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irina [24]
3 years ago
10

hi, can you check if my answers are correct? I'm not so confident in this topic (topic: Classifying Types of Chemical Reactions)

Chemistry
1 answer:
NikAS [45]3 years ago
7 0
I am pretty sure your answer are correct , from what I know. Good job!
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Emperical formula of carbon dioxide
Stells [14]

CO2 is the emperical formula of carbon dioxide

6 0
3 years ago
Read 2 more answers
If 11.5 ml of vinegar sample (d=1g/ml) is titrated with 18.5 ml of standardized Sodium hydroxide
insens350 [35]

Answer:

4.83% of acetic acid in the vinegar

Explanation:

3 0
3 years ago
According to VSEPR theory , in which fashion will the bonds and lone pairs of electrons be arranged about the central atom in th
Elina [12.6K]

Answer:

Tetrahedral, trigonal pyramidal, trigonal bipyramidal.

Explanation:

The VSPER theory states that the bonds of sharing electrons and the lone pairs of electrons will repulse as much as possible. So, by the repulsion, the molecule will have some shape.

In the ion PO₄³⁻, the central atom P has 5 electrons in its valence shell, so it needs 3 electrons to be stable. Oxygen has 6 electrons at the valence shell and needs 2 to be stable. 3 oxygens share 1 pair of electrons with P, and the two lone pair remaining in P is shared with the other O, then the central atom makes 4 bonds and has no lone pairs, the shape is tetrahedral.

In the ion H₃O⁺, the central atom O has 6 electrons in its valence shell and needs 2 electrons to be stable. The hydrogen has 1 electron, and need 1 more to be stable. The hydrogens share 1 pair of electrons with the oxygen, then it remains 3 electrons at the central atom, and the VSPER theory states that the shape will be a trigonal pyramidal.

In the AsF₅, the central atom As has 5 valence electrons, and F has 1 electron in its valence shell, so each F shares one pair of electrons with As, and there are no lone pairs in the central atom. For 5 bonds without lone pairs, the shape is trigonal bipyramidal.

7 0
3 years ago
In the haber process, ammonia is synthesized from nitrogen and hydrogen: n2 (g) + 3h2 (g) → 2nh3 (g) δg° at 298 k for this react
allsm [11]

Answer:

e- 7.25 x 10³.

Explanation:

∵ ΔG = -RTlnK,

where, ΔG is the free energy change.

R is the general gas constant (R = 8.324 J/mol.K).

K is the equilibrium constant of the reaction.

  • For the reaction: <em>N₂(g) + 3H₂(g) → 2NH₃(g),</em>

K = (PNH₃)²/(PN₂)(PH₂)³ = (0.65)²/(1.9)(1.6)³ = 5.43 x 10⁻².

∵ ΔG = -RTlnK.

∴ ΔG = -(8.314 J/mol.K)(298 K) ln(5.43 x 10⁻²) = 7.218 x 10³ J/mol.

3 0
3 years ago
Why are the transition metals not isoelectronic with noble gas atoms?
V125BC [204]
The atoms and ions that have the similar electron configuration are named isoelectronic. The common ions of representative elements are isoelectronic with a noble gas. When forming ions, transition metals be unable to find their valence s-sublevel electrons before they lose their d-sub level electrons.
8 0
3 years ago
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