Candle wax melts low temperature, it is not conductive to electricity, it is insoluble in water and partially soluble in solvents nonpolar, like gasoline. Than type of links are present in the candle wax?

A. Electrostatics.
B. Apolar.
C. lónicos.
D. Hydrogen bridges.

Answers

Answer 1

electrostatic and ionic are definitely not the answer because they have high melting point

hydrogen bonds are too weak and not permanent.

so the answer is apolar as it is soluble in polar solvents (water)

Answer 2

Answer:

B. Nonpolar

Explanation:

The low melting point tells you the compound is not ionic, metallic, or a network solid.

It is almost certainly a molecular solid.

It does not conduct electricity, so it is not metallic (which we have already ruled out).

It is insoluble in polar solvents (water) and soluble in nonpolar solvents (gasoline).

Since like dissolves like, the molecule is nonpolar.

The type of links must be nonpolar.


Related Questions

Select the true statement concerning voltaic and electrolytic cells. Select one: a. Voltaic cells involve oxidation-reduction reactions while electrolytic cells involve decomposition reactions. b. Voltaic cells require applied electrical current while electrolytic cells do not. . c. all electrochemical cells, voltaic and electrolytic, must have spontaneous reactions. d. Electrical current drives nonspontaneous reactions in electrolytic cells.

Answers

Answer:

Electrical current drives nonspontaneous reactions in electrolytic cells.

Explanation:

Electrochemical cells are cells that produce electrical energy from chemical energy.

There are two types of electrochemical cells; voltaic cells and electrolytic cells.

A voltaic cell is an electrochemical cell in which electrical energy is produced from spontaneous chemical process while an electrolytic cell is an electrochemical cell where electrical energy is produced from nonspontaneous chemical processes. Current is needed to drive these nonspontaneous chemical processes in an electrolytic cell.

Answer:

electrolytic cells generate electricity through a non-spontaneous reaction while voltaic cells absorb electricity to drive a spontaneous reaction.

Explanation:

Answer via Educere/ Founder's Education

Determine the oxidation state for each of the elements below. The oxidation state of ... silver ... in ... silver oxide Ag2O ... is ... ___ . The oxidation state of sulfur in sulfur dioxide SO2 is ___ . The oxidation state of iron in iron(

Answers

Answer:

The oxidation state of silver in [tex]\rm Ag_2O[/tex] is [tex]+1[/tex].

The oxidation state of sulfur in [tex]\rm SO_2[/tex] is [tex]+4[/tex].

Explanation:

The oxidation states of atoms in a compound should add up to zero.

Ag₂O

There are two silver [tex]\rm Ag[/tex] atoms and one oxygen [tex]\rm O[/tex] atom in one formula unit of [tex]\rm Ag_2O[/tex]. Therefore:

[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].

The oxidation state of oxygen in most compounds (with the exception of peroxides and fluorides) is [tex]-2[/tex]. Silver oxide [tex]\rm Ag_2O[/tex] isn't an exception. Therefore:

[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times (-2) = 0\end{aligned}[/tex].

Solve this equation for the (average) oxidation state of [tex]\rm Ag[/tex]:

[tex]\text{Oxidation state of $\rm Ag$} = 1[/tex].

SO₂

Similarly, because there are one sulfur [tex]\rm S[/tex] atom and two oxygen [tex]\rm O[/tex] atoms in each [tex]\rm SO_2[/tex] molecules:

[tex]\begin{aligned}&\rm 1\times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].

The oxidation state of [tex]\rm O[/tex] in [tex]\rm SO_2[/tex] is also [tex]-2[/tex], not an exception, either.

Therefore:

[tex]\begin{aligned}&\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times (-2) = 0\end{aligned}[/tex].

Solve this equation for the oxidation state of [tex]\rm S[/tex] here:

[tex]\text{Oxidation state of $\rm S$} = 4[/tex].

