Answer:
Since it is given that the mass of a jelly bean is less than the mass of a gum drop.
So, when we counted out 10 of each kind of candy and measured the mass of each kind of candy, the mass of the jellybeans would be less than the mass of the gumdrops.
This is because mass of jelly bean is less and even if we take take or more jelly beans then also their total mass will remain less than the total mass of same number of gum drops
Explanation:
Explanation:
During a chemical reaction, the atoms of the original substances gain, lose or share their electrons with those of the substances with which they are reacting. The reaction creates new substances made up of a new combination of atoms and a different configuration of electrons.
Answer:
-88.66 kJ/mol
Explanation:
The expressions of heat capacity (Cp,m) for C(s) and for H₂(g) are:
C(s): Cp,m/(J K-1 mol-1) = 16.86 + (4.77T/10³) - (8.54x10⁵/T²)
H₂(g): Cp,m/(J K-1 mol-1) = 27.28 + (3.26T/10³) + (0.50x10⁵/T²)
Cp = A + BT + CT⁻²
For the Kirchoff's Law:
ΔHf = ΔH°f + 
Where ΔH°f is the enthalpy at 298 K, T1 is 298 K, T2 is the temperature given (373 K), and DCp is the variation of Cp (products less reactants). ΔH°f for ethene is -84.68 kJ/mol and the reaction is:
2C(s) + 3H₂(g) → C₂H₆
So, DCp:
dA = A(C₂H₆) - [2xA(C) + 3xA(H₂)] = 14.73 - [2x16.86 + 3x27.28] = -100.83
dB = B(C₂H₆) - [2xB(C) + 3xB(H₂)] = 0.1272 - [2x4.77x10⁻³ + 3x3.26x10⁻³] = 0.10788
dC = C(C₂H₆) - [2xC(C) + 3xC(H₂)] = 0 - (2x(-8.54x10⁵) + 3x0.50x10⁵) = 15.58x10⁵
dCp = -100.83 + 0.10788T + 15.58x10⁵T⁻²
= -3796.48 J/mol = -3.80 kJ/mol (solved by a graphic calculator)
ΔHf = -84.68 - 3.80
ΔHf = -88.66 kJ/mol
Physical properties can be observed or measured without changing the composition of matter. These are properties such as mass, weight, volume, and density. Remember that density is a physical property.
Explanation:
These elements are rare because:
<u>Helium fuses into the carbon by the combination of three helium nuclei (Z = 2) and one carbon nucleus (Z = 6), therefore bypassing elements with Z= 3, 4 and 5 which are lithium, beryllium, and boron respectively. Therefore, the fusion processes in cores of the stars do not form these three elements. </u>
