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alexdok [17]
3 years ago
14

Which of the following is generally FALSE? Group of answer choices Elements in the same group have the same valence electron con

figuration. Completely empty and completely full shells are very stable. Atomic radius of elements in a group decreases as you go from top to bottom. First ionization energy of elements in the same group decreases as you go from top to bottom. All of these statements are false.
Chemistry
1 answer:
Whitepunk [10]3 years ago
3 0

Answer: Option (c) is the correct answer.

Explanation:

Generally, when we move from top to bottom in a group then there occur increase in the number of electrons due to which there will also occur increase in the number of shells.

As atomic size is the distance between the nucleus and valence shell of an atom. Hence, more is the number of shells present in an atom more will be its atomic radius.

Thus, we can conclude that the statement atomic radius of elements in a group decreases as you go from top to bottom, is generally FALSE.

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Answer:

118 elements have been discovered so far

Explanation:

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2 years ago
In a solution the___is the liquid that the____is added to.
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Answer:

In a solution the solvent is the liquid that the solute is added to.

Explanation:

Solute + Solvent = Solution.

For example: An aqueous solution of 10 g of NaCl in 100 g of water

NaCl → solute → what you dissolved

H₂O → solvent → where you dissolve

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What way do S waves move
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4 0
10 months ago
I need help with 1,2,3, and 4
Schach [20]

Answer:

  • Problem 1: 1.85atm
  • Problem 2: 110mL
  • Problem 3: 290 mL
  • Problem 4: 1.14 atm

Explanation:

Problem 1

<u>1. Data</u>

<u />

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

<u>2. Formula</u>

Since the temeperature is constant you can use Boyle's law for idial gases:

          PV=constant\\\\P_1V_1=P_2V_2

<u>3. Solution</u>

Solve, substitute and compute:

         P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2

        P_2=3.25atm\times755mL/1325mL=1.85atm

Problem 2

<u>1. Data</u>

<u />

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

<u>2. Formula</u>

You assume that the temperature does not change, and then can use Boyl'es law again.

          P_1V_1=P_2V_2

<u>3. Solution</u>

This time, solve for V₂:

           P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2

Substitute and compute:

        V_2=548mmHg\times 125mL/625mmHg=109.6mL

You must round to 3 significant figures:

        V_2=110mL

Problem 3

<u>1. Data</u>

<u />

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

<u>2. Formula</u>

At constant pressure, Charle's law states that volume and temperature are inversely related:

         V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

The temperatures must be in absolute scale.

<u />

<u>3. Solution</u>

a) Convert the temperatures to kelvins:

  • T₁ = 25 + 273.15K = 298.15K

  • T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

        \dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml

You must round to two significant figures: 290 ml

Problem 4

<u>1. Data</u>

<u />

a) P = 865mmHg

b) Convert to atm

<u>2. Formula</u>

You must use a conversion factor.

  • 1 atm = 760 mmHg

Divide both sides by 760 mmHg

       \dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}

<u />

<u>3. Solution</u>

Multiply 865 mmHg by the conversion factor:

    865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer

3 0
3 years ago
3. Determine the moles of sodium, Na, containing 7.9x1024 atoms.​
-BARSIC- [3]

Answer:

12.7mol Na.

Explanation:

Hello there!

In this case, according to the concept of mole, which stands for the amount of substance, we can recall the concept of Avogadro's number whereby we understand that one mole of any substance contains 6.022x10²³ particles, for the given atoms of sodium, we can calculate the moles as shown below:

7.9x10^{23}atoms*\frac{1mol}{6.022x10^{23}atoms} \\\\

Thus, by performing the division we obtain:

12.7molNa

Regards!

8 0
3 years ago
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