Under certain conditions, the substance mercury(II) oxide can be broken down to form mercury and oxygen. If 32.2 grams of mercury(II) oxide react to form 29.8 grams of mercury, how many grams of oxygen must simultaneously be formed

Answer:

0.796 M

Explanation:

Step 1: Given data

Gravimetric concentration (Cg): 4.78 g%g

Density of the solution (ρ): 1.00 g/mL

Step 2: Calculate the volumetric concentration of the solution (Cv)

We will use the following expression.

Cv = Cg × ρ

Cv = 4.78 g%g × 1.00 g/mL = 4.78 g%mL

Step 3: Calculate the molarity of the solution (M)

The volumetric concentration is 4.78 g%mL, that is, there are 4.78 g of acetic acid per 100 mL of solution. We can calculate the molarity using the following expression.

M = mass solute / molar mass solute × liters of solution

M = 4.78 g / 60.05 g/mol × 0.1 L = 0.796 M

Answer:

pH = 12.66

Explanation:

The HIO3 reacts with NaOH as follows:

HIO3 + NaOH → H2O + NaIO3

The moles of HIO3 and NaOH added are:

Moles HIO3:

0.0100L * (0.100mol / L) = 0.00100 moles HIO3

Moles NaOH:

0.05000L * (0.100mol / L) = 0.00500 moles NaOH

As moles NaOH > Moles HIO3, the moles of NaOH that remain are:

0.00500mol - 0.00100mol = 0.00400 moles NaOH.

After the reaction you will have only NaOH and NaIO3. As NaIO3 is a salt, the pH of the solution is determined by only NaOH. Its concentration is:

Moles NaOH: 0.00500 moles NaOH

Volume: 50.0mL + 50.0mL + 10.0mL = 110.0mL = 0.110L

Molarity: 0.0455M NaOH = [OH-]

pOH = -log [OH-] = 1.34

pH = 14 - pOH

pH is the measure of the hydrogen or the hydronium ion in an aqueous solution. The pH of the mixture containing sodium hydroxide is 12.66.

pH is the potential of the hydrogen and is given as a negative log of the hydrogen concentration in the aqueous solution.

The balanced chemical reaction can be shown as:

[tex]\rm HIO_{3} + NaOH \rightarrow H_{2}O + NaIO_{3}[/tex]

Moles of iodic acid are calculated as:

[tex]\begin{aligned} \rm moles &= \rm molarity \times volume\\\\&= 0.100 \;\rm M \times 0.0100 \;\rm L \\\\&= 0.00100\;\rm moles\end{aligned}[/tex]

Moles of sodium hydroxide are calculated as:

[tex]\begin{aligned} \rm moles &= \rm molarity \times volume\\\\&= 0.100 \;\rm M \times 0.05000 \;\rm L \\\\&= 0.00500\;\rm moles\end{aligned}[/tex]

The remaining moles of sodium hydroxide are 0.00500mol - 0.00100mol = 0.00400 moles.

The pH of the mixture will be determined by sodium hydroxide, as sodium iodate is a salt.

The molar concentration of sodium hydroxide is calculated as:

[tex]\begin{aligned} \rm M &= \rm \dfrac{moles}{volume}\\\\&= \dfrac{0.00500}{0.110}\\\\&= 0.0455 \;\rm M\end{aligned}[/tex]

pH is calculated as:

[tex]\begin{aligned} \rm pOH &= \rm -log[OH^{-}] = 1.34\\\\\rm pH &= \rm 14 - pOH\\\\\rm pH &= 12.66\end{aligned}[/tex]

Therefore, 12.66 is the pH of the mixture.

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Answer:

72.96 of water produce by 45.4 L of oxygen at STP.

Explanation:

[tex]H_2+\frac{1}{2}O_2\rightarrow H_2O[/tex]

1 mole of oxygen=22.4 L at STP

[tex]\frac{1}{2}[/tex]\mole of oxygen=22.4/2=11.2 L

11.2 L of oxygen required to produce water=1 mole

1 L of oxygen required to produce water=1/11.2 mole

45.4 L of oxygen required to produce water=[tex]\frac{1}{11.2}\times 45.4[/tex]

45.4 L of oxygen required to produce water=[tex]\frac{45.4}{11.2}[/tex]moles

1 mole of water=18 g

[tex]\frac{45.4}{11.2}[/tex]moles of water=[tex]18\times \frac{45.4}{11.2}[/tex]

[tex]\frac{45.4}{11.2}[/tex]moles of water=72.96 g

Hence, 72.96 of water produce by 45.4 L of oxygen at STP.