When titrating a strong acid with a strong base, after the equivalence point is reached, the pH will be determined exclusively by: Select the correct answer below:
A) hydronium concentration
B) hydroxide concentration
C) conjugate base concentration
D) conjugate acid concentration

Answers

Answer:

B) hydroxide concentration

Explanation:

Hello,

In this case, since we are talking about strong both base and acid, since the base is the titrant and the acid the analyte, once the equivalence point has been reached, some additional base could be added before the experimenter realizes about it, therefore, since the titrant is a strong base, it completely dissociates in hydroxide ions and metallic ions which allows us to compute the pOH of the solution by known the hydroxide ions concentration.

After that, due to the fact that the pH is related with the pOH as shown below:

pH=14-pOH

We can directly compute the pH.

Best regards.

The insoluble salts below are put into 0.10 M hydrochloric acid solution. Do you expect their solubility to be more, less, or about the same as in a pure water solution?
1. Zinc sulfide
2. Silver chloride
3. Lead iodide
4. Silver hydroxide

Answers

Answer:

1. Zinc sulfide : about the same solubility, no common ion is found.

2. Silver chloride : less solubility due to the presence of chloride ions provided by the 0.10 M hydrochloric acid.

3. Lead iodide  : about the same solubility, no common ion is found.

4. Silver hydroxide : about the same solubility, no common ion is found.

Explanation:

Hello,

In this case, we first must remember that adding a common ion (which is related with the dissolving solid) decreases the solubility of the insoluble solid due to the fact Le Chatelier's principle states the reaction will shift leftwards (reactants) to reestablish equilibrium, therefore, we have:

1. Zinc sulfide : about the same solubility, no common ion is found.

2. Silver chloride : less solubility due to the presence of chloride ions provided by the 0.10 M hydrochloric acid.

3. Lead iodide  : about the same solubility, no common ion is found.

4. Silver hydroxide : about the same solubility, no common ion is found.

Best regards.

A chemist prepares a solution of sodium chloride by measuring out 25.4 grams of sodium chloride into a 100. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's sodium chloride solution. Be sure your answer has the correct number of significant digits.

Answers

Answer:

The concentration in mol/L =  4.342 mol/L

Explanation:

Given that :

mass of sodium chloride = 25.4 grams

Volume of the volumetric flask = 100 mL

We all know that the molar mass of sodium chloride NaCl = 58.5 g/mol

and number of moles = mass/molar mass

The number of moles of sodium chloride = 25.4 g/58.5 g/mol

The number of moles of sodium chloride = 0.434188 mol

The concentration in mol/L = number of mol/ volume of the solution

The concentration in mol/L = 0.434188 mol/ 100 × 10⁻³ L

The concentration in mol/L =  4.34188 mol/L

The concentration in mol/L =  4.342 mol/L

Find the pH of these buffer solutions using the information provided: 1L solution containing 80g of lactic acid (MW

Answers

Answer:

pH of the solution is 2.0

Explanation:

The lactic acid is a weak acid that is in equilibrium with water as follows:

Lactic acid + H2O ⇄ Lactate + H₃O⁺

And Ka for lactic acid: 1.38x10⁻⁴

Ka = 1.38x10⁻⁴ = [Lactate] [H₃O⁺] / [Lactic acid]

Initial concentration of lactic acid is (MW: 112.06g/mol):

80g * (1mol / 112.06g) / 1L = 0.714M

The equilibrium concentration of the species in the equilibrium are:

[Lactate] = X

[H₃O⁺] = X

[Lactic acid] = 0.714-X

Replacing in Ka expression:

1.38x10⁻⁴ = [X] [X] / [0.714-X]

9.8532x10⁻⁵ -  1.38x10⁻⁴X = X²

9.8532x10⁻⁵ -  1.38x10⁻⁴X - X² = 0

Solving for X:

X = -1.0x10⁻². False solution, there is no negative concentrations

X = 9.86x10⁻³M. Right solution.

