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3 years ago

Decide which element probably has a boiling point most and least similar to the boiling point of cesium.

Chemistry

1 answer:

natka813 [3]3 years ago

5 0

Answer:

Take a look at the attachment below

Explanation:

Take a look at the periodic table. As you can see, Rubidium is the closest element to Cesium, and happens to have the closest boiling point to Cesium, with only a difference of about 30 degrees.

Respectively, you would think that fluorine should have the least similarity to Cesium with respect to it's boiling point, considering it is the farthest away from the element out of the 4 given. This is not an actual rule, there are no fixed trends of boiling points in the periodic table, there are some but overall the trends vary. However in this case fluorine does have the least similarity to Cesium with respect to it's boiling point, a difference of about 1,546.6 degrees.

Hope that helps!

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Consider the titration of 100.0 mL of 0.100 M acetic acid with 0.100 M NaOH. CH3CO2H(aq) + OH-(aq) → CH3CO2-(aq) + H2O(ℓ) Ka for

allsm [11]

Answer:

a) pH = 5.70

b) pH = 8.22

c)pH = 11.68

Explanation:

a)NaOH (aq)+CH3COOH(aq) ------> CH3COONa(aq) + H2O(aq)

1mol of NaOH react with 1mol of CH3COOH

No of mole in 90ml of 0.1M NaOH

= (0.1mol/1000ml)×90ml

= 0.009

No of mol in 100ml of CH3COOH

= (0.1/1000)×100

= 0.01

No of mol of CH3COOH after addition

= 0.01-0.009

= 0.001

Total volume = 100ml + 90ml

= 190ml

Final molarity of CH3COOH =( 0.0025/190) ×1000

= 0.00526M

Concentration of CH3COONa formed =( 0.0075/190) ×1000

= 0.0474M

Ka of CH3COOH = 1.8 × 10^-5

pka = -log(ka)

pKa = 4.75

Applying Henderson equation

pH = pKa + log ( [A-]/[HA])

pH = 4.75 + log ( 0.0474/0.00526)

= 5.70

At equivalencepoint point ,

No of moles of CH3COOH = 0

No of moles of CH3COO- = 0.01 mol

Total volume = 200ml

molarity of CH3COO- = 0.01/2

= 0.0050M

CH3COO- (aq) + H2O(l) <---------> CH3COOH(aq) + OH-(aq)

Kb = [ CH3COOH] [ OH- ] / [ CH3COO- ]

= 1.8 × 10^-5

[ CH3COOH ] = X

[ OH-] = X

[ CH3COO-] = 0.0050 - X

5.6 × 10^-10 = X^2/ (0.0050 - X)

we can assume , 0.0050 - X = 0.0050

5.6 × 10^-10 = X^2/0.0050

X = 1.67 × 10^6

[OH-] = 1.67 × 10^-6

pOH = 5.78

pH = 14 - pOH

pH = 14 -5.78

pH = 8.22

c) No of mol of OH from excess 10ml of NaOH = (0.1mol /1000ml)×10ml = 0.001mol

No of mol of OH- from hydrolysis of CH3COO- = (1.67×10^-6/1000)×200= 3.34×10^-7mol

Second one is negligible

So, no of mol OH- = 0.001mol

Total volume = 100ml + 110ml

= 210ml

[OH-] =( 0.001/210)×1000

= 0.0048M

pOH = -log[OH-]

= - log (0.0048)

= 2.32

pH = 14 - 2.32

pH = 11.68

4 0

3 years ago

What happens if you increase a current in a electromagnetic

Nezavi [6.7K]

Answer:

If you increase current through an electromagnet, the magnetic field around it will increase

Hope this helps!

7 0

3 years ago

Does solubility have any unit

Contact [7]

Answer:

Solubility may be stated in various units of concentration such as molarity, molality, mole fraction, mole ratio, mass (solute) per volume (solvent) and other units. The extent of solubility ranges widely, from infinitely soluble (without limit) ( miscible) such as ethanol in water, to poorly soluble, such as silver chloride in water.

Explanation:

7 0

3 years ago

Read 2 more answers

What is the first step in creating a unit conversion factor?

NNADVOKAT [17]

Determining the units.

5 0

3 years ago

When 10.0 grams of sulfur reacts with fluorine gas at a pressure of 2.69 atmosphere in a 5.00 L container at 0.00 degrees Celsiu

Gwar [14]

Answer:

74.1%

Explanation:

Based on the reaction:

S₈ + 16F₂ → 8SF₄

1 mole of sulfur reacts with 16 moles of F₂ to produce 8 moles of SF₄

To solve this question we must find the moles of each reactant in order to find the moles of SF₄. Thus, we can find the theoretical mass produced. Percent yield is:

Percent yield = Actual yield (25.0g) / Theoretical yield * 100

Moles S₈: 256.52g/mol

10.0g * (1mol / 256.52g) = 0.0390 moles

Moles F₂:

PV = nRT

PV/RT = n

Where P is pressure in atm, V is volume in liters, R is gas constant and T is absolute temperature (0°C = 273.15K)

2.69atm*5.00L / 0.082atmL/molK*273.15K = n

0.600 moles = n

For a complete reaction of 0.600 moles F₂ are required:

0.600mol F₂ * (1mol S₈ / 8 mol F₂) = 0.075 moles S₈

As there are just 0.0390 moles, S₈ is limiting reactant.

The theoretical moles and mass of SF₄ -Molar mass: 108.07g/mol- is:

0.0390 moles S₈ * (8mol SF₄ / 1mol S₈) = 0.312 moles SF₄ * (108.07g) =

33.7g

Percent yield = 25.0g / 33.7g * 100

= 74.1%

6 0

3 years ago