How many moles of n2 (g) are present in 1. 00 l of n2 (g) at 100. °c and 1 atm?


______ moles

• The balanced equation for the reaction is given by:

• From the above reaction we can see that:

1 mole of ethanol (CH3Ch2OH) produces 1 mol of acetic acid(CH3COOH)

so . x mole of ethanol will produce 0.9mol of acetic acid ....(cross multiply)

xmol ethanol * 1mol acetic = 1mol ethamol* 0.9molacetic

∴ xmol ethanol = 1*0.9 /1

= 0.90 mol

• This means that, 0.90 mol of ethanol, is needed to produce 0.9mol acetic acid,.

Hope it helps you!

-miraculousfanx-

The pressure of PCI₃ in the equilibrium mixture is approximately 0.000954 atm.

To solve this problem, we can use the ideal gas law and the relationship between partial pressures and equilibrium constant (Kp) to find the pressure of PCI₃.

Given the balanced equation for the given reaction is:

PCI₃ (g) + Cl₂ (g) -> PCI₅ (g)

Let

P₁ = P(PCI₃) = equilibrium partial pressure of PCI₃

Given:

P(PCI₅) = 0.224 atm

P(Cl₂) = 0.284 atm,

Kp = equilibrium constant

Kp = [P(PCI₅) * P(Cl₂)) / (P(PCI₃)]

Substituting the given values:

66.7 = (0.224 atm * 0.284 atm) / P₁

Now we can solve for P₁:

P₁ = (0.224 atm * 0.284 atm) / 66.7

P₁ = 0.000954 atm

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The percent yield of C₂H₅Cl if the reaction produces 204g of C₂H₅Cl is 110.9%.

The quantity of moles of a product created in relation to the amount of reactant consumed during a chemical reaction is measured in chemistry as yield, also known as reaction yield. The amount of product produced using all the moles of the limiting reactant is known as the theoretical yield. To locate the desired product, examine the reaction.

The actual yield is simply calculated by dividing it by the theoretical yield and multiplying the result by 100. The difference between theoretical and real yield is that theoretical yield is the amount of product you actually received.

C₂H₆ + Cl₂ --> C₂H₅Cl + HCl

mol of C₂H₆ = 138/30

                    = 4.6

Cl₂ moles = 203/71

                = 2.85

C₂H₅Cl moles = Cl₂ moles = 2.85

Theoretical yield = 2.85 x 64.51

                            = 183.85 gm

% yield = ( 204/183.85) x100

           = 110.9 %

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Answer:

water vapor (gas) is becoming liquid

The Ksp (solubility product constant) for PbBr2 at the given temperature is approximately 4.7 x 10^-6.

The solubility product constant (Ksp) is a measure of the extent to which a sparingly soluble salt dissociates into its ions in a solution. For the compound PbBr2, the equilibrium equation is:

PbBr2(s) ⇌ Pb2+(aq) + 2Br-(aq)

The Ksp expression for this equilibrium is:

Ksp = [Pb2+][Br-]²

Given that the molar solubility of PbBr2 is 2.17 x 10^-3 M, we can assume that the concentration of Pb2+ and Br- ions in the saturated solution is equal to the solubility value.

Thus, [Pb2+] = 2.17 x 10^-3 M and [Br-] = 2 x (2.17 x 10^-3 M) = 4.34 x 10^-3 M (since the stoichiometry of PbBr2 is 1:2).

Plugging these values into the Ksp expression:

Ksp = (2.17 x 10^-3 M) * (4.34 x 10^-3 M)²

≈ 4.7 x 10^-6

Therefore, the Ksp for PbBr2 at the given temperature is approximately 4.7 x 10^-6.

The Ksp for PbBr2 at the given temperature is approximately 4.7 x 10^-6. This value represents the equilibrium constant for the dissociation of PbBr2 into its constituent ions, Pb2+ and Br-, in a saturated solution.

