Give two examples of pure substances and two examples of mixtures

Physics

2 answers:

nikdorinn [45]3 years ago

7 0

Answer:

1 kerosene is the pure substance, salt and water is the mixture substance

lions [1.4K]3 years ago

5 0

Answer:

pure substansubdiamond nd sulphur. mixture:salt nd oil ,oil nd water

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Today's topic

babunello [35]

Answer:

While a body is said to be in motion if it changes its position with respect to immediate surroundings.

A body is said to be in uniform motion if it covers equal distances in equal interval of time.

A body is said to be in non-uniform motion if it covers unequal distances in equal interval of time or vice-versa

4 0

2 years ago

What product is obtained from the aldol condensation of cyclohexanone?

Evgesh-ka [11]

Answer:

First product is FCH-OH chemically known as 2-[2-furyl(hydroxyl)methyl]-Second product is FCH i.e (2E)-2-[2-furyl-methylene]-cyclohexanone

Explanation:

Please see the attached image for complete chemical reaction of aldol condensation of cyclohexanone

Aldol Condensation is a form of electrophilic substitution reaction in which the alpha carbon in enols or enolate anions is substituted by an electrophile to form carbon-carbon bond. Cyclohexanone also known as the first ketone consists of two alpha-carbons and four potential substitutions i.e alpha-hydrogens but none of the hydrogen on the ring is substituted. Ketones such as cyclohexanone are much more acidic than their parent hydrocarbon.

First product is FCH-OH chemically known as 2-[2-furyl(hydroxyl)methyl]-cyclohexanone that further undergoes dehydration resulting into FCH i.e (2E)-2-[2-furyl-methylene]-cyclohexanone

Based on the explanations above, the compound formed is shown in the image.

6 0

3 years ago

A capacitor is charged until it holds 5.0 j of energy. it is then connected across a 10-kω resistor. in 13.6 ms , the resistor d

prohojiy [21]

Answer:

The capacite is C=5.32 uF using the equations of voltage and energy in capacitance

Explanation:

The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J

Using:

$V_{t}= V_{o}e^{\frac{-t}{R*C} }$

Voltage in this case is the energy dissipated so

$E_{t}= E_{o}e^{\frac{-t}{R*C} }$

$\frac{\sqrt{E_t} }{\sqrt{E_o} } = e^{\frac{-t}{R*C} }$

$\frac{\sqrt{3 J} }{\sqrt{5J} } = e^{\frac{-13.6ms}{10kw*C} }$

Using the equation to find capacitance

$ln 0.775= e^{\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\ln(0.775)= ln * e^{\frac{-13.6 x10^{3s} }{10x10^{3}*C }} \\\\ln(0.775)= {\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\C= \frac{-13.6 x10^{-3} }{10x10x^{3}*ln(0.775) }$