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.
Answer:
The 18th-century German physicist Daniel Gabriel Fahrenheit originally took as the zero of his scale the temperature of an equal ice-salt mixture
Explanation:
The hybridization for C in acetylene, HCCH, or C₂H₂ is 'sp'.
Discussion:
There are three different forms of hybridization -
- sp- The first occurs when two carbon atoms are triple linked.
- sp₂- When two carbon atoms are double-bonded to one another, this is known as sp₂.
- sp₃- When a single bond joins two carbon atoms, this is known as sp₃.
In the case of acetylene(HCCH or C₂H₂):
- The carbon atom requires additional electrons to establish four bonds with hydrogen and other carbon atoms in the synthesis of C₂H₂. One 2s₂ pair is consequently transferred to the vacant 2pz orbital. Each carbon has two sp hybrid orbitals after the 2s orbital in each atom combines with one of the 2p orbitals.
- As a result of the atoms' symmetrical alignment in a single plane, C₂H₂ possesses a linear molecular structure. Due to their lower electronegative nature than Hydrogen atoms, all Carbon atoms are situated near the center of the Lewis structure of C₂H₂.
H-C≡C-H
Therefore, it is concluded from the above discussion that the hybridization type of acetylene is 'sp'.
Learn more about hybridization here:
brainly.com/question/14140731
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6.7× 10³ g/mol*L ×(1Kg÷1000g) × (1L÷10dL)= 0.67