As [H₃O⁺] = X

[H₃O⁺] = 9.86x10⁻³M

and pH = -log [H₃O⁺] = -log 9.86x10⁻³M

pH = 2.0

pH of the solution is 2.0

g Does a reaction occur when aqueous solutions of barium hydroxide and aluminum sulfate are combined

Answers

Answer:

3BaO + Al₂(SO₄)₃  →  Al₂O₃+ 3BaSO₄

Explanation:

Yes! A reactiin occurs between barium hydroxide and auminium sulphate.

barium sulfate (BaSO4) and aluminum hydroxide (Al(OH)3) are the products obtained in this reaction.

The reaction is given by the equation below;

3BaO + Al₂(SO₄)₃  →  Al₂O₃+ 3BaSO₄

A mixture of 50ml of 0.1M HCOOH and 50ml of 0.05M NaOH is equivalent to

Answers

Answer:

d) a solution that is 0.025M in HCOOH and 0.025M in HCOONa

Explanation:

The reaction of a weak acid (HOOH) with NaOH is as follows:

HCOOH + NaOH → HCOONa + H₂O

Based on the reaction, 1 mole of the acid reacts with 1 mole of the base (Ratio 1:1).

The initial moles of both species are:

HCOOH: 0.050L × (0.1mol / L) = 0.0050 moles of HCOOH

NaOH: 0.050L × (0.05 mol / L) = 0.0025 moles NaOH

After the reaction, all NaOH reacts with HCOOH producing HCOONa (Because moles of NaOH < moles HCOOH).

Final moles:

HCOOH: 0.0050 moles - 0.0025 moles (After reaction) = 0.0025 moles

HCOONa: Moles HCOONa = Initial Moles NaOH: 0.0025 moles

As volume of the mixture is 100mL (50 from the acid + 50 from NaOH), molarity of both HCOOH and HCOONa is:

0.0025 moles / 0.100L = 0.025M of both HCOOH and HCOONa

Thus, the initial mixture is equivalent to:

d) a solution that is 0.025M in HCOOH and 0.025M in HCOONa

1500 L has how many significants figures

Answers

Answer:

It has 2

Explanation:

The significant figures are 1 and 5!

Hope this helps:)

A chemist fills a reaction vessel with 0.978 g aluminum hydroxide AlOH3 solid, 0.607 M aluminum Al+3 aqueous solution, and 0.396 M hydroxide OH− aqueous solution at a temperature of 25.0°C.
Under these conditions, calculate the reaction free energy ΔG for the following chemical reaction:
Al(OH)3(s) = A1+ (aq) +30H (aq)
Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.
KJ

Answers

Answer: [tex]\Delta G^{0}[/tex] = 168.12 kJ

Explanation: Gibbs Free Energy, at any time, is defined as the enthalpy of the system minus product of temperature and entropy of the reaction, i.e.:

[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]

Enthalpy is defined as internal heat existent in the system. It is calculated as:

[tex]\Delta H^{0} = \Sigma H^{0}_{product} - \Sigma H^{0}_{reagent}[/tex]

Using Enthalpy Formation Table:

[tex]\Delta H^{0} = [3*(-299.9)+(-524.7)] - (-1277)[/tex]

[tex]\Delta H^{0} = 62,6 kJ[/tex]

Entropy is the degree of disorder in the system. It is found by:

[tex]\Delta S^{0} = \Sigma S^{0}_{products} - \Sigma S^{0}_{reagents}[/tex]

Calculating:

[tex]\Delta S^{0} = (-321.7) + 3(-10.8) - 0[/tex]

[tex]\Delta S^{0} = -354.1J[/tex]

And so, Gibbs Free energy will be:

[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]

[tex]\Delta G^{0} = 62600 - [298.(-354.1)][/tex]

[tex]\Delta G^{0} = 168121.8 J[/tex]

Rounding to the nearest kJ:

[tex]\Delta G^{0}[/tex] = 168.12 kJ

Ammonia, methane, and phosphorus trihydride are three different compounds with three different boiling points. Rank their boiling points in order from lowest to highest.
A. CH4< NH3 < PH3
B. NH3 < PH3< CH4
C. CH4 < PH3 < NH3
D. NH3 < CH4< PH3
E. PH3< NH3 < CH4

Answers

Answer:

B. NH3 < PH3< CH4

Explanation:

Hello,

In this case, taking into account that the boiling point of ammonia, methane and phosphorous trihydrate are -33.34 °C , -161.5 °C  and -87.7 °C , clearly, methane has the lowest boiling point (most negative) and ammonia the greatest boiling point (least negative), therefore, ranking is:

B. NH3 < PH3< CH4

Best regards.

