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IgorC [24]
2 years ago
6

When heating a flammable or volatile solvent for a recrystallization, which of these statements are correct? More than one answe

r may be correct.
1)You should not use an open flame to heat the solvent.
2)You should heat the solvent in a stoppered flask to keep vapor away from any open flames.
3)You should ensure that no one else is using an open flame near your experiment.
Chemistry
1 answer:
otez555 [7]2 years ago
3 0

Explanation:

A volatile substance is defined as the substance which can easily evaporate into the atmosphere due to weak intermolecular forces present within its molecules.

Whereas a flammable substance is defined as a substance which is able to catch fire easily when it comes in contact with flame.

Hence, when we heat a flammable or volatile solvent for a recrystallization then it should be kept in mind that should heat the solvent in a stoppered flask to keep vapor away from any open flames so that it won't catch fire.

And, you should ensure that no one else is using an open flame near your experiment.

Thus, we can conclude that following statements are correct:

  • You should heat the solvent in a stoppered flask to keep vapor away from any open flames.
  • You should ensure that no one else is using an open flame near your experiment.
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Explain two ways that photosynthesis helps a predator like a wolf or<br> tiger.
Sindrei [870]

Answer: Photosynthetic organisms take light energy and use it to make their own "food". In this process they use carbon dioxide and light to make sugars and oxygen. This is great for other animals that utilize this oxygen and eat the plants for food.

HOPE THIS HELPS

4 0
2 years ago
Based on a kc value of 0.200 and the given data table, what are the equilibrium concentrations of xy, x, and y, respectively? ex
creativ13 [48]
Missing in your question :
Concentration by (M):
                   Xy:             y:            X
initial          0.2            0.3         0.3
change       +X            -X            -X
equilibrim  (0.2+x)    (0.3-x)     (0.3-x)
according to Kc formula: when Kc = 0.2
Kc = [XY]/[X]*[Y]
0.2 = (0.2+x) / (0.3-x)*(0.3-x)
0.2=(0.2+x) / (0.3-x)^2 by solving this equation
0.2*(0.3-x)^2 = 0.2+x
0.2* (0.09-0.6x+x^2)= 0.2 +x
0.0018 - 0.12 X +0.2X^2 = 0.2 + X
0.2X^2 -1.12 X -0.1982 = 0
∴X= 0.17 
∴[XY] = 0.2 + 0.17 = 0.37 m
∴[X] = 0.3 - 0.17 =0.13 m
∴[y] = 0.3 - 0.17 = 0.13 m
  


5 0
2 years ago
Consider a triprotic acid like phosphoric acid or citric acid. which expression correctly describes the relative magnitudes of t
BaLLatris [955]
The triprotic acid like H₃PO₄ contains three protons H⁺ so it ionized through three steps, the acidity strength of first proton higher than second proton higher than third one so Ka1 > Ka2 > Ka3 But
For pKa which equals to -log Ka ... the higher the value of Ka, the smaller the value of pKa. so the correct answer will be:
pKa1 < pKa2 < pKa3
7 0
2 years ago
Calculate the moles of camphor dissolved in 1.32 L of phenol. The molar mass of camphor is 152 g/mol and the molarity of phenol
LUCKY_DIMON [66]

Answer:

moles of camphor = 0.0522 moles

Explanation:

<u>Data:</u>

MW camphor= 152 g/mol

V solution = 1.32 L

M solution = 6.01 M

moles solute = ?

  • To calculate the moles of camphor, you must first know the grams of solute (camphor) that exist in the solution, this is calculated from the molarity equation:

M=\frac{solute mass}{solution volume}

  • From there the grams of the solute (camphor) are cleared:

solutemass=M*solutionvolume=6.01M*1.32L=7.9332g

  • Then by means of the molecular weight (MW) equation the moles can be obtained:

MW=\frac{mass}{moles}

moles=\frac{mass}{MW} =\frac{7.9332g}{152\frac{g}{mol} } =0.0522 moles

6 0
2 years ago
A flask of volume 2.0 liters, provided with a stopcock, contains oxygen at 20 oC, 1.0 ATM (1.013X105 Pa). The system is heated t
leonid [27]

Answer:

1.27 atm is the final pressure of the oxygen in the flask (with the stopcock closed).

2.6592 grams of oxygen remain in the flask.

Explanation:

Volume of the flask remains constant = V = 2.0 L

Initial pressure of the oxygen gas = P_1=1.0 atm

Initial temperature of the oxygen gas = T_1=20^oC =293.15 K

Final pressure of the oxygen gas = P_2=?

Final temperature of the oxygen gas = T_2=100^oC =373.15 K

Using Gay Lussac's law:

\frac{P_1}{T_1}=\frac{P_2}{T_2}

P_2=\frac{P_1\times T_2}{T_1}=\frac{1 atm\times 373.15 K}{293.15 K}=1.27 atm

1.27 atm is the final pressure of the oxygen in the flask (with the stopcock closed).

Moles of oxygen gas = n

P_1V_1=nRT_1 (ideal gas equation)

n=\frac{P_1V_1}{RT_1}=\frac{1 atm\times 2.0 L}{0.0821 atm l/mol K\times 293.15 K}=0.08310 mol

Mass of 0.08310 moles of oxygen gas:

0.08310 mol × 32 g/mol = 2.6592 g

2.6592 grams of oxygen remain in the flask.

6 0
2 years ago
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