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3 years ago

Explain why a lithium ion (Li+) is smaller than a lithium atom (Li).

Chemistry

2 answers:

bonufazy [111]3 years ago

7 0

Answer:

Lithium ions are smaller than lithium atoms. This is because the lithium atom has 2 shells, while the ion just has one shell. Bromide ions are larger than bromine atoms, due to increased inter-electronic respulsion.

Explanation:

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Vitek1552 [10]3 years ago

5 0

Li+ lost an electron, meaning it got more positive, but also smaller

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Be sure to answer all parts. Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after

Fantom [35]

Answer:

(a) pH = 12.73

(b) pH = 10.52

Explanation:

The net balanced reaction equation is:

KOH + HBr ⇒ H₂O + KBr

The amount of KOH present is:

n = CV = (0.1000 molL⁻¹)(30.00 mL) = 3.000 mmol

(a) The amount of HBr added in 9.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(9.00 mL) = 0.900 mmol

This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:

(3.000 mmol) - (0.900 mmol) = 2.100 mmol KOH

After the addition of HBr, the volume of the KOH solution is 39.00 mL. The concentration of KOH is calculated as follows:

C = n/V = (2.100 mmol) / (39.00 mL) = 0.0538461 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0538461) = 1.2688

pH = 14 - pOH = 14 - 1.2688 = 12.73

(b) The amount of HBr added in 29.80 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(29.80 mL) = 2.980 mmol

This amount of HBr will neutralize an equivalent amount of KOH, leaving the following amount of KOH:

(3.000 mmol) - (2.980 mmol) = 0.0200 mmol KOH

After the addition of HBr, the volume of the KOH solution is 59.80 mL. The concentration of KOH is calculated as follows:

C = n/V = (0.0200 mmol) / (59.80 mL) = 0.0003344 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0003344) = 3.476

pH = 14 - pOH = 14 - 3.476 = 10.52

(0.1000 molL⁻¹)(38.00 mL) = 3.800 mmol

This amount of HBr will neutralize all of the KOH present. The amount of HBr in excess is:

(3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr

After the addition of HBr, the volume of the analyte solution is 68.00 mL. The concentration of HBr is calculated as follows:

C = n/V = (0.800 mmol) / (68.00 mL) = 0.01176 M HBr

The pH of the solution can then be calculated:

pH = -log[H⁺] = -log(0.01176) = 1.93

4 0

3 years ago

Calculate the mass fraction of sodium chloride in the solution if 20 g of it is dissolved in 300 ml of water.

MA_775_DIABLO [31]

The mass fraction of sodium chloride is 0.0625

<h3>What is the mass fraction of sodium chloride in the solution?</h3>

The mass fraction of sodium chloride is the ratio of the mass of sodium chloride to the total mass of the solution.

The mass fraction of sodium chloride is determined as follows;

mass of sodium chloride = 20 g

mass of water = volume * density

density of water = 1 g/mL

volume of water = 300 mL

mass of water = 300 mL * 1 g/mL

mass of water = 300 g

total mass of solution = 20 + 300 = 320 g

mass fraction of sodium chloride = 20/320

mass fraction of sodium chloride = 0.0625

Learn more about mass fraction at: brainly.com/question/14783710

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1 year ago

Question 5 of 10

pychu [463]

The answer is D. S2O6

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3 years ago

Which of the following is NOT a true statement?

Artyom0805 [142]

Answer:

c is not a true statement

8 0

3 years ago

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What does Hess's law state can be done in order to be able to react solid magnesium with oxygen gas safely (that is, without exp

aleksley [76]

Answer:

C. The reaction can be broken down and performed in steps

Explanation:

Hess's Law of Constant Heat Summation states that irrespective of the number of steps followed in a reaction, the total enthalpy change for the reaction is the sum of all enthalpy changes corresponding to all the steps in the overall reaction. The implication of this law is that the change of enthalpy in a chemical reaction is independent of the pathway between the initial and final states of the system.

To obtain MgO safely without exposing magnesium to flame, the reaction sequence shown in the image attached may be carried out. Since the enthalpy of the overall reaction is independent of the pathway between the initial and final states of the system, the sum of the enthalpy of each step yields the enthalpy of formation of MgO.

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3 years ago