All categories

2 years ago

State whether the following liquids are miscible with hexane (C6H14) by entering "yes" or "no".

1 answer:

3 0

Benzene, C6H6 is miscible with hexane (C6H14) but iron (IIl) nitrate, and Sulfuric acid, H2SO4Fe(NO3)3 are not miscible with hexane (C6H14)

Since organic or non polar dissolves in nom polar solvent.here both benzene and hexane (C6H14) are non polar solvents.But iron (IIl) nitrate, Fe(NO3)3 and Sulfuric acid, H2SO4 are inorganic and polar, therefore cannot dissolve with non polar solvent like hexane.

Miscible is a flowery heard that oil and water aren't very miscible substances, while seltzer and orange juice are miscible and scrumptious.Miscible beverages are also described as liquids that can blend to form a homogeneous solution. Miscible drinks generally blend with out limit, meaning they may be soluble at all quantities.

Miscible approach the materials blend completely. If substances are miscible, they're also absolutely soluble in each other no matter the order of creation. for example, tetrahydrofuran and water are miscible.

Learn more about miscible liquids here:-brainly.com/question/10423831

#SPJ1

You might be interested in

You start with 1 L of CO2 at standard temperature and pressure in a closed container. If you raise the temperature of the gas, t

pychu [463]

Answer:

Increase

Explanation:

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

If the initial temperature and pressure is standard,

Pressure = 1 atm

Temperature = 273.15 K

then we increase the temperature to 400.0 K, The pressure will be,

1 atm / 273.15 K = P₂/400.0K

P₂ = 1 atm × 400.0 K / 273.15 K

P₂ = 400.0 atm. K /273.15 K

P₂ = 1.46 atm

Pressure is also increase from 1 atm to 1.46 atm.

8 0

3 years ago

A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal

SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻× $\frac{1molI_{3}^-}{2molS_{2}O_{3}^-}$ = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻× $\frac{2molCe^{4+}}{1molI_{3}^-}$ = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)× $\frac{140,116g}{1mol}$ = 0,0625 g of Ce(IV)

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = 1,812 wt%

I hope it helps!

5 0

3 years ago

I need the answers of these

Anni [7]

Answer:

Hope this helps :D

Explanation:

Metal: Aluminum and Copper

Non-Metal: Hydrogen and Flourine

Acid: Sulfuric Acid and Phosphoric Acid

Alkali Metals: Hydrogen and Lithium

Compounds: Water and Carbon Dioxide

Elements: Carbon and Oxygen

4 0

2 years ago

Which of the following chemical formulas represents a common acid?

wariber [46]

HCI is one of the most common acids out of the following

5 0

3 years ago

When a 0.245-g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises 1.643 ∘C . When a 0.260-g sample

sveticcg [70]

Answer:

The heat of combustion per moles of caffeine is 4220 kJ/mol

Explanation:

Step 1: Data given

⇒ When benzoic acid sample of 0.245 grams is burned the temperature rise is 1.643 °C

⇒ When 0.260 gram of caffeine is burned, the temperature rise is 1.436 °C

⇒ Heat of combustion of benzoic acid = 26.38 kJ/g

Step 2: Calculate the heat released: for combustion of benzoic acid

0.245 g benzoic acid * 26.38 kJ/g = 6.4631 kJ

Step 3: Calculate the heat capacity of the calorimeter:

c = Q/ΔT

Q = 6.4631 kJ / 1.643°C = 3.934 kJ/ °C

Step 4: Calculate moles of a 0.260 g sample of caffeine:

Moles caffeine = Mass caffeine / Molar mass caffeine

0.260 grams/ 194.19 g/mol = 0.0013389 moles

Step 5: Calculate heat released: for combustion of caffeine

Q = c * ΔT

Q = 3.934 kJ/°C * 1.436 °C = 5.65 kJ

Step 6: Calculate the heat of combustion per mole of caffeine

5.65 kJ / 0.0013389 moles = 4219.9 kJ/mol ≈ 4220 kJ/mol

The heat of combustion per moles of caffeine is 4220 kJ/mol

4 0

4 years ago