Answer:

Explanation:

Target organ toxins are chemicals that can cause adverse effects or disease states manifested in specific organs of the body. Toxins do not affect all organs in the body to the same extent due to their different cell structures.

Answer:

Molecules of O₂F₂ = mass of O₂F₂ × (1 mole O₂F₂ / 70 g O₂F₂) × (6.02 × 10²³ molecules / one mole of O₂F₂)

Explanation:

The Avogadros constant gives the the number of specified entities in one mole of a substance. One mole of any substance contains 6.02 × 10²³ particles. Therefore, one mole of O₂F₂ contains 6.02 × 10²³ molecules.

Also, the molar mass of a substance is the mass in grams of one mole of that substance. It is obtained by summing the relative atomic masses of all the atoms of the elements in the substance. For O₂F₂, the molar mass = (2 × 16 + 2 × 19) g/mol = 70 g/mol

Converting to molecules of O₂F₂;

To convert from grams of a substance to molecules of that substance, multiply by the ratio of one mole and mass of one mole, and then by the number of molecules per mole.

Molecules of A = mass of A × (1 mole / mass of one mole) × (6.02 × 10²³ molecules / 1 mole)

Therefore,Molecules of O₂F₂ = mass of O₂F₂ × (1 mole O₂F₂ / 70 g O₂F₂) × (6.02 × 10²³ molecules /one mole of O₂F₂)

Answer:

By how many times would you expect Al2(SO4)3 to depress the F.P of water compared to sucrose C12H22011 ?.

Explanation:

The freezing point of a pure solvent decreases further by adding a nonvolatile solute.

This is called depression in freezing point.

When an ionic solute is dissolved then the depression in the freezing point is proportional to the number of ions present in the solution.

In aluminum sulfate, there are five ions formed as shown below:

[tex]Al_2(SO_4)_3(aq)->2Al^3^+(aq)+3SO_4^2^-(aq)[/tex]

But sucrose is a covalent compound and it does not undergo dissociation.

Hence, aluminum sulfate decreases the freezing point of water by five times compared to sucrose.

Explanation:

Answer:

0.64 m

Explanation:

Given that;

1L = 0.001 cubic metre

Then;

263 L = 263 L × 0.001 cubic metre/1L

= 0.263 cubic metre

Volume of a cube = l^3

l= 3√V

l= 3√0.263 cubic metre

l= 0.64 m

As the gas expands on the surrounding, work is done by the system.

Therefore, W= -279J

Absorbtion of heat,q= +216J

∆U=q+W = (216-279)J= -63J

Answer:

66.7%

Explanation:

Step 1: Write the balanced equation

CHCl₃ + Cl₂ ⇒ CCl₄ + HCl

Step 2: Calculate the theoretical yield of CCl₄ from 11.9 g of CHCl₃

According to the balanced equation, the mass ratio of CHCl₃ to CCl₄ is 119.38:153.82.

11.9 g CHCl₃ × 153.82 g CCl₄/119.38 g CHCl₃ = 15.3 g CCl₄

Step 3: Calculate the percent yield of CCl₄

Given the experimental yield of CCl₄ is 10.2 g, we can calculate the percent yield using the following expression.

%yield = (exp yield/theo yield) × 100%

%yield = (10.2 g/15.3 g) × 100% = 66.7%

Answer:

this answer is for first one

Explanation:

A higher evalations, where the atmospheric pressure is much lower, the boiling point increase with increased pressure up to the critical point where the gas and liquid properties become identical

Answer:

The volume of the gas is fixed and will not change.

Explanation:

The volume of the gas will not change because there is no change in temperature. Temperature increases the volume of gases enclosed in a container.

Answer:

Explanation:

1mol of [tex]O_2=2*16{gr\over{mol}}=32{gr\over{mol}}\\\\15.7gr->15.7gr*{1\over{32{gr\over{mol}}}}=0.491mol~of~O_2[/tex]

as 1mol of molecular oxigen reacts with 4 mol of aluminium

1 mol of O2         -----------------------------> 4 mol of Al

0.491 mol of O2  ------------------------------>     x

[tex]x={0.491*4\over{1}}~mol~of~Al=1.9625~mol~of~Al[/tex]

Answer:

41264 g of CO₂

Explanation:

Combustion reaction is:

C₃H₈ + 5O₂ →  3CO₂  +  4H₂O

1 mol of propane react to 5 moles of oxygen in order to proudce 3 moles of carbon dioxide and 4 moles of water.

