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zubka84 [21]
3 years ago
6

CO_2 sublimes readily at 25°C. Which properties are usually associated with a compound that undergoes this kind of change?

Chemistry
1 answer:
antoniya [11.8K]3 years ago
4 0

Explanation:

Sublimation is defined as a process in which solid state of a substance directly changes into vapor or gaseous state without undergoing liquid phase.

For example, naphthalene balls show sublimation at room temperature.

As this process does not cause any change in chemical composition of a substance. Hence, it is known as a physical process.

Similarly, when CO_{2} sublimes readily at 25^{o}C. This shows change in physical state of carbon dioxide is taking place, i.e, from solid to gaseous phase.

Thus, we can conclude that when CO_{2} sublimes readily at 25^{o}C then it means physical properties are usually associated with a compound that undergoes this kind of change.

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In this chemical Formula for Ammonia, the Subscripts indicate what?
zubka84 [21]

Answer:

B

Explanation:

B. There are two atoms of Nitrogen and two atoms of Hydrogen combined to make Ammonia.

7 0
3 years ago
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Please help me get the answer to bothe a and b​
neonofarm [45]

Answer:

If a metal and metal solution react, the more reactive metal will displace the less reactive metal from solution. If the metal in solution you start with is formed from a more reactive metal than the metal to be added, no reaction will occur.

3 0
2 years ago
Which solution has the same boiling point as 0.25 mol CaCl2 dissolved in 1000 g water?
Rufina [12.5K]
The boiling point of water at 1 atm is 100 degrees celsius. However, when water is added with another substance the boiling point of it rises than when it is still a pure solvent. This called boiling point elevation, a colligative property. The equation for the boiling point elevation is expressed as the product of the ebullioscopic constant (0.52 degrees celsius / m) for water), the vant hoff factor and the concentration of solute (in terms of molality). 

 ΔT(CaCl2) = i x K x m = 3 x 0.52 x 0.25 = 0.39 °C
<span> ΔT(Sucrose) = 1 x 0.52 x 0.75 = 0.39 </span>°C<span>
</span><span> ΔT(Ethylene glycol) = 1 x 0.52 x 1 = 0.52 </span>°C<span>
</span><span> ΔT(CaCl2) = 3 x 0.52 x 0.50 = 0.78 </span>°C<span>
</span><span> ΔT(NaCl) = 2 x 0.52 x 0.25 = 0.26 </span>°C<span>
</span>
Thus, from the calculated values, we see that 0.75 mol sucrose dissolved on 1 kg water has the same boiling point with 0.25 mol CaCl2 dissolved in 1 kg water.

5 0
3 years ago
If the enantiomeric excess of a mixture is 75%, what are the % compositions of the major and minor enantiomer?
Maru [420]

Let us say that R is the major enantiomer, while S is the minor enantiomer, therefore the formula for enantiomeric excess (ee) is:

ee = (R – S) * 100%

 

Let us further say that the fraction of R is x (R = x), and therefore fraction of S is 1 – x (S = 1 – x), therefore:

75 = (x – (1 – x)) * 100

75 = 100 x – 100 + 100 x

200 x = 175

x = 0.875

 

Summary of answers:

R = major enantiomer = 0.875 or 87.5%

<span>S = minor enantiomer = (1 – 0.875) = 0.125 or 12.5%</span>

7 0
4 years ago
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