Balance the equation <br> NaOH + CH3COOH

A solution is 40.0% by volume benzene (C6H6) in carbon tetrachloride at 20C. The vapor pressure of pure benzene at this temperat

Alexandra [31]

Answer:

84.30 mm Hg

Explanation:

In 100 cm³ of solution we have: 40 cm³ C6H6 and 60 cm³ CCl4. Given the densities we can calculate their masses and number of moles, and since by Raoult´s law

Ptotal = XAPºA + XBPºB

where XA= mol fraction =na/(na +nb) and PºA vapor pressure pure of pure component A

m C6H6 = 40 cm³ x 0.87865 g/cm³ = 35.146 g

mol C6H6 = 35.146 g/ 78.11 g/mol = 0.45 mol

mass CCl4 = 60 cm³ x 1.5940 g/cm³ = 95.640 g

mol CCl4 = 95.640 g / 153.82 g/mol = 0.62 mol

mol tot = 1.07

XC6H6 = 0.45/ 1.07 = 0.42 XCCl4 = 0.62/1.07 =0.58

Ptot (mmHg) = 0.42 x 74.61 + .58 x 91.32 = 84.30 mmHg

6 0

3 years ago

Unit for mass, volume and density

Kazeer [188]

The mass of water is expressed in grams (g) or kilograms (kg), and the volume is measured in liters (L), cubic centimeters (cm3), or milliliters (mL). Density is calculated by the dividing the mass by the volume, so that density is measured as units of mass/volume, often g/mL.

5 0

4 years ago

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One way to increase the rate of a reaction occurring in solution is to stir the solution. How does this increase the rate of the

riadik2000 [5.3K]

<span>A. It increases the number of collisions between molecules.

I hope this helps :)</span>

7 0

3 years ago

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Please help will give points<br>Match the type of thermal energy with the letters on the drawing

Katen [24]

A: radiation

B: convection

C: conduction

7 0

3 years ago

Gaseous methane (CH₄) reacts with gaseous oxygen gas (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O) If 0.3

AveGali [126]

Answer : The percent yield of $CO_2$ is, 68.4 %

Solution : Given,

Mass of $CH_4$ = 0.16 g

Mass of $O_2$ = 0.84 g

Molar mass of $CH_4$ = 16 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the moles of $CH_4$ and $O_2$ .

$\text{ Moles of }CH_4=\frac{\text{ Mass of }CH_4}{\text{ Molar mass of }CH_4}=\frac{0.16g}{16g/mole}=0.01moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{0.84g}{32g/mole}=0.026moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that

As, 2 mole of $O_2$ react with 1 mole of $CH_4$

So, 0.026 moles of $O_2$ react with $\frac{0.026}{2}=0.013$ moles of $CH_4$

From this we conclude that, $CH_4$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CO_2$

From the reaction, we conclude that

As, 2 mole of $O_2$ react to give 1 mole of $CO_2$

So, 0.026 moles of $O_2$ react to give $\frac{0.026}{2}=0.013$ moles of $CO_2$

Now we have to calculate the mass of $CO_2$

$\text{ Mass of }CO_2=\text{ Moles of }CO_2\times \text{ Molar mass of }CO_2$