Predict whether the compounds are soluble or insoluble in water

Answers

Answer 1
Answer:

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.

Answer 2
Answer:

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: KNO_3, AgNO_3, and CuBr_2.

As per this,

  • Nitrate (NO^{3-) salts are often soluble in water. KNO_3 is potassium nitrate.
  • The majority of nitrate (NO^{3-) salts, including silvernitrate, are soluble in water, including AgNO_3.
  • The majority of bromide (Br^-) salts, including copper(II) bromide, are water soluble.

Insoluble:

  • Lead(II) chloride, or PbCl_2, is an exception and is regarded as being insoluble in water among the chloride (Cl^-) salts.
  • Barium sulphate, also known as BaSO4, is an exception to the rule of most sulphate (SO^{4-) compounds being soluble in water.

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.


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Cobalt chloride Select one: a. Forms a single hydrate which may be pink or blue
b. Is colorless in the presence of water
c. Does not exist as a hydrate
d. Forms different hydrates which have different colors

Answers

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.

In the reaction C zH4 + H2 - e) +4 CHo the carbon atoms are a) oxidized b) reduced c) cannot be determined

Answers

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.

      C_(2)H_(4) + H_(2) \rightarrow C_(2)H_(6)

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.

of all the hydrogen nuclei in the ocean, 0.0156 how much deuterium could be obtained from 1.0 gal of ordinary tap water

Answers

Answer:

Poop Butt.

Explanation: Poop Butt.

The number-average molecular weight of a polypropylene is 663419 g/mol. Compute the degree of polymerization (Key in your answer as the whole number only. Avoiding round-off error by using all the numbers during intermediate steps.

Answers

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:

DP_n=\frac {Total\ molecular\ weight\ of\ the\ polymer}{Molecular\ weight\ of\ the\ monomer}

DP_n=\frac {663419\ g/mol}{42.08\ g/mol}

Degree of polymerization = 15766

Please help me out asap. I’ll give brainliest

Answers

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.  

The rate constant doubles when the temperature is increased from 45.0 C to 73.0 C. What is the activation energy for this reaction? (R=8.314 J/Kmol)?

Answers

Answer:

Ea =22542.6

Explanation:

The rate constant k is affected by the temperature and this dependence may be represented by the Arrhenius equation:

k=Ae^{-(E_a)/(RT) }

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

ln(k_2)/(k_1) = (E_a)/(R)[(1)/(T_1)-(1)/(T_2)  ]

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