Explain your reasoning. Match each explanation to the appropriate blanks in the sentences on the right.
1. the atomic radius decreases
2. the number of gas molecules decreases
3. molar mass and structure complexity decreases
4. structure complexity decreases
5. molar mass decreases
6. each phase (gas, liquid, solid) becomes more ordered
A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as______.
B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as_______.
C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as_______.

Answers

Answer:

A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases.

B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases.

C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases.

Explanation:

Hello.

In this case, we can understand a higher entropy when more disorder is present and a lower entropy when less disorder is present, thus:

A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases since iodine has the greatest molar mass (254 g/mol) and fluorine the least molar mass (38 g/mol).

B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases since hydrogen peroxide weights 34 g/mol as well as hydrogen sulfide but the peroxide has more bonds (more complex, higher entropy).

C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases since diamond has a well-ordered structure and amorphous carbon has a very disordered one.

Best regards.

Solid cesium bromide has the same kind of crystal structure as CsCl which is pictured below: If the edge length of the unit cell is 428.7 pm, what is the density of CsBr in g/cm3.

Answers

Answer:

[tex]\mathbf {density \ d =4.4845 \ g/cm^3}[/tex]

Explanation:

Let recall the crystal structure of CsBr obtains a BCC structure. In a BCC structure, there exist only two atom per cell.

The density d of CsBr in g/cm³ can be calculated by using the formula:

[tex]\mathtt{ density \ d = \dfrac{z \times molar\ mass \ (M)}{ edge \ length \ (a) \ \times avogadro's \ number \ (N)}}[/tex]

where;

z = 1 mole of CsBr

edge length = 428.7 pm = (4.287 × 10⁻⁸)³ cm

molar mass of CsBr = 212.81 g/mol

avogadro's number = 6.023 × 10²³

[tex]\mathtt{ density \ d = \dfrac{1 \times 212.81}{(4.287 \times 10^{-8})^3 \times 6.023 \times 10^{23}}}[/tex]

[tex]\mathtt{ density \ d = \dfrac{ 212.81}{47.4540533}}[/tex]

[tex]\mathbf {density \ d =4.4845 \ g/cm^3}[/tex]

A piece of solid Fe metal is put into an aqueous solution of Cu(NO3)2. Write the net ionic equation for any single-replacement redox reaction that may be predicted. Assume that the oxidation state of in the resulted solution is 2 . (Use the lowest possible coefficients for the reaction. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank. If no reaction occurs, leave all boxes blank and click on Submit.)

Answers

Answer:

Fe(s) + Cu^2+(aq) ---> Fe^2+(aq) + Cu(s)

Explanation:

The ionic equation shows the actual reaction that took place. It excludes the spectator ions. Spectator ions are ions that do not really participate in the reaction even though they are present in the system.

For the reaction between iron and copper II nitrate, the molecular reaction equation is;

Fe(s) + Cu(NO3)2(aq)----> Fe(NO3)2(aq) +Cu(s)

Ionically;

Fe(s) + Cu^2+(aq) ---> Fe^2+(aq) + Cu(s)

When the nuclide bismuth-210 undergoes alpha decay:
The name of the product nuclide is_____.
The symbol for the product nuclide is_____
Fill in the nuclide symbol for the missing particle in the following nuclear equation.
_____ rightarrow 4He+ 234Th
2 90
Write a balanced nuclear equation for the following:
The nuclide radium-226 undergoes alpha emission.

Answers

Explanation:

An atom undergoes alpha decay by losing a helium atom.

