Answer:
a, g, c
Explanation:
The conversion of the stable cyclopentane into Trans-1, 2dibromocyclopentane will require three step reactions.
The first is to convert the compound into a cyclopentene, through the addition of Bromine water under heat and photons (light). So option A is the first in the order. This will generate 1 bromocyclopentane through halogenation of the alkane. Secondly, a hot and strong base should be added like the NaOEt, EtOH to remove the added bromine and one atom of hydrogen from the resulting 1 bromocyclopentane in the previous reaction. This will yield cyclopentene, thus making the compound more electrophilic. So option g is required. Thirdly, bromine molecules will be added (C) to take up their places at the two electrophilic regions of the compound to produce Trans-1, 2dibromocyclopentane.
Answer:
0.106 mol (3s.f.)
Explanation:
To find the number of moles, divide the mass of glucose (in grams) by its Mr. Glucose has a chemical formula of C6H12O6. To find the Mr, add all the Ar of all the atoms in C6H12O6.
Ar of C= 12, Ar of H= 1, Mr of O= 16
These Ar values can be found on the periodic table.
Mr of glucose= 6(12)+ 12(1) + 6(16)= 180
Moles of glucose
= mass ÷ mr
= 19.1 ÷ 180
= 0.106 mol (3 s.f.)
The charge on Pb in Pb(SO3)2 is Lead (IV) Sulfite.
Answer:
- There will be 1.23 moles of helium in the balloon at STP
Explanation:
1) <u>Initial conditions of the helium gas</u>:
- V = 20.0 liter
- p = 1.50 atm
- T = 25.0 °C = 25.0 + 273.15 K = 298.15 K
2) <u>Ideal gas equation</u>:
- pV = n RT
- p, V, and T are given above
- R is the Universal constant = 0.0821 atm-liter / ( K - mol)
- n is the unknown number of moles
3) <u>Solve for n</u>:
- n = 1.50 atm × 20.0 liter / (0.0821 atm-liter /k -mol ×298.15K)
4) <u>At STP:</u>
- STP stands for standard pressure and temperature.
- The amount (number of moles) of the gas will not change because the change of pressure and temperature, so the number of moles reamain the same: 1.23 mol.
Considering the ideal gas law, a sample weighing 9.49 g occupies 68.67 L at 353 K and 2.00 atm.
Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P× V = n× R× T
In this case, you know:
- P= 2 atm
- V= ?
- n=
being 2g/mole the molar mass of H2, that is, the amount of mass that a substance contains in one mole. - R= 0.082
- T= 353 K
Replacing:
2 atm× V = 4.745 moles× 0.082× 353 K
Solving:
V = (4.745 moles× 0.082× 353 K)÷ 2 atm
<u><em>V= 68.67 L</em></u>
Finally, a sample weighing 9.49 g occupies 68.67 L at 353 K and 2.00 atm.
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