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3 years ago

All organic compound soluble in WATEe and conduct electrically

Chemistry

1 answer:

Delvig [45]3 years ago

4 0

Explanation:

Not all organic substances are soluble in water and will not conduct electricity. Only polar organic compounds are soluble in water and will conduct electricity.

The general rule of solubility is that like dissolves likes.
Polar compounds will only dissolve other polar compounds too. Water is a polar solvent.
Non-polar compounds will only dissolve in non-polar compounds too.
Since water is a polar organic compound, only polar organic compounds will dissolve in it.
Not all organic compounds are polar.
When a compound dissolves in water, it produces ions that makes it able to conduct electricity.
Fats and oil are insoluble in water

learn more:

Inorganic compounds brainly.com/question/5047702

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What happens to the motion of molecules when temperature decreases

Juliette [100K]

Answer:

As the temperature of a solid, liquid or gas increases, the particles move more rapidly.

4 0

3 years ago

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Suppose that 0.48 g of water at 25∘C∘C condenses on the surface of a 55-gg block of aluminum that is initially at 25∘C∘C. If the

GarryVolchara [31]

Answer : The final temperature of the metal block is, $25^oC$

Explanation :

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminum = 55 g

$m_2$ = mass of water = 0.48 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminum = $25^oC$

$T_2$ = temperature of water = $25^oC$

$c_1$ = specific heat of aluminum = $0.900J/g^oC$

$c_2$ = specific heat of water= $4.184J/g^oC$

Now put all the given values in equation (1), we get

$55g\times 0.900J/g^oC\times (T_{final}-25)^oC=-[0.48g\times 4.184J/g^oC\times (T_{final}-25)^oC]$

$T_{final}=25^oC$

Thus, the final temperature of the metal block is, $25^oC$

6 0

3 years ago

When 1.5 grams of magnesium burns, how many grams of magnesium oxide will be formed?

SashulF [63]

The balanced chemical reaction for this would be written as:

2Mg + O2 = 2MgO

We use this reaction and the amount of the reactant given to calculate for the amount of magnesium oxide that is produced. We do as follows:

1.5 g Mg (1 mol / 24.31 g) ( 2 mol MgO / 2 mol Mg ) (40.30 g /1 mol ) = 2.49 g MgO produced

6 0

2 years ago

How many moles of chromium III nitrate are produced When chromium reacts with 0.85 moles of lead for nitrate to produce chromium

Pepsi [2]

0.85 moles formula units of lead nitrate will produce 0.57 moles formula units of chromium (III) nitrate.

<h3>Explanation</h3>

Typically, the oxidation state of Pb in lead nitrate tend to be +2. In other words, Pb in lead nitrate tends to exist as $\text{Pb}^{2+}$ ions. The formula for a nitrate ion is ${\text{NO}_3}^{-}$ . The charge on each of the nitrate ion is -1. The charge on the two ions should balance. As a result, each $\text{Pb}^{2+}$ ion in lead nitrate would pair up with two ${\text{NO}_3}^{-}$ ions. The formula for lead nitrate will be $\text{Pb}({\text{NO}_3})_2$ . Each formula unit of lead nitrate will contain one $\text{Pb}^{2+}$ ion and two ${\text{NO}_3}^{-}$ ions.

The "III" in the name "chromium (III) nitrate" is a Roman Numeral. It indicates that the oxidation state of Cr in chromium (III) nitrate is +3. The Cr in that compound will exist as $\text{Cr}^{3+}$ . Similarly, each $\text{Cr}^{3+}$ will pair up with three ${\text{NO}_3}^{-}$ ions. The formula for chromium (III) nitrate will be $\text{Cr}(\text{NO}_3})_3$ . Each formula unit of chromium (III) nitrate will contain one ${\text{NO}_3}^{-}$ ion and three ${\text{NO}_3}^{-}$ ions.

0.85 moles formula units of lead nitrate will contain 0.85 × 2 = 1.7 moles of ${\text{NO}_3}^{-}$ ions. Those nitrate ions will end up in 1.7 / 3 = 0.57 moles formula units of chromium (III) nitrate. As a result, the reaction will produce 0.57 moles formula units of chromium (III) nitrate.

7 0

3 years ago

Calculate number of moles of 4.4g of CO2 and 5.6L of NH3 . Pls help . Its urgent

kupik [55]

Moles of CO₂ = mass / molecular weight
Moles of CO₂ = 4.4 / (12 + 16 x 2)
Moles of CO₂ = 0.1 mol

Each mole of gas occupies 22.4 L at STP. Therefore,
Moles of NH₃ = 5.6 / 22.4
Moles of NH₃ = 0.25 mol

4 0

3 years ago

All organic compound soluble in WATEe and conduct electrically ​

All organic compound soluble in WATEe and conduct electrically