So when bismuth undergoes alpha decay, we have;

²¹⁰₈₃Bi --> ⁴₂He + X

Mass number;

210 = 4 + x

x = 206

Atomic number;

83 = 2 + x

x = 81

The element is Thallium. The symbol is Ti.

For the second part;

X --> ⁴₂He + ²³⁴₉₀Th

Mass number;

x = 4 + 234 = 238

Atomic Number;

x = 2 + 90 = 92

The balanced nuclear equation is;

²³⁸₉₂U --> ⁴₂He + ²³⁴₉₀Th

What type of bond would form between two atoms of phosphorus? A. Single covalent bond B. Single ionic bond C. Triple covalent bond D. Double covalent bond

Answers

Answer:

A double bond is formed when two pairs of electrons are shared between the two participating atoms. It is represented by two dashes (=). It is represented by two dashes (=). Double covalent bonds are much stronger than a single bond, but they are less stable

Explanation:

whts the ph of po4 9.78

Answers

Answer:

4.22

Explanation:

We know from the question, that the pOH of the solution is 9.78. Now the pOH is defined as -log [OH^-].

If the pOH of a solution is given, one may obtain the pH of such solution from the formula;

pH + pOH =14

Hence we can write;

pH = 14-pOH

pH = 14 - 9.78 = 4.22

Hence the pH of the solution is 4.22.

Calculate the concentration of H3O+ in a solution that contains 5.5 × 10-5 M OH- at 25°C. Identify the solution as acidic, basic, or neutral.

Answers

Explanation:

To calculate [H3O+] in the solution we must first find the pH from the [ OH-]

That's

pH + pOH = 14

pH = 14 - pOH

To calculate the pOH we use the formula

pOH = - log [OH-]

And [OH-] = 5.5 × 10^-5 M

So we have

pOH = - log 5.5 × 10^ - 5

pOH = 4.26

Since we've found the pOH we can now find the pH

That's

pH = 14 - 4.26

pH = 9.74

Now we can find the concentration of H3O+ in the solution using the formula

pH = - log H3O+

9.74 = - log H3O+

Find the antilog of both sides

H3O+ = 1.8 × 10^ - 10 M

The solution is basic since it's pH lies in the basic region.

Hope this helps you

243
Am
95
1. The atomic symbol of americium-243 is shown. Which of the following is correct?
• A. The atomic mass is 243 amu, and the atomic number is 95.
B. The atomic mass is 338 amu, and the atomic number is 95.
• C. The atomic mass is 95 amu, and the atomic number is 243.
D. The atomic mass is 243 amu, and the atomic number is 338.​

Answers

Answer:

A. The atomic mass is 243 amu, and the atomic number is 95.

The half-life of radium-226 is 1620 years. What percentage of a given amount of the radium will remain after 900 years

Answers

Answer:

68%

Explanation:

Since we need a percentage we can use any number we want for our initial value.

5(1/2)^900/1620 = 3.40

(3.40 / 5)*100 = 68%

To make sure lets use a different initial amount

1(1/2)^900/1620 = 0.68

(0.68/1) * 100 = 68%

The percentage of radium that will remain after 900 years is 68%.

To solve this question, we'll assume the initial amount of radium-226 to be 1.

Now, we shall proceed to obtaining the percentage of radium-226 that will after 900 years. This can be obtained as illustrated below:

Step 1

Determination of the number of half-lives that has elapsed.

Half-life (t½) = 1620 years

Time (t) = 900 years

Number of half-lives (n) =?

[tex]n = \frac{t}{t_{1/2}}\\\\n = \frac{900}{1620}\\\\n = \frac{5}{9}[/tex]

Step 2:

Determination of the amount remaining

Initial amount (N₀) = 1

Number of half-lives (n) = 5/9

Amount remaining (N) =?

[tex]N = \frac{N_{0} }{2^{n}}\\\\N = \frac{1}{2^{5/9}}[/tex]

N = 0.68

Step 3

Determination of the percentage remaining.

Initial amount (N₀) = 1

Amount remaining (N) = 0.68

Percentage remaining =?

