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Molodets [167]
4 years ago
10

The most negative electron affinity is most likely associated with which type of atoms?

Chemistry
2 answers:
Llana [10]4 years ago
7 0

Explanation:

elements are based on electrical conductivity

Goshia [24]4 years ago
3 0

Answer:

small nonmetal atoms

Explanation:

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To test the quality of water sources in her area, Shameka collects five samples from each water source and tests the quality of
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Answer:

The answer is "Verification of results ".

Explanation:

The scientific system consisted inside an iterative or cyclical process which experimental results and technically confirm a theory. There are four things which scientific design aims people explain:  

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Its first stage of analysis involves collecting the samples taken for the current context, that are the focus of discussion, and then testing the samples taken to determine the absorbance of soft and hard salts throughout the water. It is required for water quality to be determined. Therefore, the collection and testing of samples falls under the scientific design category of classification, whereby Shamekha tried to verify the findings.

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3 years ago
The table below gives some characteristics of element Y.
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The ________ ion is represented by the electron configuration [ar]3d2.
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The most unreactive substances are:
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3 years ago
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A flask of volume 2.0 liters, provided with a stopcock, contains oxygen at 20 oC, 1.0 ATM (1.013X105 Pa). The system is heated t
leonid [27]

Answer:

1.27 atm is the final pressure of the oxygen in the flask (with the stopcock closed).

2.6592 grams of oxygen remain in the flask.

Explanation:

Volume of the flask remains constant = V = 2.0 L

Initial pressure of the oxygen gas = P_1=1.0 atm

Initial temperature of the oxygen gas = T_1=20^oC =293.15 K

Final pressure of the oxygen gas = P_2=?

Final temperature of the oxygen gas = T_2=100^oC =373.15 K

Using Gay Lussac's law:

\frac{P_1}{T_1}=\frac{P_2}{T_2}

P_2=\frac{P_1\times T_2}{T_1}=\frac{1 atm\times 373.15 K}{293.15 K}=1.27 atm

1.27 atm is the final pressure of the oxygen in the flask (with the stopcock closed).

Moles of oxygen gas = n

P_1V_1=nRT_1 (ideal gas equation)

n=\frac{P_1V_1}{RT_1}=\frac{1 atm\times 2.0 L}{0.0821 atm l/mol K\times 293.15 K}=0.08310 mol

Mass of 0.08310 moles of oxygen gas:

0.08310 mol × 32 g/mol = 2.6592 g

2.6592 grams of oxygen remain in the flask.

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3 years ago
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