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sertanlavr [38]
3 years ago
15

Unknown element x is a metal that ionically bonds to sulfur.is the formula, x3s feasible? why or why not?a)it is feasible. the t

hree metallic ions each receive an electron from sulfur.b)no, it is not feasible. sulfur cannot give three electrons to the metal since the sulfide ion has a -2 charge.c)no, it is not feasible. the metallic ion x only has one valence electron and the sulfide ion receives two electrons.d)no, it is not feasible. three metallic ions cannot provide the exact number of electrons that one sulfur needs for the ionic bond.
Chemistry
2 answers:
nasty-shy [4]3 years ago
8 0
The correct answer is:
d) No, it is not feasible. three metallic ions cannot provide the exact number of electrons that one sulfur needs for the ionic bond.
Because Sulfur is divalent so it need to gain 2 electrons from metal so if we have 3 metals they can't provide only two electrons only.
ad-work [718]3 years ago
4 0

Answer:

D)

Explanation:

No, it is not feasible. Three metallic ions cannot provide the exact number of electrons that one sulfur needs for the ionic bond. One sulfide ion would receive exactly two electrons for the ionic bond. This is impossible to get from three metallic ions.

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Answer:

Electrical conductivity

Explanation:

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3 0
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In many modern cars the fuel pump is usually electric and located inside the fuel tank. The pump creates positive pressure in the fuel lines, pushing the gasoline to the engine. The higher gasoline pressure raises the boiling point.
5 0
3 years ago
How many L of 3.0 M H2SO4 solution can be prepared by using 100.0 mL OF 18 M H2SO4?
max2010maxim [7]

Answer: A volume of 600 mL of 3.0 M H_{2}SO_{4} solution can be prepared by using 100.0 mL OF 18 M H_{2}SO_{4}.

Explanation:

Given: V_{1} = ?,        M_{1} = 3.0 M\\

V_{2} = 100.0 mL,       M_{2} = 18 M

Formula used to calculate the volume is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute the values into above formula as follows.

M_{1}V_{1} = M_{2}V_{2}\\3.0 M \times V_{1} = 18 M \times 100.0 mL\\V_{1} = \frac{18 M \times 100.0 mL}{3.0M}\\= 600 mL

Thus, we can conclude that a volume of 600 mL of 3.0 M H_{2}SO_{4} solution can be prepared by using 100.0 mL OF 18 M H_{2}SO_{4}.

6 0
3 years ago
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Mekhanik [1.2K]

Answer:B

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5 0
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Why is it important to make sure that the endpoint is noted accurately in titration?
andrezito [222]
There should be no air bubbles, as a bubble bursting can affect the volume measurement and disrupt the accuracy of the experiment.
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