Percentage remaining = N/N₀ × 100

Percentage remaining = 0.68/1 × 100

Percentage remaining = 68%

Therefore, the percentage amount of radium-226 that remains after 900 years is 68%

Learn more: https://brainly.com/question/10406952

One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of silver nitrate solution. Any chloride anions in solution will combine with the silver cations to produce bright white silver chloride precipitate. Suppose an EPA chemist tests a sample of groundwater known to be contaminated with nickel(II) chloride, which would react with silver nitrate solution like this:

Answers

Answer:

6.5 mg/L.

Explanation:

Step one: write out and Balance the chemical reaction in the Question above:

NiCl2 + 2AgNO3 =====> 2AgCl + Ni(NO3)2.

Step two: Calculate or determine the number of moles of AgCl.

So, we are given that the mass of AgCl = 3.6 mg = 3.6 × 10^-3 g. Therefore, the number of moles of AgCl can be calculated as below:

Number of moles AgCl = mass/molar mass = 3.6 × 10^-3 g / 143.32. = 2.5118 × 10^-5 moles.

Step three: Calculate or determine the number of moles of NiCl2.

Thus, the number of moles of NiCl2 = 2.5118 × 10^-5/ 2 = 1.2559 × 10^-5 moles.

Step four: detemine the mass of NiCl2.

Therefore, the mass of NiCl2 = number of moles × molar mass = 1.2559 × 10^-5 moles × 129.6 = 1.6 × 10^-3 g.

Step five: finally, determine the concentration of NiCl2.

1000/ 250 × 1.6 × 10^-3 g. = 6.5 mg/L.

Because of movements at the Mid-Atlantic Ridge, the Atlantic Ocean widens by about 2.5 centimeters each year. Explain which type of plate boundary causes this motion.

Answers

Answer:

A divergent plate boundary  

Explanation:

At a divergent boundary, the plates pull away from each other and generate new crust.

 

Answer:

Because the ocean becomes larger, this is a divergent plate boundary. Divergent plates cause the ocean floor to expand, making the ocean larger.

Explanation:

PLATO ANSWER

Find the density if the volume is 15 mL and the mass is 8.6 g. (5 pts)
Find the volume if the density is 2.6 g/mL and the mass is 9.7 g.(5 pts)
Find the mass if the density is 1.6 g/cm3 and the volume is 4.1 cm3 (5 pts)
Find the density if the initial volume of water is 12.8 mL, the final volume is 24.6 mL and the mass of the object is 4.3 g. Make a drawing to show the water displacement using a graduated cylinder. (gdoc, gdraw)

Answers

Answer:

[tex]\large \boxed{\text{0.57 g/mL; 3.7 mL; 6.6 g; 0.366 g/mL}}[/tex]

Explanation:

1. Density from mass and volume

[tex]\text{Density} = \dfrac{\text{mass}}{\text{volume}}\\\\\rho = \dfrac{m}{V}\\\\\rho = \dfrac{\text{8.6 g}}{\text{15 mL}} = \text{0.57 g/mL}\\\text{The density is $\large \boxed{\textbf{0.57 g/mL}}$}[/tex]

2. Volume from density and mass

[tex]V = \text{9.7 g}\times\dfrac{\text{1 mL}}{\text{2.6 g}} = \text{3.7 mL}\\\\\text{The volume is $\large \boxed{\textbf{3.7 mL}}$}[/tex]

3. Mass from density and volume

[tex]\text{Mass} = \text{4.1 cm}^{3} \times \dfrac{\text{1.6 g}}{\text{1 cm}^{3}} = \textbf{6.6 g}\\\\\text{The mass is $\large \boxed{\textbf{6.6 g}}$}[/tex]

4. Density by displacement

Volume of water + object = 24.6 mL

Volume of water                = 12.8 mL

Volume of object               = 11.8 mL

[tex]\rho = \dfrac{\text{4.3 g}}{\text{11.8 mL}} = \text{0.36 g/mL}\\\text{The density is $\large \boxed{\textbf{0.36 g/mL}}$}[/tex]

Your drawing showing water displacement using a graduated cylinder should resemble the figure below.

