Answer: 15.3 grams C
Explanation: 1 mole is 6.02x10^23 atoms. We can find the moles of C in 7.675 x 10^23 atoms of C by dividing:
(7.675 x 10^23 atoms C)/(6.02x10^23 atoms C/mole) = 1.275 moles C
The molar mass of carbon is 12g/mole. So the mass of 7.675 x 10^23 atoms is (1.275 moles C)*(12 g/mole C) = 15.3 grams.
Answer: Out of the given options acetic acid is in household vinegar.
Explanation:
Household vinegar is the one that is commonly used in our home while cooking a number of dishes.
The common name of vinegar is acetic acid and its chemical formula is 
.
For example, vinegar (acetic acid) is used while making noodles.
Thus, we can conclude that out of the given options acetic acid is in household vinegar.
Answer:
Iodide> Bromide > chloride > flouride
Explanation:
During a nucleophilic substitution reaction, a nucleophilie replaces another in a molecule.
This process may occur via an ionic mechanism (SN1) or via a concerted mechanism (SN2).
In either case, the ease of departure of the leaving group is determined by the nature of the C-X bond. The stronger the C-X bond, the worse the leaving group will be in nucleophilic substitution. The order of strength of C-X bond is F>Cl>Br>I.
Hence, iodine displays the weakest C-X bond strength and it is thus, a very good leaving group in nucleophillic substitution while fluorine displays a very high C-X bond strength hence it is a bad leaving group in nucleophilic substitution.
Therefore, the ease of the use of halide ions as leaving groups follows the trend; Iodide> Bromide > chloride > flouride
Answer:
D 30
Explanation:
whenever there is a reaction, the amount of atoms don't change, because matter can't be created or destroyed. Add the carbon and oxygen atoms together
Answer:
a) 0 J
b) -2.67x10² J
c) -2.09x10³ J
Explanation:
For an isothermic expansion (with constant temperature) the work (W) is :
W = -pΔV, where p is the pressure and ΔV the volume variation. The minus signal is used because the compression is positive and ΔV is negative (Vf < Vi).
a) In vacuum, the relative pressure is 0 atm, so the work:
W = -0x(5.7 - 1.3)
W = 0 J
b) For a constant pressure of 0.60 atm
W = -0.6atmx(5.7 - 1.3)L = -2.64 L.atm
1 L.atm = 101.3 J
W = -2.64x101.3 = -2.67x10² J
c) For a pressure of 4.7 atm
W = -4.7atmx(5.7 - 1.3) L = - 20.68 atm.L
1 atm.L = 101.3 J
W = -20.68x101.3 = -2.09x10³ J
