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3 years ago

When the elements in group 1 are considered in order from top to bottom, each successive element at standard pressure has

2 answers:

6 0

When the elements in group 1 are considered in order from top to bottom, each successive element at a standard pressure has a lower melting point and boiling point. This is because melting and boiling points decrease when the increasing atomic number. The correct answer will be D.

Talja [164]3 years ago

4 0

Answer is: d. a lower melting point and boiling point.

Alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

Bond between these elements is metallic bond.

Both the melting and boiling points decrease down the group 1, because the decrease in the strength of each metallic bond.

From top to bottom, atomic radius increase and the distance between the nucleus and delocalized electrons increases, so metallic bond is weaker (attractions fall).

Metallic bond is formed between electrons and positively charged metal ions.

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A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

sammy [17]

<u>Answer:</u> The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

<u>Calculation for freezing point of solution:</u>

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

<u>Calculation for boiling point of solution:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

5 0

3 years ago

What is the maximum number of moles of N-acetyl-p-toluidine can be prepared from 70. milliliters of 0.167 M p-toluidine hydrochl

Kitty [74]

Answer:

$\large \boxed{\text{0.012 mol}}$

Explanation:

We will need a balanced equation with moles, so let's gather all the information in one place.

CH₃C₆H₄NH₂·HCl + (CH₃CO)₂O ⟶ CH₃C₆H₄NHCOCH₃ + junk

V/mL: 70.

c/mol·L⁻¹: 0.167

For simplicity in writing , let's call p-toluidine hydrochloride A and N-acetyl-<em>p</em>-toluidine B.

The equation is then

A + Ac₂O ⟶ B + junk

1. Moles of A

$\text{Moles of A} = \text{70. mL A}\times \dfrac{\text{0.167 mmol A}}{\text{1 mL A}}= \text{12 mmol A}$

2. Moles of B

The molar ratio is 1 mol B:1 mol A

Moles of B = moles of A = 12 mmol = 0.012 mol

$\text{You can prepare $\large \boxed{\textbf{0.012 mol}}$ of N-acetyl-p-toluidine. }$

3 0

3 years ago

Under what conditions do you expect hydrogen gas to deviate from ideal gas behavior? Explain your answer.

Ostrovityanka [42]

Hydrogen is a non-polar gas with very weak intermolecular forces of attraction. Hydrogen will deviate from the ideal gas behavior at high pressure.

4 0

3 years ago

The weight of a table as measured by a student is 20 kg.The

miskamm [114]

(20/23) times 100 is 87%

now the answer asks for percentage ERROR

100-87 =13

so answer is C

7 0

2 years ago

Read 2 more answers

If the actual yield of a reaction is 37.6 g while the theoretical yield is 112.8 g what is the percent yield

Zigmanuir [339]

<h2>Hello!</h2>

The answer is:

The percent yield of the reaction is 32.45%

To calculate the percent yield, we have to consider the theoretical yield and the actual yield. The theoretical yield as its name says is the yield expected, however, many times the difference between the theoretical yield and the actual yield is notorious.

We are given that:

$ActualYield=37.6g\\TheoreticalYield=112.8g$

Now, to calculate the percent yield, we need to divide the actual yield by the theoretical and multiply it by 100.

So, calculating we have:

$PercentYield=\frac{ActualYield}{TheoreticalYield}*100\\\\PercentYield=\frac{37.6g}{112.8g}*100=0.3245*100=32.45(percent)$

Hence, we have that the percent yield of the reaction is 32.45%.

Have a nice day!

8 0

3 years ago

Read 2 more answers