 

How does the spontaneity of the process below depend on temperature? PCl5(g)+H2O(g)→POCl3(g)+2HCl(g) ΔH=−126 kJ mol−1, ΔS=146 J K−1mol−

Answers

The given question is incomplete, the complete question is:

How does the spontaneity of the process below depend on temperature? PCI5(9)+H2O(g)POCI3(g) +2HCI(g) -126 kJ mol1, AS = 146 J K-'mol1 ΔΗ Select the correct answer below: nonspontaneous at all temperatures spontaneous at all temperatures spontaneous at high temperatures and nonspontaneous at low temperatures spontaneous at low temperatures and nonspontaneous at high temperatures

Answer:

The correct answer is spontaneous at all the temperatures.

Explanation:

Gibbs Free energy is an essential relation that determines the spontaneity of any reaction, that is, ΔG = ΔH - TΔS

When ΔG is less than zero, that is, negative, the reaction is considered to be in spontaneous state. Based on the given information, ΔH = -126 kJ/mol

= -126000 J/mol, it is negative

ΔS = 146 J/K/mol, it is positive

Now, ΔG = ΔH-TΔS

= (-ve) - T (+ve), Thus, when ΔH, is -ve, ΔS is +ve, -TΔS is -ve, the ΔG will be -ve. Therefore, reaction will be spontaneous at all the temperatures.  

If the theoretical yield of a reaction is 332.5 g and the percent yield for the reaction is 38 percent, what's the actual yield of product in grams? \

A. 8.74 g
B. 12616 g
C. 116.3 g
D. 126.4 g

Answers

Answer: D - 126.4g

Explanation:  

% Yield = Actual Yield/Theoretical Yield

38% = Actual Yield/332.5

38/100 = Actual Yield/332.5

(.38)(332.5) = 126.35 g = 126.4 g Actual Yield

Answer:

is D. the correct answer

Explanation:

I'm not sure if it is. Please let me know if I'm mistaking.

A 1.0 L buffer solution is 0.250 M HC2H3O2 and 0.050 M LiC2H3O2. Which of the following actions will destroy the buffer?

A. adding 0.050 moles of NaOH
B. adding 0.050 moles of LiC2H3O2
C. adding 0.050 moles of HC2H3O2
D. adding 0.050 moles of HCl
E. None of the above will destroy the buffer.

Answers

Answer:

D

Explanation:

Addition of 0.05 M HCl, will react with all of the C2H3O2- from LiAc which will give 0.05 M more HAc. So there will be no Acetate ion left to make the solution buffer. Hence, the correct option for the this question is d, which is adding 0.050 moles of HCl.

The action that destroys the buffer is option c. adding 0.050 moles of HCl.

What is acid buffer?

It is a solution of a weak acid and salt.

Here, The buffer will destroy at the time when either HC2H3O2 or NaC2H3O2 should not be present in the solution.

The addition of equal moles of HCl finishly reacts with equal moles of NaC2H3O2. Due to this,  there will be only acid in the solution.

Since

moles of HC2H3O2 = 1*0.250 = 0.250

moles of NaC2H3O2 = 1*0.050 = 0.050.

moles of HCl is added = 0.050

Now

The reaction between HCl and NaC2H3O2

[tex]HCl + NaC_2H_3O_2 \rightarrow HC_2H_3O_2 + NaCl[/tex]

Now

BCA table is

            NaC2H3O2  HCl       HC2H3O2

Before 0.050 0.050 0.250

Change -0.050 -0.050 +0.050

After 0 0 0.300

Now, the solution contains the acid (HC2H3O2 ) only.

Therefore addition of 0.050 moles of HCl will destroy the buffer.

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Write a balanced chemical equation for the base hydrolysis of methyl butanoate with NaOH. (Use either molecular formulas or condensed structural formulas, but be consistent in your equation.)

