Answer:
The polar compounds are soluble in water while non polar are insoluble in water.
Explanation:
Solvent is the that part of solution which is present in large proportion and have ability to dissolve the solute. In simplest form it is something in which other substance get dissolve. The most widely used solvent is water, other examples are toluene, acetone, ethanol, chloroform etc.
Water is called universal solvent because of high polarity all polar substance are dissolve in it. Hydrogen is less electronegative while oxygen is more electronegative and because of difference in electronegativity hydrogen carry the partial positive charge while oxygen carry partial negative charge.
Water create electrostatic interaction with other polar molecules. The negative end of water attract the positive end of polar molecules and positive end of water attract negative end of polar substance and in this way polar substance get dissolve in it.
Example:
when we stir the sodium chloride into water the cation Na⁺ ions are surrounded by the negative end of water i.e oxygen and anion Cl⁻ is surrounded by the positive end of water i.e hydrogen and in this way all salt is get dissolved.
The chemicals that can dissolve in a certain solvent to create a homogenous mixture known as a solution are said to be soluble chemicals. The compounds that are soluble are: , , and .
As per this,
Insoluble:
Thus, these are the classification of the compounds as per their solubility.
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Your question seems incomplete, the prpbable complete question is:
Predict whether the following compounds are soluble or insoluble in water. Soluble Insoluble PbCl2, BaSO4, KNO3, AgNO3, and CuBr2.
b. Is colorless in the presence of water
c. Does not exist as a hydrate
d. Forms different hydrates which have different colors
Answer:Forms different hydrates which have different colors
Explanation:
CoCl2 in its anhydrous form is blue in colour. This anhydrous compound could absorb moisture, first forming the purple dihydrate and absorbs more water molecules to form the hexahydrate. Hence various hydrates of cobalt II chloride have different colours as stated above. Equations of reaction for the formation of the two hydrates are attached.
Answer: Option (b) is the correct answer.
Explanation:
Reduction is defined as the process in which there occurs gain of hydrogen. Whereas oxidation is defined as the process in which there occurs loss of hydrogen.
As the given reaction is as follows.
Since, hydrogen is being added in this chemical reaction. It means that reduction is taking place and carbon atom is reduced.
Thus, we can conclude that in the given reaction carbon atoms are reduced.
Answer:
Poop Butt.
Explanation: Poop Butt.
Answer:
15766
Explanation:
The average molecular weight of the polypropylene = 663419 g/mol.
The mass of the repeating unit , monomer which is propylene = 42.08 g/mol
The degree of polymerization is:
Degree of polymerization = 15766
Answer:
I belive the answer is A but the image quality is hard to tell.
Explanation:
As pressure increases the higher the melting point of rock becomes making it harder to melt, thats why molten rocks brought to the surface melt because of the change in pressure.
Answer:
Ea =22542.6
Explanation:
The rate constant k is affected by the temperature and this dependence may be represented by the Arrhenius equation:
where the pre-exponential factor A is assumed to be independent of temperature, R is the gas constant, and T the temperature in K. Taking the natural logarithm of this equation gives:
ln k = ln A - Ea/(RT)
or
ln k = -Ea/(RT) + constant
or
ln k = -(Ea/R)(1/T) + constant
These equations indicate that the plot of ln k vs. 1/T is a straight line, with a slope of -Ea/R. These equations provide the basis for the experimental determination of Ea.
now applying the above equation in the problem
we can write that
solve for Ea:
Ea = R[Ln(k2/k1)] / [(1/T1) - (1/T2)]
but k_2 = 2 k_1, hence:
Ea = (8.314 J/moleK)[ln(2)] / [(1/273+45) - (1/273+73)]
Ea =22542.6
Answer:
The activation energy for this reaction is 22.6 kJ/ mol
Explanation:
Step 1: Data given
Rate constant doubles when Temperature goes from 45.0 °C to 73.0 °C
R = 8.314 J/K*mol
Step 2: Calculate the activation energy
Log (k2/k1) = Ea / 2.303R *((1/T1) - (1/T2))
⇒ with k1 = initial rate constant
⇒ with k2 = rate constant after doubled = 2k1
⇒ T1 = initial temperature = 45.0 °C = 318 Kelvin
⇒ T2 = Final temperature = 73.0 °C = 346 Kelvin
log (2) = Ea / (2.303*8.314) *((1/318) - (1/346))
log(2) = Ea / (2.303*8.314) * 0.00025448
Ea = 22649 J/mol = 22.6 kJ/mol
The activation energy for this reaction is 22.6 kJ/ mol