2Co + O2 = 2CO2
In this reaction 10.8 mole of carbon dioxide was produced .calculate the number of moles of carbon monoxide used in this reaction to produce such number of moles of carbon dioxide

Answer:

Explanation:

We can use the ideal gas law to solve for the temperature of the neon gas:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

First, let's convert the volume from milliliters to liters:

V = 1520 mL = 1.52 L

Next, let's rearrange the ideal gas law to solve for T:

T = PV/nR

Now we can plug in the values and solve:

T = (325 mmHg)(1 atm/760 mmHg)(1.52 L)/(0.250 mol)(0.0821 L·atm/(mol·K))

T = 89.6 K

To convert to degrees Celsius, we subtract 273.15:

T = 89.6 K - 273.15 = -183.6 °C

Therefore, the temperature of the neon gas is 89.6 K or -183.6 °C.

The amount of heat required is 246.165 Joules.



We can use the following method to figure out how much energy it will take to raise the temperature of benzene:

Q = mcΔT

Where Q = the amount of energy needed (in Joules).

m = the number of grams of benzene.

c = the amount of heat that benzene can hold (in J/g*K).

T = temperature change (in Kelvin)

But since the volume of benzene is given to us, we need to find the mass of Benzene by:

Density = mass/volume

mass = density x volume

mass = 0.876 g/cm3 x 50.00 mL = 43.8 g

Change in temperature = 28.75 - 25.52 = 3.23 K

Now, we can use the given numbers to fill in the formula for Q:

Q = mcΔT

Q = 43.8 g x 1.74 J/g*K x 3.23 K

Q = 246.165 J

So, it takes 238.92 Joules of heat energy to raise the temperature of 50 mL of benzene from 25.52°C to 28.75°C.

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Answer:

[tex]\Large \boxed{\boxed{\textsf{heat required = 246.2 J (4 s.f)}}}[/tex]

Explanation:

The heat energy required to raise the temperature of the reaction, i.e, the amount of heat energy released in the reaction (noting the rise in temperature, this is an exothermic reaction), can be calculated with the following calorimetry equation:

[tex]\Large \boxed{\textsf{$q=mc\Delta T$}} \sf \,, where:\\ \\\bullet q = quantity\,of\,heat\,released\,(measured\,in\,joules)\\ \bullet m = solvent\,of\,mass\,(measured\,in\,g\,or\,kg)\\\bullet c = specific\,heat\,capacity\,of\,solution\\\bullet \Delta T = change\,in\,temperature\,of\,solution[/tex]

To solve this, we know that:

[tex]\textsf{$\bullet$ m = 43.8 g (1 mL = 1 cm$^3$, using density = g/cm$^3$)}\\\textsf{$\bullet$ c = 1.74 J/g/K}\\\textsf{$\bullet \Delta T$ = 3.23 ($\Delta T$ is the same whether Kelvin or Celsius is used)}[/tex]

Inputting these values into the formula:

[tex]\large \textsf{$q = (43.8)(1.74)(3.23)$}\\ \\\large \textsf{$\therefore q=246.2$ J}\\ \\ \\\Large \boxed{\boxed{\textsf{$\therefore$ heat required = 246.2 J (4 s.f)}}}[/tex]

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Copper is the limiting reactant, with n(Copper) = 0.969 mol being less than n(Nitric acid). It produces 17.44 grammes of water. As the outcome is negative, there is no excess Nitric acid. The reaction uses up all of the Nitric acid.

The method described in the following steps can balance a redox equation: (1) Split the equation into two equal halves. (2) Make each half-mass reaction's and charge equal. (3) Ensure that the quantity of electrons going to each half-reaction is the same. The half-reactions should be combined.

n(Nitric acid) = (246 mL) x (6.60 mol/L) = 1.6236 mol

Since n(Copper) = 0.969 mol is less than n(Nitric acid), Copper is the limiting reactant.

m(Water) = n(Water) x M(Water) = 0.969 mol x 18.015 g/mol = 17.44 g

Therefore, 17.44 grams of Water is formed.

n(Nitric acid) needed = n(Copper) x 2 = 1.938 mol

The excess amount of Nitric acid is:

n(Nitric acid) excess = n(Nitric acid) initial - n(Nitric acid) needed

= 1.6236 mol - 1.938 mol

= -0.3144 mol

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Difluoroethane (DFE) is a cheap, commonly obtainable volatile chemical that is safe for recreational inhalation.