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While balancing the chemical equations, the sub scripts aren't changed because by changing sub script we actually change the compound formed, hence coefficients are changed so the compound remains the same.

for example : ( for understanding )

in balancing the given reaction,

H2 + O2 =》H2O

if we change Oxygen's sub script in " H2O "

we get,

H2O + O2 =》H2O2

but here the compound formed isn't water, it is "H2O2" (hydrogen peroxide), so we can't change the sub script.

but by changing the coefficient, we get :

2 H2 + O2 =》2 H2O

Answer:

You will have to mix 240 mL of 75% acid with 300 mL of 30% acid to produce 540 mL of 50% acid

Explanation:

Answer:

Complex permanent tissue is defined as a collection of structurally dissimilar cells performing a common function or set of functions.

its functions are:

1.help in the transportation of organic material, water, and minerals up and down the plants.

2.it helps in transportation of food from leaves to other parts of plants.

have a great dayyyyyy

3C⁻⁻ on reaction with HI gives I⁻⁻⁻ as the main products and not H⁻ and C₂H₅I.

When 3C⁻⁻ is reacted with HI, the reaction product obtained is I⁻⁻⁻ as the main product. The C₂H₅I and H⁻ are not produced in significant quantities and cannot be considered the main product.The 3C⁻⁻ compound reacts with HI in the presence of a solvent to produce hydrogen gas, H⁻, C₂H₅I, and I⁻⁻⁻. The primary product obtained is I⁻⁻⁻ because it is stable and has a higher energy than C₂H₅I and H⁻.However, the reaction can be controlled to obtain C₂H₅I and H⁻ as the primary products by changing the reaction conditions. The reaction must be carried out in anhydrous conditions and at a low temperature so that the reaction proceeds in the desired direction.

3C⁻⁻ on reaction with HI gives I⁻⁻⁻ as the main products and not H⁻ and C₂H₅I. However, the reaction can be controlled to obtain C₂H₅I and H⁻ as the primary products by changing the reaction conditions.

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The weight percent of copper in bornite is approximately 63.316%.

The weight percent of copper (Cu) in bornite (Cu5FeS4) can be calculated by considering the atomic masses of copper (Cu), iron (Fe), and sulfur (S) and using the formula:

\(\[\text{{Weight percent of Cu}} = \frac{{\text{{Atomic mass of Cu}} \times \text{{Number of Cu atoms}}}}{{\text{{Formula mass of Cu5FeS4}}}} \times 100\%\]\)

Given that the atomic mass of Cu is 63.546 g/mol, the atomic mass of Fe is 55.845 g/mol, the atomic mass of S is 32.065 g/mol, and the formula mass of Cu5FeS4 is 501.849 g/mol, we can substitute these values into the formula:

\(\[\text{{Weight percent of Cu}} = \frac{{5 \times 63.546}}{{501.849}} \times 100\%\]\)

Simplifying the calculation gives:

\(\[\text{{Weight percent of Cu}} = 63.316\%\]\)

Therefore, the weight percent of copper in bornite is approximately 63.316%.

To calculate the gross value of the mining operation, we multiply the weight of the ore body (20,000,000 tons) by the cost per ton ($85):

\(\[\text{{Gross value}} = \text{{Weight of ore body}} \times \text{{Cost per ton}}\]\)

\(\[\text{{Gross value}} = 20,000,000 \times 85 = \$1,700,000,000\]\)

The expenses for the mining operation can be calculated by multiplying the weight of the ore body (20,000,000 tons) by the cost per ton ($85):

\(\[\text{{Expenses}} = \text{{Weight of ore body}} \times \text{{Cost per ton}}\]\)

\(\[\text{{Expenses}} = 20,000,000 \times 85 = \$1,700,000,000\]\)

The net value (profit or loss) of the mining operation can be obtained by subtracting the expenses from the gross value:

\(\[\text{{Net value}} = \text{{Gross value}} - \text{{Expenses}}\]\)\(\[\text{{Net value}} = \$1,700,000,000 - \$1,700,000,000 = \$0\]\)

Therefore, the net value of this mining operation is zero, indicating that there is neither profit nor loss.

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Answer: The density of kerosene is \(2.7g/cm^3\)

Explanation:

Density is defined as the mass contained per unit volume.