In a combustion reaction, our reactant reacts to oxygen and the products are always CO₂ and water.

We have the volume of propane but we need moles of it, so we need to apply density.

Density = mass / volume so mass = density . volume.

Density of propane is: 493 g/L

Mass of propane is 493 g/L . 27.9L = 13754.7 g

We convert mass to moles: 13754.7 g . 1 mol/ 44g = 312.6 moles

According to reaction, 1 mol of propane can produce 3 moles of CO₂

Our 312.6 moles will produce 312.6 . 3 = 937.8 moles

We convert moles to mass: 937.8 mol . 44 g/mol = 41264 g

Answer:

The answer is in the explanation.

Explanation:

A solution is defined as the homogeneous mixture of a solute (In this case, NaCl) and the solvent (water).

To prepare 1L of the solution, the student can weigh the 3g of NaCl in the volumetric flask but need to add slowly water to dissolve the NaCl (That is very soluble in water). When all NaCl is dissolved the student must transfer the solution to the 1L volumetric flask. Then, you must add more water to the beaker until "Clean" all the solute of the beaker to transfer it completely to the volumetric flask.

Answer:

A = 65.46 u

Explanation:

Given that,

The composition of zinc is as follows :

Zn-64 = 48.63%

Zn-66 = 27.90%

Zn-67 = 4.10%

Zn-68 = 18.75%

Zn-70 = .62%

We need to find the  average atomic mass of the given element. It can be solved as follows :

[tex]A=\dfrac{48.63\times 64+27.90\times 66+4.1\times 67+18.75\times 68+0.62\times 70}{100}\\A=65.46\ u[/tex]

So, the average atomic mass of zinc is 65.46 u.

Answer: The molarity of each of the given solutions is:

(a) 1.38 M

(b) 0.94 M

(c) 1.182 M

Explanation:

Molarity is the number of moles of a substance present in liter of a solution.

And, moles is the mass of a substance divided by its molar mass.

(a) Moles of ethanol (molar mass = 46 g/mol) is as follows.

[tex]Moles = \frac{mass}{molar mass}\\= \frac{28.5 g}{46 g/mol}\\= 0.619 mol[/tex]

Now, molarity of ethanol solution is as follows.

[tex]Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.619 mol}{4.50 \times 10^{2} \times 10^{-3}L}\\= 1.38 M[/tex]

(b) Moles of sucrose (molar mass = 342.3 g/mol) is as follows.

[tex]Moles = \frac{mass}{molar mass}\\= \frac{21.6 g}{342.3 g/mol}\\= 0.063 mol[/tex]

Now, molarity of sucrose solution is as follows.

[tex]Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.063 mol}{0.067 L} (1 mL = 0.001 L)\\= 0.94 M[/tex]

(c) Moles of sodium chloride (molar mass = 58.44 g/mol) are as follows.

[tex]Moles = \frac{mass}{molar mass}\\= \frac{6.65 g}{58.44 g/mol}\\= 0.114 mol[/tex]

Now, molarity of sodium chloride solution is as follows.

[tex]Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.114 mol}{0.0962 L}\\= 1.182 M[/tex]

Thus, we can conclude that the molarity of each of the given solutions is:

(a) 1.38 M

(b) 0.94 M

(c) 1.182 M

Answer is 9

pKb=−logK

b=−log10^-5=5

pOH=pKb+log[acid]*[salt]

Substitute values in the above expression.

pOH=5+log0.1*0.1=5

Hence, the pH of the solution is pH=14−pOH=14−5=9

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Answer:

Density and rigidity

Answer:

The full amount (5.00 g) will be dissolved in 1 L of water at 25°C.

Explanation:

The molecular weight (MW) of Vanillin (C₈H₈O₃) is calculated from the chemical formula as follows:

MW(C₈H₈O₃) = (12 g/mol x 8) + (1 g/mol x 8) + (16 g/mol x 3) = 152 g/mol

If 0.070 mol of C₈H₈O₃ are soluble per liter of water at 25°C, the maximum mass that can be dissolved in 1 L is:

0.070 mol x 152 g/mol = 10.64 g

Since 5.00 g is lesser than the maximum amount that can be dissolved (10.64 g), the added amount will be completely dissolved in 1 L of water at 25°C.