Answers

Explanation:

C5H10O2 + NaOH = C2H5COONa + C2H5OH

your result are : sodium propanoate and ethanol

A balanced chemical equation represents atoms and their numbers with their charge. The balanced equation for base hydrolysis is C₅H₁₀O₂ + NaOH → C₂H₅COONa + C₂H₅OH.

What is hydrolysis?

Base hydrolysis is the splitting of the ester linkage by the basic molecule. As the result the acidic ester portion makes the salt, and also alcohol is produced as the by-product.

The base hydrolysis of methyl butanoate is shown as,

C₅H₁₀O₂ + NaOH → C₂H₅COONa + C₂H₅OH

Here, sodium propanoate and ethanol are produced by the splitting of methyl butanoate in the presence of the base (NaOH).

Therefore, C₅H₁₀O₂ + NaOH → C₂H₅COONa + C₂H₅OH is balanced reaction.

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15. Calculate the critical angle of glass and water combination. Show your calculation. 16. What is the critical angle for the interface between Mystery A and glass

Answers

Answer:

15. Critical angle of glass and water combination, θ = 62.45°

16. Critical angle for the interface between Mystery A and glass, θ = 37.93°

Note; The question is incomplete. The complete question is as follows:

Medium Air Water Glass Mystery A Mystery B Table-2 Speed (m/s) 1.00 C 0.75 c 0.67 0.41 c 0.71 c n 1.00 1.33 1.50 Index of Refraction n of a given medium is defined as the ratio of speed of light in vacuum, c to the speed of light in a medium, v. n = c/v

Table-4: Incident Angle (degrees) Reflected Angle Refracted angle (degrees) (degrees) % Intensity of reflected ray 0 10 20 30 40 50 N/A N/A N/A 30 40 50 0 11.3 22.7 34.2 46.3 59.5 N/A N/A N/A 0.67 1.22 3.08 % Intensity of refracted ray 100 100 100 99.33 98.78 96.92

When rays travel from a denser medium to a less dense medium, we can define a critical angle of incidence θ such that refracted angle θ₂ = 90°. Applying Snell's law: Critical angle θ = sin-1(n₂/n₁).

When the angle of incidence is greater than the critical angle, 100% of the light intensity is reflected. This is called total internal reflection because all the light is reflected.

15. Calculate the critical angle of glass and water combination. Show your calculation.

16. What is the critical angle for the interface between Mystery A and glass?

Explanation:

15.  Applying Snell's law; Critical angle θ = sin-1(n₂/n₁).

where n₂,refractive index of water = 1.33, n₁, refractive index of glass = 1.50 since glass is denser than water

θ = sin-1(1.33/1.50)

θ = 62.45°

Critical angle of glass and water combination, θ = 62.45°

16.  Refractive index of mystery A , n = c/v

where v = 0.41 c

therefore, n = c / 0.41 c = 2.44

Critical angle for the interface between Mystery A and glass, θ = sin-1(n₂/n₁).

where n₂,refractive index of glass = 1.50, n₁, refractive index of mystery A = 2.44 since mystery A is denser than glass as seen from its refractive index

θ = sin-1(1.50/2.44)

θ = 37.93°

Critical angle for the interface between Mystery A and glass, θ = 37.93°

While balancing a chemical equation, we change the _____ to balance the number of atoms on each side of the equation.

Answers

Answer:

While balancing a chemical equation, we change the coefficient  to balance the number of atoms on each side of the equation

Explanation:

While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.

What is chemical equation?

To summarize in chemistry terms, a chemical equation depicts the initial chemicals, or reactants, on the left-hand side and the final compounds, or products, just on right-hand side, divided by an arrow. In the chemical equation, the number of atoms in each element as well as the total charge are the same on opposite of the equation's sides.

Chemical equations are used in chemistry to depict chemical processes by writing the reactants and products in terms of their corresponding chemical formulas. While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.

Therefore, while balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.

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Zeros laced at the end of the significant number are...

Answers

Answer:

Zeros located at the end of significant figures are significant.

Explanation:

Hope it will help :)

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