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mojhsa [17]
3 years ago
15

Predict the shape of the molecule....

Chemistry
1 answer:
Gemiola [76]3 years ago
4 0

Answer: trigonal bipyramidal

Explanation:

Number of electron pairs = \frac{1}{2}[V+N-C+A]

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

SbCl_5 :

In the given molecule, antimony is the central atom and there are five chlorine as monovalent atoms.

The number of electron pairs are 5 that means the hybridization will be sp^3d and geometry of the molecule will be trigonal bipyramidal.

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25 g of a chemical is dissolved in 75 g of water. What is the percent by mass concentration of this chemical
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Answer:

300

Explanation:

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3 years ago
Explain how you would find the solubility of a solute
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Answer:

Solubility indicates the maximum amount of a substance that can be dissolved in a solvent at a given temperature. Such a solution is called saturated. Divide the mass of the compound by the mass of the solvent and then multiply by 100 g to calculate the solubility in g/100g .

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In part 1 of lab 2 you will make and standardize a solution of naoh(aq). suppose in the lab you measure the solid naoh and disso
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Which elements make up 95 percent (by weight) of the human body?
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Hope this helps.
3 0
3 years ago
Be sure to answer all parts. A concentration cell consists of two Sn/Sn2+ half-cells. The electrolyte in compartment A is 0.23 M
kolbaska11 [484]

Answer:

Part A. The half-cell B is the cathode and the half-cell A is the anode

Part B. 0.017V

Explanation:

Part A

The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).

So, the half-cell B is the cathode and the half-cell A is the anode.

Part B

By the Nersnt equation:

E°cell = E° - (0.0592/n)*log[anode]/[cathode]

Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.

E°cell = 0 - (0.0592/2)*log(0.23/0.87)

E°cell = 0.017V

3 0
4 years ago
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