\(Density=\frac{mass}{Volume}\)

Given : Mass of kerosene = mass of beaker with kerosene - mass of empty beaker = 60 g - 20 g = 40 g

Volume of kersone = \(15cm^3\)

Putting in the values we get:

\(Density=\frac{40g}{15cm^3}\)

\(Density=2.7g/cm^3\)  

Thus the density of kerosene is \(2.7g/cm^3\)

Answer:

12113

Explanation:

step by step:no

Sharon, a gymnast, has a body fat percentage of 7%. This is considered a very low body fat percentage, and is often associated with athletes and fitness competitors. Maintaining such a low body fat percentage requires strict diet and exercise regimes, and can have potential health risks.

Body fat percentage is the proportion of fat to total body weight. For athletes like Sharon, having a low body fat percentage is often desirable as it can improve performance and appearance.

A body fat percentage of 7% is considered very low, and is often only achieved by bodybuilders, fitness competitors, and other elite athletes.

However, maintaining such a low body fat percentage requires strict diet and exercise regimes, which can have potential health risks. Extremely low body fat levels can lead to hormonal imbalances, decreased immunity, and reproductive issues in women.

Therefore, it is important for athletes like Sharon to balance their desire for a low body fat percentage with maintaining overall health and well-being.

In conclusion, Sharon's body fat percentage of 7% is very low and reflects her dedication to fitness and athletics.

However, achieving and maintaining such a low body fat percentage can come with potential health risks and requires careful attention to diet and exercise.

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Answer:

m= 85.7g

V= 32.4mL

d= m/V

d= 2.65g/mL

Answer: The reactants

Explanation: These are the substances that go into a chemical reaction and start it.

If you placed nails in the beaker of cupper precipitate then the displacement reaction takes place where Iron replaces the Copper in Copper sulfate to form Iron sulfate.

Displacement reaction is defined as the type of chemical reaction in which the atom of the more reactive elements displaces the atom of the less reactive elements from its compound. This reaction occurs in both metals and nonmetals. This is also known as a replacement reaction or substitution reaction. In this reaction, one element that is in the form of a free element replaces the other element from its compound to make a new element and a compound. There is one free element and one compound instead of both being compounds.

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Answer:

there it is fella tried on ma own consciousness

The volume of ethane gas, C₂H₆ in cm³ that will react to produce 40 cm³ of CO₂ is 20 cm³

2C₂H₆ + 7O₂ —> 4CO₂ + 6H₂O

From the balanced equation above,

4 cm³ of CO₂ were obtained from 2 cm³ of C₂H₆

From the balanced equation above,

4 cm³ of CO₂ were obtained from 2 cm³ of C₂H₆

Therefore,

40 cm³ of CO₂ will be obtain from = (40 × 2) / 4 = 20 cm³ of C₂H₆

Thus, 20 cm³ of C₂H₆ are required for the reaction.

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To combine ions to form ionic compounds, we need the combine in such a way that it gets neutral charge.

We can combine each anion with each cation to get the 4 compounds we need.

To combine SO₄²⁻ with Pb⁴⁺ we first find the Least Common Multiple of their charges, 2 and 4.

They have the factor 2 in common, so the LCM is 4. This is the final charge of each that will cancel out.

To get 4+, we only need 1 Pb⁴⁺.

To get 4-, we need 2 SO₄²⁻.

So, the formula is:

Pb(SO₄)₂

To combine SO₄²⁻ with NH₄⁺ is easier because one of them has single charge. In this case, we can simply pick one of the multiple charge ion and the same amount that will cancel its charge of the single charged one.

So, we picke 1 SO₄²⁻, ending with 2-.

And we picke 2 NH₄⁺, ending with 2+.

The formula:

(NH₄)₂SO₄

To combine C₂H₃O₂⁻ with Pb⁴⁺ we do the same, because the anion is single charged.

Pick 1 Pb⁴⁺, ending with 4+.

Pick 4 C₂H₃O₂⁻, ending with 4-.

The formula:

Pb(C₂H₃O₂)₄

To combine C₂H₃O₂⁻ with NH₄⁺, both have same charge, so we just need one of each and their charges will cancel out.

The formula:

NH₄C₂H₃O₂

So, the formulas are:

Pb(SO₄)₂

(NH₄)₂SO₄

Pb(C₂H₃O₂)₄

NH₄C₂H₃O₂

Answer:

If you follow the periodic table and its orbitals, you'll see that the element that has the electron configuration of 1s²2s²2p³ is Nitrogen. It's period, group, and atomic number are 2, 15, and 7 respectively. The other 2 are Neon, who's period, group, and atomic number are 2, 18, and 10 and the last one is Fluorine, who's period, group, and atomic number are 2, 17, and 9.

