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lidiya [134]
4 years ago
15

Sodium tends lose a single electron in natural settings. Based on what you know, what are two other elements that tend to do the

same thing?
Chemistry
1 answer:
Oksi-84 [34.3K]4 years ago
6 0

Answer:

Lithium and Sodium

Explanation:

Losing and electron in natural setting is characterizes of elements in group one. These are elements known as the alkaline earth metals. They are the most electropositive elements on the periodic table.

These elements ionize by losing an electron yin their outermost shell to attain the configuration of the nearest noble gas. These elements are usually found in combined and rarely seen in uncombined state principally due to their very reactive nature.

Sodium naturally would ionize by losing one electron. Other elements capable of this even at a better rate because they are more electropositive are potassium and lithium. Both are also group one alkaline metals

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The parameter pH is a measure of the substance's acidity or basicity. It is quantitatively equal to the negative logarithm of the concentration of H+ ions. So, the lower the pH, the more acidic the substance, Otherwise, the higher the pH the more basic the substance. The pH range runs from 1 to 14, with 7 being neutral.

So, if we are asked to distinguish which of those have the lowest pH, we have to know the moles of H+ ions. Since all of them have a concentration of 0.1 M, concentration is not a factor. Thus, we just have to identify the strongest acid among the list. That is easy to answer because you only have to remember 7 strong acid occurring in nature: HCl, H₂SO₄, HNO₃, HClO₄, HClO₃, HBr, and HI. Since only HNO₃ is included in the list among the choices, the answer would be letter D. 
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Perform the following conversions:
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Answer :

The conversion used for the temperature from Fahrenheit to degree Celsius is:

^oC=\frac{5}{9}\times (^oF-32)

The conversion used for the temperature from degree Celsius to Kelvin is:

K=^oC+273

(a) 1.06^oF

^oC=\frac{5}{9}\times (^oF-32)

^oC=\frac{5}{9}\times (1.06^oF-32)

^oC=-17.188

The temperature in degree Celsius is, -17.188^oC

K=^oC+273

K=-17.188+273

K=255.812

The temperature in Kelvin is, 255.812 K

(b) 3410^oC

^oC=\frac{5}{9}\times (^oF-32)

3410^oC=\frac{5}{9}\times (^oF-32)

^oF=6106

The temperature in degree Fahrenheit is, 6106^oF

K=^oC+273

K=3410+273

K=3683

The temperature in Kelvin is, 3683 K

(c) 6.1\times 10^3K

K=^oC+273

6.1\times 10^3K=^oC+273

^oC=5827

The temperature in degree Celsius is, 5827^oC

^oC=\frac{5}{9}\times (^oF-32)

5827^oC=\frac{5}{9}\times (^oF-32)

^oF=10456.6

The temperature in degree Fahrenheit is, 10456.6^oF

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