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ipn [44]
2 years ago
14

if lithium loses an electron to become li+ what is the average atomic mass of the lithium ion? explain how you arrived at your a

nswer
Chemistry
2 answers:
Pavlova-9 [17]2 years ago
8 0
The mass of an electron 1/1823 amu's = 0.0005 amu's... the average mass of litium is 6.941 amu's. by losing an electron, the mass of average atomic mass of the Lithium doesn't really change much. 
Aliun [14]2 years ago
4 0

<u>Answer:</u> The average atomic mass of the lithium ion remains the same as lithium atom which is 6.94 amu

<u>Explanation:</u>

Mass number is defined as the sum of number of protons and number of neutrons present in an atom. It is represented as A.

A = Mass number = Number of neutrons + Number of protons.

An ion is formed when a neutral atom looses or gains electrons.

  • When an atom looses electrons, it results in the formation of positive ion known as cation.
  • When an atom gains electrons, it results in the formation of negative ion known as anion.

So, the loss or gain of electrons does not effect the mass number of an atom. Thus, when lithium atom looses an electron to form lithium ion (Li^+), the average atomic mass of the ion remains the same as the neutral atom.

Average atomic mass of lithium atom = 6.94 amu

Hence, the average atomic mass of lithium ion is 6.94 amu.

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Answer:

A group of atoms bonded together by different chemical reactions. It is the smallest fundamental unit of a chemical compound.

Explanation:

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2 years ago
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A gas occupies a volume at 34.2 mL at a temperature of 15.0 C and a pressure of 800.0 torr. What will be the volume of this gas
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The answer is 34.1 mL.
Solution:
Assuming ideal behavior of gases, we can use the universal gas law equation
     P1V1/T1 = P2V2/T2
The terms with subscripts of one represent the given initial values while for terms with subscripts of two represent the standard states which is the final condition.
At STP, P2 is 760.0torr and T2 is 0°C or 273.15K. Substituting the values to the ideal gas expression, we can now calculate for the volume V2 of the gas at STP:
     (800.0torr * 34.2mL) / 288.15K = (760.0torr * V2) / 273.15K
     V2 = (800.0torr * 34.2mL * 273.15K) / (288.15K * 760.0torr)
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3 years ago
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A piece of wood has a labeled length value of 63.2 cm. You measure its length three times and record the following data: 63.1 cm
Ulleksa [173]

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Accepted value is true value.

Measured values is calculated value.

In the question given Accepted value (true value) = 63.2 cm

Given Measured(calculated values) = 63.1 cm , 63.0 cm , 63.7 cm

1) Percent error (%) for first measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.1 cm

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Percent error = \frac{\left | 63.2 - 63.1 \right |}{63.2}\times 100

Percent error = \frac{0.1}{63.2}\times 100

Percent error = 0.00158\times 100

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2) Percent error (%) for second measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.0 cm

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Percent error = \frac{\left | 63.2 - 63.0 \right |}{63.2}\times 100

Percent error = \frac{0.2}{63.2}\times 100

Percent error = 0.00316\times 100

Percent error = 0.316 %

3) Percent error (%) for third measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.7 cm

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Percent error = \frac{\left | 63.2 - 63.7 \right |}{63.2}\times 100

Percent error = \frac{\left | -0.5 \right |}{63.2}\times 100

Percent error = \frac{(0.5)}{63.2}\times 100

Percent error = 0.00791\times 100

Percent error = 0.791 %

Percent error for each measurement is :

63.1 cm = 0.158%

63.0 cm = 0.316%

63.7 cm = 0.791%




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3 years ago
What is the role of NaCl and EDTA in DNA isolation?
alexandr402 [8]
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