Answer:

In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 the Tert butyl occupies the axial and the bromine occupies equatorial positions. Compound 1 reacts faster than compound 2.

Explanation:

In cyclic organic compounds, substituents may occupy the axial or equatorial positions. The axial positions are aligned parallel to the symmetry axis of the ring while the equatorial positions are around the plane of the ring.

Bulky substituents have more room in the equatorial than in the axial position. This means that compound 1 is more stable than compound 2.

This is clear on the basis of stability of the molecules because compound 1 will react faster than compound 2 since the bulky tertiary butyl group in compound 1 occupy equatorial and not axial positions.

Answer:

An unknown compound's 2,4-DNP product melting range is between 144-146'C. It does not give a silver mirror on the Tollens test and is slow to react to the chromic acid test. What would this compound be

Explanation:

From the given data it is clear that the unknown compound gives positive test with 2,4-DNP reagent.

That means it has a carbonyl group.Either aldehyde or ketone.

It does not give a silver mirror on the Tollens test and is slow to react to the chromic acid test.

That means aldehyde is absent.

So, the other carbonyl group that is ketone is present in the given unknown compound.

Answer:

These batteries are constructed of several single cells connected in series each cell produces approximately 2.1 volts. ... A battery cell consists of two lead plates a positive plate covered with a paste of lead dioxide and a negative made of sponge lead, with an insulating material (separator) in between.

Answer:

C. Products have higher potential energy, and the reaction is endothermic.

Explanation:

a) Since this is a double displacement reaction, we write the balanced equation as

[tex]2AgNO_3(aq) + CaCl_2(aq) \\ \rightarrow 2AgCl(s) + Ca(NO_3)_2(aq)[/tex]

b) Next we find the number of moles of AgNO3 in the solution.

[tex](0.005\:\text{L})(0.500\:M\:AgNO_3) \\ = 0.0025\:\text{mol}\:AgNO_3[/tex]

Next, use the molar ratio to find the necessary amount of CaCl2 to react with the AgNO3:

[tex]0.0025\:\text{mol}\:AgNO_3× \left(\dfrac{1\:\text{mol}\:CaCl_2}{2\:\text{mol}\:AgNO_3} \right)[/tex]

[tex]= 0.00125\:\text{mol}\:CaCl_2[/tex]

The volume of 0.500 M solution of CaCl2 necessary to react all of the given AgNO_3 is then

[tex]V = \dfrac{0.00125\:\text{mol}\:CaCl_2}{0.500\:\text{M}\:CaCl_2}[/tex]

[tex]= 0.0025\:\text{L} = 2.5\:\text{mL}\:CaCl_2[/tex]

c) The theoretical yield can then be calculated as

[tex]0.0025\:\text{mol}\:AgNO_3 × \left(\dfrac{2\:\text{mol}\:AgCl}{2\:\text{mol}\:AgNO_3} \right)[/tex]

[tex]= 0.0025\:\text{mol}\:AgCl[/tex]

Converting this amount of AgCl into grams, we get

[tex]0.0025\:\text{mol}\:AgCl × \left(\dfrac{143.32\:\text{g}\:AgCl}{1\:\text{mol}\:AgCl} \right)[/tex]

[tex]= 0.358\:\text{g}\:AgCl[/tex]

Answer:

hydrogen, oxygen, chlorine, ammonia

Explanation:

Air is a mixture of gases. When we say "clean" air here, we are referring to air that does not contain pollutant gases.

Some components of air such as water vapour, methane, CO2, and N2O are greenhouse gases. They are known to contribute towards global warming.

Some gases such as SO2 and NO2 contribute towards acid rain. The oxides of nitrogen are particularly involved in the formation of photochemical smog.

The halogens are known to lead to the depletion of the ozone layer and radon is a radioactive gas.

Hence, hydrogen, oxygen, chlorine, ammonia have no negative environmental impact hence they are found in clean air.

Answer:

La temperatura final del sistema es 1029,346 °C.

