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3 years ago

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a clear liquid is poured into a beaker containing another clear liquid. A cloudy yellow substance forms, as if out of nowhere, b

ut the rest of the liquid remains clear. what happened?

1 answer:

konstantin123 [22]3 years ago

8 0

A precipitate forms from a double displacement reaction or metathesis.

Explanation:

If given two clear solutions and upon reacting a cloudy/insoluble substance forms with the rest of the liquid being clear, a double displacement reaction has been carried out. The insoluble cloudy substance is called precipitate.

the driving force for the bulk of double displacement reactions is the formation of precipitates.
from careful observations, a solubility chart has been developed. If the compounds reacting are known, using the chart, a chemist can predict whether a precipitate will form or not.

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For each of the following compounds, state whether it is ionic or covalent, and if it is ionic, write the symbols for the ions i

kupik [55]

Answer :

Covalent compound : It is defined as the compound which is formed by the sharing of electrons between the atoms forming a compound.

The covalent compound are usually formed when two non-metals react.

Ionic compound : It is defined as the compound which is formed when electron gets transferred from one atom to another atom.

All the polyatomic ions always form an ionic compound.

Polyatomic ions : It is a charged species that composed of two or more atoms and these charged species are bonded by the covalent bond.

For the given options:

Option A: $KClO_4$

This compound is formed by the combination of potassium, $K^+$ which is a metal and $ClO_4^-$ ion which is a polyatomic ion. Thus, it will form an ionic compound.

Option B: $Mg(C_2H_3O_2)_2$

This compound is formed by the combination of magnesium, $Mg^{2+}$ which is a metal and $C_2H_3O_2^{-}$ ion which is a polyatomic ion. Thus, it will form an ionic compound.

Option C: $H_2S$

Hydrogen and sulfur, both are non-metals and they will form a covalent compound.

Option D: $Ag_2S$

This compound is formed by the combination of silver, $Ag^{+}$ which is a metal and sulfur, $S^{2-}$ which is a non-metal. Thus, it will form an ionic compound.

Option E: $N_2Cl_4$

Nitrogen and chlorine, both are non-metals and they will form a covalent compound.

Option F: $Co(NO_3)_2$

This compound is formed by the combination of cobalt, $Co^{2+}$ which is a metal and $NO_3^{-}$ ion which is a polyatomic ion. Thus, it will form an ionic compound.

6 0

2 years ago

From the graph of Density vs. Concentration, created in Graph 1, what was the relationship between the concentration of the suga

USPshnik [31]

The graph is not given in the question, so, the required graph is attached below:

Answer:

According to the graph, the relationship between the density of the sugar solution and the concentration of the sugar solution is directly proportional to each other as they both are increasing exponentially.

The graph shows that, the density of sugar solution will increase with the increase in concentration of sugar in the solution.

8 0

2 years ago

A sample contains 2.2 g of the radioisotope niobium-91 and 15.4 g of its daughter isotope, zirconium-91. how many half-lives hav

dybincka [34]

Answer: 3

Explanation: This is a radioactive decay and all the radioactive process follows first order kinetics.

Equation for the reaction of decay of $_{19}^{40}\textrm{K}$ radioisotope follows:

$Moles of zirconium=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{15.4}{91}=0.17moles$

$Moles of niobium=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{2.2}{91}=0.024moles$

$_{41}^{91}\textrm{Nb}\rightarrow _{40}^{91}\textrm{Zr}+_{+1}^0e$

By the stoichiometry of above reaction,

1 mole of $_{40}^{91}\textrm{Zr}$ is produced by 1 mole $_{41}^{91}\textrm{Nb}$

So, 0.17 moles of $_{40}^{91}\textrm{Zr}$ will be produced by = $\frac{1}{1}\times 0.17=0.17\text{ moles of }_{40}^{91}\textrm{Nb}$

Amount of $_{82}^{212}\textrm{K}$

decomposed will be = 0.17 moles

Initial amount of $_{40}^{91}\textrm{Nb}$ will be = Amount decomposed + Amount left = (0.17 + 0.024)moles =0.194 moles

$a=\frac{a_o}{2^n}$

where,

a = amount of reactant left after n-half lives = 0.024

$a_o$ = Initial amount of the reactant = 0.194

n = number of half lives= ?

Putting values in above equation, we get:

$0.024=\frac{0.194}{2^n}$

$n=3$

Therefore, 3 half lives have passed.

3 0

2 years ago

Read 2 more answers

The Kp for the reaction below is 1.49 × 108 at 100.0°C:

zhenek [66]

In an equilibrium mixture of the three gases, PCO = PCl2 = 2.22 × 10-4 atm. The partial pressure of the product, phosgene (COCl2), is kp=(COCl2)/(CO)(Cl2) which is $1.48*10^=x/(2.24*10-4)^2$ . So, the correct answer is <span>A) 7.34.</span>

8 0

2 years ago

Read 2 more answers

Be sure to answer all parts. Consider the following balanced redox reaction (do not include state of matter in your answers): 2C

timurjin [86]

Answer:

The specie which is oxidized is:- $CrO_2^-$

The specie which is reduced is:- $ClO^-$

Explanation:

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

$X\rightarrow X^{n+}+ne^-$

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^{n+}+ne^-\rightarrow X$

For the given chemical reaction:

$2CrO_2^- + 6ClO^- + 2H_2O\rightarrow 2CrO_4^{2-} + 3Cl_2 + 4OH^-$

The half cell reactions for the above reaction follows:

Oxidation half reaction: $CrO_2^- + 2H_2O + 4OH^-\rightarrow CrO_4^{2-} + 4H_2O + 3e^-$

Reduction half reaction: $2ClO^- + 4H_2O + 2e^-\rightarrow Cl_2 + 2H_2O + 4OH^-$

Thus, the specie which is oxidized is:- $CrO_2^-$

The specie which is reduced is:- $ClO^-$

8 0

3 years ago