Answer:
The answer to your question is: a. pentose
Explanation:
a) a pentose is a monosaccharide that has 5 carbons in its structure, if we look at the question the chemical formula only has 5 carbons, then this is the right answer.
b) an oligosaccharide is a group of 2 or more monosaccharides (till 5 or 6) then, it will have 10 or more carbons, this is not the right answer
c) a triose is a monosaccharide that has only three carbons, this is not the right answer.
d) a hexose is a monosaccharide that has 6 carbons, of course this answer is wrong.
e) a polysaccharide is a group of 6 or more monosaccharides, it will have 20 or more carbons, this answer is wrong.
Answer:
Explanation:
Hello!
In this case, since the heat of vaporization is related with the energy required by a substance to undergo the phase transition from liquid to gas, we can compute such amount of energy as shown below:
In such a way, since the enthalpy of vaporization is given as well as the mass, we compute the energy as shown below:
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Answer:
Br
Explanation:
Bromine atoms tend to gain just one electron to get to a full octet, as Bromine is in Group VII.
Answer:
6.82 kg
Explanation:
Given that the amount of water is 15L and we know that the density of water is ≈ 1kg/L. The mass of water is given by mass = volume x density, i.e,
mass = 15 x 1 = 15 kg. Also the specific heat capacity of water is 4.186 KJ/kg.
The sublimation enthalpy of dry ice is 571 KJ/kg.
Now, the amount of heat lost by water is entirely used up for the sublimation (conversion from soild to gas) of dry ice. And the heat (Q) lost by water is given as : Q = mCΔT, where m is the mass of water, C the specific heat capacity of water and ΔT the change in temperature.
Here, Q = 15 x 4.186 x (90 - 28) = 3892.98 KJ.
This amount of heat is taken up by the dry ice for its sublimation. Also the energy taken by dry ice (Q') for its sublimation is given by: Q' = m'L', where m' is the mass of dry ice, L' is the latent heat of sublimation (i.e, the amount of heat required per kg of a substance to sublime) of dry ice amd L' = 571 KJ/kg.
Now, Q' =m'L' = heat lost by water = 3892.98KJ.
And, m'L' = m' x 571 KJ/kg = 3892.98 KJ. (Dividing with 571)
Therefore, m' = 6.82 kg.
