Here are the solutions:
For NaCl, there would be one electron transferred from the sodium atom, now producing a cation to the chlorine atom resulting in a chlorine anion. Forming ionic bond.
For CaS, there would be 2 electrons transferred from an atom of Ca to S, this can be obtained by simply looking at the ionic charges and or combining capacities of Ca.
For BaO, likewise, 2 electrons that are valence electrons are transferred from Ba to Oxygen.
In the ionic compound of KBr, the atom of K, potassium donated its one valence electron to Br, in need of it to achieve a stable octet, and produce the chemical bond.
Finally, LiF, a single valence electron is transferred from a metal atom of Li to F, for both of the atoms that would form the ionic bond to achieve a stable octet, and or full electron shell.
Answer:
6.52 × 10^14 Hz
i don't know if that's right tbh
The molarity of the NaOH solution is 0.03 M
We'll begin by calculating the mole of the KHP
- Mass = 0.212 g
- Molar mass = 204.22 g/mol
- Mole of KHP =?
Mole = mass /molar mass
Mole of KHP = 0.212 / 204.22
Mole of KHP = 0.001 mole
Next, we shall determine the molarity of the KHP solution
- Mole of KHP = 0.001 mole
- Volume = 50 mL = 50/1000 = 0.05 L
- Molarity of KHP =?
Molarity = mole / Volume
Molarity of KHP = 0.001 / 0.05
Molarity of KHP = 0.02 M
Finally , we shall determine the molarity of the NaOH solution
KHP + NaOH —> NaPK + H₂O
From the balanced equation above,
- The mole ratio of the acid, KHP (nA) = 1
- The mole ratio of base, NaOH (nB) = 1
From the question given above, the following data were obtained:
- Volume of acid, KHP (Va) = 50 mL
- Molarity of acid, KHP (Ma) = 0.02 M.
- Volume of base, NaOH (Vb) = 35 mL
- Molarity of base, NaOH (Mb) =?
MaVa / MbVb = nA / nB
(0.02 × 50) / (Mb × 35) = 1
1 / (Mb × 35) = 1
Cross multiply
Mb × 35 = 1
Divide both side by 35
Mb = 1 / 35
Mb = 0.03 M
Thus, the molarity of the NaOH solution is 0.03 M
Complete question:
See attached photo
Learn more about titration: brainly.com/question/25866669
Answer:
Explanation:
The movement of the electrons is illustrated in the picture attached to this answer. It is a four-step reaction mechanism.
First STEP: The first step involves the transfer of an electron from sodium to form a radical anion.
Second STEP: This radical anion then removes a proton/hydrogen from ammonia in a bid to neutralize itself (hence the hydrogen becomes bonded to the anion).
Third STEP: The sodium (from NaNH₂ formed) transfers an electron again to produce a vinyl carbanion.
Fourth STEP: The carbanion then removes a proton/hydrogen from ammonia (like in the second step) to form a neutral trans-alkene.
NOTE: The circled numbers denote each step while the mechanism on the left represents the use of any alkyl group (R and R') while the mechanism on the right assumes both alkyl groups are methyl. Hence, 2-butyne started the reaction and the final product was trans-2-butene.
Answer:
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Explanation: