Answer:
d= 14.007 amu
Explanation:
Abundance of N¹⁴ = 99.63%
Abundance of N¹⁵ = 0.37%
Atomic mass of N¹⁴ = 14.003 amu
Atomic mass of N¹⁵ = 15.000 amu
Average atomic mass = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (14.003 × 99.63)+(15.000× 0.37) /100
Average atomic mass = 1395.12 + 5.55 / 100
Average atomic mass = 1400.67/ 100
Average atomic mass = 14.007 amu.
Explanation:
Ions are always formed when metals and non-metals interact because metals are electropositive. They willing release electrons to non-metals that are electronegative.
This activity results in charge separation. The transfer of electrons from one specie to another is what results in an ionic bond and the precedence of charged particles.
Between non-metals, the electrons are jointly shared. Therefore, there is no charge separation.
Answer:
A water molecule consists of two hydrogen atoms bonded to an oxygen atom, and its overall structure is bent. This is because the oxygen atom, in addition to forming bonds with the hydrogen atoms, also carries two pairs of unshared electrons. All of the electron pairs—shared and unshared—repel each other.
Explanation:
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
Answer: A volume of 59 mL of 0.220 M HBr solution is required to produce 0.0130 moles of HBr.
Explanation:
Given: Moles = 0.0130 mol
Molarity = 0.220 M
Molarity is the number of moles of solute present in liter of a solution.

Substitute the values into above formula as follows.

As 1 L = 1000 mL
So, 0.059 L = 59 mL
Thus, we can conclude that a volume of 59 mL of 0.220 M HBr solution is required to produce 0.0130 moles of HBr.