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kipiarov [429]
3 years ago
13

The following reaction takes place at high temperatures.

Chemistry
1 answer:
monitta3 years ago
3 0

Answer:

Cr_2O_3  (s) + 2Al  (l) ⇒ 2Cr  (l)  + Al_2O_3  (l)

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3.0 L of a gas has a temperature of 78 C find the volume of the gas at standard temperature
jeka57 [31]

Answer:

= 2.33 L

Explanation:

V1/T1 = V2/T2

V1 = 3.0 L

T1 = 78 + 273

   = 351 K

At s.t.p the temperature is 273 K and pressure is 1 atm.

V2 = ?

T2 = 273 K

V2 = V1T2/T1

    = (3.0 ×273)/351

    = 2.33 L

Explanation:

ik it is confusing but that is what i got

4 0
3 years ago
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[?] describes how small molecules can be selectively removed from a colloidal suspension while retaining large molecules.
IceJOKER [234]
Dialysis is correct……………
7 0
2 years ago
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Earthquakes cause other natural disasters, such as the disaster in this photo from a major earthquake that hit Alaska in 1964. W
Mashcka [7]

On March 27, 1964, Southeastern Alaska experienced a magnitude 9.2 earthquake. The earthquake caused a major tectonic tsunami, which hit the Southeastern coast of Alaska, the Pacific Coast of British Columbia, and the west coast of the United States. During the earthquake, there were also large landslides and submarine slumps.

3 0
3 years ago
Imine formation from an aldehyde and an amine proceeds reversibly under slightly acidic conditions. The reaction is reversible d
Lostsunrise [7]

Answer:

Imine can be isolated from the reaction mixture as water is continuously removed from the reaction chamber

Explanation:

In this reaction, a non -aqueous solvent  is not used (not mentioned in the question). Thus, we can say that there is continuous removal water under suitable reacting conditions and hence the imine formed is left behind.

5 0
3 years ago
You want to prepare 500.0 mL of 1.000 M KNO3 at 20°C, but the lab (and water) temperature is 24°C at the time of preparation. Ho
timama [110]

Explanation:

As per the given data, at a higher temperature, at 24^{o}C, the solution will occupy a larger volume than at 20^{o}C.

Since, density is mass divided by volume and it will decrease at higher temperature.

Also, concentration is number of moles divided by volume and it decreases at higher temperature.

At 20^{o}C, density of water=0.9982071 g/ml  

Therefore, \frac{concentration}{density} will be calculated as follows.

                 = \frac{C_{1}}{d_{1}}

                 = \frac{1.000 mol/L}{0.9982071 g/ml}

                 = 1.0017961 mol/g  

At 24^{o}C, density of water = 0.9972995 g/ml

Since, \frac{concentration}{density} = \frac{C_{2}}{d_{2}}

                           = \frac{C_{2}}{0.9972995}

Also,             \frac{C_{1}}{d_{1}} = \frac{C_{2}}{d_{2}}

so,                   1.0017961 mol/g = \frac{C_{2}}{0.9972995}

                      C_{2} = 1.0017961 \times 0.9972995

                                  = 0.9990907 mol/L

Therefore, in 500 ml, concentration of KNO_{3} present is calculated as follows.

             C_{2} = \frac{concentration}{volume}

               0.9990907 mol/L = \frac{concentration}{0.5 L}  

               concentration = 0.49954537 mol

Hence, mass (m'') = 0.49954537 mol \times 101.1032 g/mol = 50.5056 g       (as molar mass of KNO_{3} = 101.1032 g/mol).

Any object displaces air, so the apparent mass is somewhat reduced, which requires buoyancy correction.

Hence, using Buoyancy correction as follows,

                      m = m''' \times \frac{(1 - \frac{d_{air}}{d_{weights}})}{(1 - \frac{d_{air}}{d})}}

where,          d_{air} = density of air = 0.0012 g/ml

                     d_{weight} = density of callibration weights = 8.0g/ml                      

                     d = density of weighed object

Hence, the true mass will be calculated as follows.

           True mass(m) = 50.5056 \times \frac{(1 - \frac{0.0012}{8.0})}{(1 - (\frac{0.0012}{2.109})}

             true mass(m) = 50.5268 g

                                  = 50.53 g (approx)

Thus, we can conclude that 50.53 g apparent mass of KNO_{3} needs to be measured.

8 0
3 years ago
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