Answer:

Explanation:

The mass of the sun is about 1.98 x 10^30 kilograms. The mass of the earth is 5.97x 10^24 kilograms.

1.98 x 10^30 / 5.97x 10^24 = 331658 times greater

Answer:

Explanation:

The mass of the sun / the mass of the earth

= 1.98 x 10^30 / 5.97x 10^24

= 3.32 x 10^5 times greater

If the density of the water is 1.00 g/cm³, the mass of the water in the pounds is 311733.7 pounds.

The backyard swimming pool can holds the water = 185 yd³

The conversion of the yd³ in to the m³ :

185 yd³ = 185 × ( 0.9144)³ m³

185 yd³ = 141.44 m³

The conversion of the m³ in to the kg :

The 1 m³ = 1000 kg

Which means that :

141.4 m³ = 141.44 × 1000 kg

141.4 m³ = 141440 kg

The conversion of the kg in to the pounds :

1 kilogram = 2.204 pounds

141440 kg = 141440 kg × 2.204 pounds

141440 kg = 311733.7 pounds

The mass of the water in the pounds is 311733.7 pounds.

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Given the freezing-point depression for a solution is 2. 5 °C and the cryoscopic constant is 4.5 °c/m, the molality of the solution is 0.56 m.

Freezing-point depression is a drop in the temperature at which a substance freezes, caused when a smaller amount of another, non-volatile substance is added.

The freezing-point depression for a solution is 2. 5 °C. We can calculate the molality of the solution using the following expression.

ΔT = Kf × b

b = ΔT / Kf = 2.5 °C / (4.5 °C/m) = 0.56 m

where,

Given the freezing-point depression for a solution is 2. 5 °C and the cryoscopic constant is 4.5 °c/m, the molality of the solution is 0.56 m.

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The molality of a solution will be 0.56 m. if the freezing point depression for a solution is 2.5° C.

Molality is the measure of the moles of any solute in a solution per unit kg of the solvent.

Given, the freezing-point depression for a solution is 2. 5 °C

The kf is 4.5° c/m

\(\rm \Delta T = Kf \times b\\\\b = \dfrac{\Delta T}{kf} \\\\b = \dfrac{2.5^ \circ C}{4.5 \circ C/m} = 0.56 m\)

Where T is freezing point depression

kf is cryoscopic constant of the solvent

b is molality of the solution

Thus, the molality of the solution is 0.56 m.

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Answer:

D

Explanation:

The weak crust of the earth does not hold together as well as strong crust.

Answer:

A and B was right in our test.

Explanation:

A was: A volcano can suddenly get taller every time it erupts

B was: Hot magma from inside Earth is released at weak points in the crust.

The standard free energy change for the reaction of 1.78 moles of CO(g) at 308 K, 1 atm would be -36.9 kJ. This reaction is product-favored under standard conditions at 308 K.

To calculate the standard free energy change (ΔG°) for the reaction, we can use the equation:

ΔG° = ΔH° - TΔS°

Given that ΔH°rxn = -206.1 kJ and ΔS°rxn = -214.7 J/K, we need to convert the values to kJ and adjust the sign for ΔS°:

ΔH°rxn = -206.1 kJ

ΔS°rxn = -214.7 J/K = -0.2147 kJ/K

Now we can substitute these values into the equation:

ΔG° = -206.1 kJ - (308 K)(-0.2147 kJ/K)

    = -206.1 kJ + 66.1736 kJ

    = -139.9264 kJ

Since the question asks for the standard free energy change for 1.78 moles of CO(g), we need to calculate the moles of CO(g) in the reaction:

1 mole of CO(g) reacts to form 1 mole of CH₄(g)

1.78 moles of CO(g) will react to form 1.78 moles of CH₄(g)

Therefore, the standard free energy change for the reaction of 1.78 moles of CO(g) at 308 K, 1 atm is -139.9264 kJ.

Based on the negative value of ΔG°, the reaction is product-favored under standard conditions at 308 K.

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