All categories

2 years ago

Which method is used to separate the following mixture (a) water + kerosene

Chemistry

1 answer:

Ad libitum [116K]2 years ago

8 0

Answer:

The separation of kerosene, oil, and water are immiscible liquids, so they can be separated funnel.

You might be interested in

A hot air balloon is filled to 1250 m3 at 27 C. At what temperature will the balloon be filled to 1600 m3 if the pressure remain

SSSSS [86.1K]

Answer:

384.2 K

Explanation:

First we convert 27 °C to K:

27 °C + 273.16 = 300.16 K

With the absolute temperature we can use <em>Charles' law </em>to solve this problem. This law states that at constant pressure:

T₁V₂=T₂V₁

Where in this case:

T₁ = 300.16 K

V₂ = 1600 m³

T₂ = ?

V₁ = 1250 m³

We input the data:

300.16 K * 1600 m³ = T₂ * 1250 m³

And solve for T₂:

T₂ = 384.2 K

7 0

2 years ago

What is the name of the process plants use to make the energy (food) they need?

scZoUnD [109]

The answer you looking for is D

8 0

2 years ago

Read 2 more answers

The specific heat of a given substance

Lelechka [254]

The specific heat is the amount of heat per unit mass required to raise the temperature to 1 degree Celsius. (This is from google)

7 0

2 years ago

If a car is traveling 100 km/h and comes to a stop in 3 minutes,what is acceleration of a passenger who is using vehicle restrai

Rama09 [41]

We are given –

Final velocity of car is, v= 0
Initial velocity of car is, u= 100 km/hr
Time taken, t is = 3 minutes or 180 sec

Here–

$\qquad$ $\pink{\bf \longrightarrow Initial\: velocity = 100 \:km/hr}$

$\qquad$ $\sf \longrightarrow Initial\: velocity = \dfrac{ 100 \times 1000}{3600} \:m/s$

$\qquad$ $\pink{ \bf \longrightarrow Initial\: velocity = 27.78\: m/s}$

Now –

$\qquad$ ____________________________

$\qquad$ $\purple{\bf \longrightarrow Acceleration = \dfrac{Final\: Velocity -Initial \:Velocity }{Time}}$

$\qquad$ $\purple{\bf \longrightarrow Acceleration = \dfrac{v -u}{t}}$

$\qquad$ $\sf \longrightarrow Acceleration = \dfrac{(0- 27.78)}{1800}$

$\qquad$ $\sf \longrightarrow Acceleration =\cancel{ \dfrac{- 27.78}{1800}}$

$\qquad$ $\purple{\bf \longrightarrow Acceleration = -0.15 \: m/s^2}$

$\qquad$ _______________________________

6 0

2 years ago

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

3 years ago

Which method is used to separate the following mixture (a) water + kerosene​

Which method is used to separate the following mixture (a) water + kerosene