of KCl as follows:

m_{r} = m_{p}

Where <em>r</em> is for reactants and <em>p </em>is for products

m_{KNO_{3}} + m_{KCl} = m_{p}

14 g + m_{KCl} = 27.6 g

m_{KCl} = 27.6 g - 14 g = 13.6 g

Therefore, the unknown amount of potassium chloride is 13.6 grams.           

I hope it helps you!

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3 years ago
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A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4
san4es73 [151]

Answer:

83.20 g of Na3PO4

Explanation:

1 mole of Na3PO4 contains 3 moles of Na+.

Mole of Na ion to be prepared = Molarity x volume

                 = 0.700 x 725/1000

                     = 0.5075 mole

If 1 mole of Na3PO4 contains 3 moles of Na ion, then 0.5075 Na ion will be contained in:

           0.5075/3 x 1 = 0.1692 mole of Na3PO4

mole of Na3PO4 = mass/molar mass = 0.1692

Hence, mass of Na3PO4 = 0.1692 x molar mass

                                     = 0.1692 x 163.94

                                        = 83.20 g.

83.20 g of Na3PO4 will be needed.

3 0
3 years ago
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Calculate the molality of 75.0 grams of MgCl2 (molar mass=95.21 g/mol) dissolved in 500.0 g of solvent.
nordsb [41]

<u>Answer:</u> The molality of magnesium chloride is 1.58 m

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

Where,

m_{solute} = Given mass of solute (magnesium chloride) = 75.0

M_{solute} = Molar mass of solute (magnesium chloride) = 95.21 g/mol  

W_{solvent} = Mass of solvent = 500.0 g

Putting values in above equation, we get:

\text{Molality of }MgCl_2=\frac{75.0\times 1000}{95.21\times 500.0}\\\\\text{Molality of }MgCl_2=1.58m

Hence, the molality of magnesium chloride is 1.58 m

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3 years ago
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the combustion of 0.1240 kg of propane in the presence of excess oxygen produces 0.3110kg of carbon dioxide what is the limittin
mote1985 [20]

Answer:

Propane

Explanation:

From the question given, we were told that 0.1240 kg of propane reacted with excess oxygen to produce 0.3110kg of carbon dioxide.

Since the reaction took place in the presence of excess oxygen, therefore, propane is the limiting reactant as all of it is used up in the presence of excess oxygen.

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When methyloxirane is treated with HBr, the bromide ion attacks the less substituted position. However, when phenyloxirane is tr
konstantin123 [22]

Answer:

See explanation and picture below

Explanation:

First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.

In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.

In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.

4 0
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