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Leno4ka [110]
3 years ago
13

Why do the gases on the periodic table tend to form negative ions?

Chemistry
1 answer:
boyakko [2]3 years ago
8 0
Because the valence shell of gases wants to become full
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Using only the periodic table, rank the elements in each set in order of increasing size: (a) Se, Br, Cl; (b) I, Xe, Ba.
mezya [45]

Answer:

A. Cl, Se, Br

B. I, Xe, Ba

Explanation:

4 0
3 years ago
If 0.092J of heat causes a 0.267 degree C temperature change, what mass of water is present?
Thepotemich [5.8K]

Answer:

0.082g

Explanation:

The following data were obtained from the question:

Heat (Q) = 0.092J

Change in temperature (ΔT) = 0.267°C

Specific heat capacity (C) of water = 4.184J/g°C

Mass (M) =..?

Thus, the mass of present can be obtained as follow:

Q = MCΔT

0.092 = M x 4.184 x 0.267

Divide both side by 4.184 x 0.267

M = 0.092 / (4.184 x 0.267)

M = 0.082g

Therefore, mass of water was present is 0.082.

6 0
3 years ago
The spectrochemical series is I < Br< < Cl^- < F
vfiekz [6]

Answer:

b. The splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.

Explanation:

The spectrochemical series is an arrangement of ligands in increasing order of their magnitude of crystal field splitting.

Ligands that occurs towards the right in the series are called strong field ligands and they tend to cause a greater magnitude of crystal field splitting. Ligands that occur towards the left hand side in the series are called weak field ligands and they tend to cause a lesser magnitude of crystal field splitting.

Since Cl^- is a weak field ligand, it causes a lesser magnitude of d orbital splitting compared to ethylenediammine (en) which causes a greater magnitude of d orbital splitting.

Hence; the splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.

6 0
3 years ago
I will put the questions in the comments that i need help with*
Paha777 [63]
The answer is a ) electromagnetic waves
7 0
3 years ago
0.254g lead(ii)ethanoate, on adding excess K2CrO4 solution, gave 0.130g of lead(ii)chromate precipitate. what is the percentage
Alborosie

Answer:

32.8%

Explanation:

All of the Pb⁺² species precipitated as lead(II) cromate, PbCrO₄ (we know this as excess K₂CrO₄ was used).

First we convert 0.130 g of PbCrO₄ into moles, using its molar mass:

  • 0.130 g ÷ 323 g/mol = 4.02x10⁻⁴ mol PbCrO₄

There's 1 Pb⁺² mol per PbCrO₄ mol, so in total 4.02x10⁻⁴ moles of Pb⁺² were in the ethanoate sample.

We <u>convert those 4.02x10⁻⁴ moles of Pb into grams</u>:

  • 4.02x10⁻⁴ mol * 207 g/mol = 0.083 g Pb

Finally we calculate the percentage composition of Pb:

  • 0.083 g Pb / 0.254 g salt * 100% = 32.8%
3 0
3 years ago
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