Explanation:

Asumamos que el sistema conformado por el cobre y el estaño no tiene interacciones con sus alrededores. Por la Primera Ley de la Termodinámica, el cobre cede calor al estaño con tal de alcanzar el equilibrio térmico. El cobre se encuentra inicialmente en su punto de fusión, mientras que el estaño está por encima de ese punto, de modo que la transferencia de calor es esencialmente sensible:

[tex]m_{Cu}\cdot c_{Cu}\cdot (T-T_{Cu}) = m_{Sn}\cdot c_{Sn}\cdot (T_{Sn}-T)[/tex]

[tex](m_{Cu}\cdot c_{Cu} + m_{Sn}\cdot c_{Sn})\cdot T = m_{Sn}\cdot c_{Sn}\cdot T_{Sn} + m_{Cu}\cdot c_{Cu}\cdot T_{Cu}[/tex]

[tex]T = \frac{m_{Sn}\cdot c_{Sn}\cdot T_{Sn}+m_{Cu}\cdot c_{Cu}\cdot T_{Cu}}{m_{Cu}\cdot c_{Cu}+m_{Sn}\cdot c_{Sn}}[/tex] (1)

Donde:

[tex]m_{Sn}[/tex] - Masa del estaño, en gramos.

[tex]m_{Cu}[/tex] - Masa del cobre, en gramos.

[tex]c_{Sn}[/tex] - Calor específico del estaño, en calorías por gramo-grados Celsius.

[tex]c_{Cu}[/tex] - Calor específico del cobre, en calorías por gramo-grados Celsius.

[tex]T_{Sn}[/tex] - Temperatura inicial del estaño, en grados Celsius.

[tex]T_{Cu}[/tex] - Temperatura inicial del cobre, en grados Celsius.

Si sabemos que [tex]m_{Cu} = 150\,g[/tex], [tex]m_{Sn} = 35\,g[/tex], [tex]c_{Cu} = 0,093\,\frac{cal}{g\cdot ^{\circ}C}[/tex], [tex]c_{Sn} = 0,060\,\frac{cal}{g\cdot ^{\circ}C}[/tex], [tex]T_{Sn} = 560\,^{\circ}C[/tex] y [tex]T_{Cu} = 1100\,^{\circ}C[/tex], entonces la temperatura final del sistema es:

[tex]T = \frac{(35\,g)\cdot \left(0,060\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (560\,^{\circ}C)+(150\,g)\cdot \left(0,093\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (1100\,^{\circ}C)}{(35\,g)\cdot \left(0,060\,\frac{cal}{g\cdot ^{\circ}C} \right)+(150\,g)\cdot \left(0,093\,\frac{cal}{g\cdot ^{\circ}C} \right)}[/tex]

[tex]T = 1029,346\,^{\circ}C[/tex]

La temperatura final del sistema es 1029,346 °C.

Answer:

Towards the big bully

Explanation:

If a big bully and a small child are involved in a thug of war, it is clear that the bully is stronger than the child and he/she will pull the rope used in the thug of war with a greater force.

By so doing, the ball attached at the centre of the rope will naturally be drawn towards the stronger bully.

Ionic bonds are the strongest type of chemical bond and, therefore, most compounds remain solid with very high melting points

The statement best describes the strength of ionic and covalent bonds is that ionic bond is the strongest bond and has very high melting point.

A chemical bond is a lasting attraction between atoms or ions that enables the formation of molecules, crystals, and other structures. The bond may result from the electrostatic force between oppositely charged ions as in ionic bonds, or through the sharing of electrons as in covalent bonds.

Types of Chemical Bonds includes-

Ionic Bonds.

Covalent Bonds.

Hydrogen Bonds.

Polar Bonds.

Generally, ionic bonds are much stronger than covalent bonds. In ionic bonds, there is complete transfer of electrons between elements to form a stable compound. While in covalent bond, there is only sharing of electrons between two elements to form a stable compound.

Therefore, The statement best describes the strength of ionic and covalent bonds is that ionic bond is the strongest bond and has very high melting point.

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Answer:

The solution is 0.450 M. How many mL are needed?

- 0.667 mL

Explanation:

Answer:

c. CH3CH2CH(CH3)CHO

Explanation:

Hello there!

In this case, according to the process for the one-step oxidation of a primary alcohol with a moderately strong oxidizing agent like pyridinium chlorochromate (PCC), whereby an aldehyde is produced, we infer that the corresponding product will be 2-methylbutanal, which matches with the choice c. CH3CH2CH(CH3)CHO according to the following reaction:

[tex]CH_3CH_2CH(CH_3)CH_2OH\rightarrow CH_3CH_2CH(CH_3)COH[/tex]

Regards!