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Yanka [14]
2 years ago
11

If an ionic compound were composed of a4+ and b−, which unit cell structure would give a neutral compound?

Chemistry
1 answer:
Ymorist [56]2 years ago
4 0
<span>biological reactions that happen within cells while reducing the complex interactions found in a whole cell. Eukaryotic and prokaryotic cells have been used for creation of these simplified environments[1]. Subcellular fractions can be isolated by ultracentrifugation to provide molecular machinery that can be used in reactions in the absence of many of the other cellular components. Cell-free biosystems can be prepared by mixing a number of purified enzymes and coenzymes. Cell-free biosystems are proposed as a new low-cost biomanufacturing platform compared to microbial fermentation used for thousands of years. Cell-free biosystems have several advantages suitable in industrial applications</span>
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When heating a flammable or volatile solvent for a recrystallization, which of these statements are correct? More than one answe
otez555 [7]

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.
3 0
2 years ago
Salt in crude oil must be removed before the oil undergoes processing in a refinery. The
irina1246 [14]

Answer:

\large \boxed{0.64 \, \%}

Explanation:

Assume you are using 1 L of water.

Then you are washing 4 L of salty oil.

1. Calculate the mass of the salty oil

Assume the oil has a density of 0.86 g/mL.

\text{Mass of oil} = \text{4000 mL} \times \dfrac{\text{0.86 g}}{\text{1 mL}} = \text{3440 g}

2. Calculate the mass of salt in the salty oil

\text{Mass of salt} = \text{3440 g} \times \dfrac{\text{5 g salt}}{\text{100 g oil}} = \text{172 g salt}

3. Calculate the mass of salt in the spent water

\text{Mass of salt} = \text{1000 g water} \times \dfrac{\text{15 g salt}}{\text{100 g water}} = \text{150 g salt}

4. Mass of salt remaining in washed oil

Mass = 172 g - 150 g = 22 g  

5. Concentration of salt in washed oil

\text{Concentration} = \dfrac{\text{22 g}}{\text{3440 g}} \times 100 \, \% = \mathbf{0.64 \, \%}\\\\\text{The concentration of salt in the washed oil is $\large \boxed{\mathbf{0.64 \, \%}}$}

3 0
2 years ago
A 2135 cm3 sample of dry air has a pressure of 98.4 kpa at 127 degrees Celsius. What is the volume of the sample if the Temperat
kykrilka [37]

the equation is p1 x v1 divided by T1 = p1 x v2 = T2 but since the pressure is kept constant you do not even need it so the equation would now be v1 divided by t1 = v2 divided by t2

2135 cm3 divided by 127 degrees celcius = x divided by 206

answer: 3460 cm3

7 0
2 years ago
Read 2 more answers
When a 1.00 L sample of water from the surface of the Dead Sea (which is more than 400 meters below sea level and much saltier t
Harlamova29_29 [7]

Answer: Molarity of MgCl_2 in the original sample was 1.96M

Explanation:

Molarity is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{\text{no of moles}}{\text{Volume in L}}

{\text {moles of solute}=\frac{\text {given mass}}{\text {molar mass}}=\frac{186g}{95g/mol}=1.96

Now put all the given values in the formula of molarity, we get

Molarity=\frac{1.96}{1.00L}

Molarity=1.96mol/L

Thus molarity of MgCl_2 in the original sample was 1.96M

4 0
2 years ago
A 0.200 M K 2SO 4 solution is produced by ________. dilution of 250.0 mL of 1.00 M K2SO4 to 1.00 L dissolving 43.6 g of K2SO4 in
Artist 52 [7]

A 0.200 M of K2SO4 solution is produced by diluting 20.0 mL of 5.00 M K2SO4 solution to 500.0 mL.  

<u>Explanation</u>:

  •  When dealing with dilution we will use the following equation:

                              M1 V1 = M2 V2

where,

                      M1 = initial concentration

                      V1 = initial volume

                      M2 = final concentration

                      V2 = final volume

  • By diluting 20.0 mL of 5.00 M K2SO4 solution to 500.0 mL, we get

                            M1 V1 = M2 V2

     20.0 mL \times    5.00 M = M2 \times 500.0 mL

                               M2 = (20.0 mL \times    5.00 M) / 500.0 mL

                              M2 =  0.200 M.

Hence A 0.200 M of K2SO4 solution is produced by diluting 20.0 mL of 5.00 M K2SO4 solution to 500.0 mL.  

3 0
2 years ago
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