Thus, It can be found in everyday household items including propellants, refrigerants, and compressed air dusters. When breathed, DFE is a central nervous system (CNS) depressant that causes a momentary feeling of euphoria.

Toxic effects are linked to prolonged or excessive usage, and rapid termination might cause withdrawal.3–5 We describe a DFE misuse case that was accompanied by skeletal fluorosis and withdrawal psychosis.

Difluoroethane is a colourless, odourless gas that is transported under its vapour pressure as a liquefied gas. Ingestion of the liquid can result in frostbite.

Thus, Difluoroethane (DFE) is a cheap, commonly obtainable volatile chemical that is safe for recreational inhalation.

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When water ionizes, it means that some of the water molecules break apart into ions, which are electrically charged particles. If you could shrink down and see the true composition of water at the molecular level, means water molecules are in a constant state of motion and interaction with each other.

Water (H₂O) is a polar molecule, meaning it has a partial positive charge on one end (hydrogen) and a partial negative charge on the other end (oxygen). This polarity allows water molecules to interact with each other through hydrogen bonding, which gives water its unique properties, such as high boiling and melting points, high heat capacity, and strong surface tension.

When water ionizes, it means that some of the water molecules break apart into ions, which are electrically charged particles. In the case of water, it can ionize into hydrogen ions (H⁺) and hydroxide ions (OH⁻);

H₂O → H⁺ + OH⁻

This ionization occurs when water molecules dissociate or separate into these charged particles due to the transfer of a proton (H⁺) between water molecules.

If you could shrink down and see the true composition of water at the molecular level, you would observe that water molecules are in a constant state of motion and interaction with each other. Some water molecules would be dissociating into hydrogen ions (H⁺) and hydroxide ions (OH⁻), while others would be recombining to form water molecules again.

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To balance the given chemical equation using the oxidation state change method, we need to follow these steps:Step 1: Write the unbalanced equationKMnO4 + KCl + H2SO4 → K2SO4 + MnSO4 + Cl2
Step 2: Identify the elements that undergo oxidation and reductionIn this equation, the oxidation state of Mn changes from +7 to +2, which means it undergoes reduction, while the oxidation state of Cl changes from -1 to 0, which means it undergoes oxidation.

Step 3: Write the half-reactionsReduction half-reaction: MnO4^- → Mn^2+Oxidation half-reaction: Cl^- → Cl2Step 4: Balance the atoms and charges in each half-reactionReduction half-reaction: 8H+ + MnO4^- → Mn^2+ + 4H2OOxidation half-reaction: 2Cl^- → Cl2 + 2e^-Step 5: Balance the electrons in each half-reactionReduction half-reaction: 5e^- + 8H+ + MnO4^- → Mn^2+ + 4H2OOxidation half-reaction: 2Cl^- → Cl2 + 2e^-Step 6: Multiply each half-reaction by a factor to equalize the number of electrons transferredReduction half-reaction: 10e^- + 16H+ + 2MnO4^- → 2Mn^2+ + 8H2OOxidation half-reaction: 14Cl^- → 7Cl2 + 14e^-Step 7: Add the balanced half-reactions together10e^- + 16H+ + 2MnO4^- + 14Cl^- → 2Mn^2+ + 8H2O + 7Cl2Step 8: Cancel out the common terms on both sides of the equation2KMnO4 + 16KCl + 8H2SO4 → 2K2SO4 + 2MnSO4 + 7Cl2 + 8H2OTherefore, the balanced equation using the oxidation state change method is:2KMnO4 + 16KCl + 8H2SO4 → 2K2SO4 + 2MnSO4 + 7Cl2 + 8H2O.

Explanation:

Use the Ideal Gas Law :

PV = n RT

STP = standard temp and pressure =   273.15 K  and  1  atm

n = number of moles

   CO2,  (using table of elements) ,

        has mole weight of  12.011 +2*15.999=~ 44 gm/mole

                 20 gm / 44gm/mole = .455 mole of CO2

    R = gas constant = .082057  L atm /(K mole)

Plug in the numbers and solve for  'V'

(1 atm ) V = .455  * .082057 * 273.15

V = 10.2 liters

Here is another way:

 ....knowing that a mole of gas occupies 22.4 L /mole at STP

        22.4 L / mole * .455 mole = ~10.2 liters