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:
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The solution needed is prepared as below
by use of the M1V1 =M2 V2 formula where
M1 = 2.25 L
v2 = 1.0M
M2 = 9.0 M
V2 =? l
make V2 the subject of the formula V2 =M1V1/M2
= 2.25 L x 1.0M/9.0 M = 0. 25 L
therefore the solution need 0.25 L of 9.0M H3PO4 and dilute it a final volume of 2.25 l
Can you upload the picture please so i can help
Answer:
(A) 4.616 * 10⁻⁶ M
(B) 0.576 mg CuSO₄·5H₂O
Explanation:
- The molar weight of CuSO₄·5H₂O is:
63.55 + 32 + 16*4 + 5*(2+16) = 249.55 g/mol
- The molarity of the first solution is:
(0.096 gCuSO₄·5H₂O ÷ 249.55 g/mol) / (0.5 L) = 3.847 * 10⁻⁴ M
The molarity of CuSO₄·5H₂O is the same as the molarity of just CuSO₄.
- Now we use the dilution factor in order to calculate the molarity in the second solution:
(A) 3.847 * 10⁻⁴ M * 6mL/500mL = 4.616 * 10⁻⁶ M
To answer (B), we can calculate the moles of CuSO₄·5H₂O contained in 500 mL of a solution with a concentration of 4.616 * 10⁻⁶ M:
- 4.616 * 10⁻⁶ M * 500 mL = 2.308 * 10⁻³ mmol CuSO₄·5H₂O
- 2.308 * 10⁻³ mmol CuSO₄·5H₂O * 249.55 mg/mmol = 0.576 mg CuSO₄·5H₂O